mm-315 === Subject: Inverse Laplace TransformIs it possible to use Taylor Series expansions to determine theinverse laplace transform of a function? If I have something like cos(s), can I expand it into a T.S. and then take the term by terminverse, add those up and call the result the inverse transform ofcos(s)?I'm not asking about cos(s) in particular-- I don't think that it has an ILT-- but just the general procedure. === Subject: Re: Quick Math Guide to core error issuesFor those of you trying to keep up with the mathematical facts in thediscussions about the error in core mathematics from a problem with adefinition, this post will outline the important ones quickly andsuccinctly.1. First the problematic definition:Algebraic integers are defined to be roots of monic polynomials withinteger coefficient e.g. x^3 + 3x + 1 or x^234 - 34x^12 + 17, wheremonic refers to the leading coefficient.My assertion is that the over hundred year old definition excludesnumbers that have to be included to keep from having contradictioni.e. mathematical inconsistency.2. The important tool I use is a polynomial:P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)The form of the polynomial allows me to factor P(m) intonon-polynomial factors, and the factorization with those factors isP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)where the a's are roots of the following cubic:a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).3. Dispute centers around what happens when I divide P(m) by f^2,which you'll note is a factor of the polynomial in the ring ofalgebraic integers.4. Mathematicians have argued that f^2 divides off as a function of mbecause if they concede that it divides off independent of m, then Ican show that only two of the roots ofa^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m)have f as a factor.5. However, it turns out that if you go to the field of algebraicnumbers you can prove that for *certain* values of m and f, the rootsof the cubic do not have f as a factor *in the ring of algebraicnumbers* which is the inconsistency.That is, for the math to be consistent, two of the roots *should* havef as a factor as long as m and f are algebraic integers, but while Ican show they do for a particular values like m=1, f=sqrt(2), thereare other values you can show they do not *in the ring of algebraicintegers* which results from the definition and its focus on monicpolynomials.Note: In the ring of algebraic integers you can't see the problem buthave to go to the field of algebraic numbers as from within the ringof algebraic integers it appears that only two of the roots have afactor that is === f.James HarrisSubject: Re: Quick Math Guide to core error issuesAn exact duplicate (unless there was a sign change) of the original post in this thread. Adjunct Assistant Professor at the University of Montana.>Yeah, but really, I swear that one time while a had my back>turned, 17 was sneakily varying a little bit. But when I>turned around really quickly to try and catch it, it had sneakily>changed back to 17 again.>>You must have been using FORTRAN, where it cheats by giving>>subroutines the actual addresses where constants are stored, instead>>of copying the constant into a temporary variable, as if it was an>>expression.> My dad tells me that an old geek practical joke, back at the time> punch-cards, was to slip a punchcard with the instruction LET 2=3> into someone's pile. Fortran would then assign a memory location to> the constant 2, and store the value 3 there...>Not in Fortran; perhaps in some version of BASIC. Blanks are not>significant in versions of Fortran before Fortran 90, and LET has no>special significance. The meaning of> LET 2 = 3>is to assign 3 to the integer variable named LET2.Then maybe it was 2=3, or some particular implementation: as heexplained it, as soon as a constant was used, the compiler wouldassign a memory location and assign the corresponding value to thatmemory location; but that allowed you to instruct the compiler tostore a different value in that memory location later on.>However, some implementations of Fortran made it possible to change the>value of a constant by passing it to a subroutine that would treat it as>a variable.Maybe that was it... but as I recall, it was supposed to happen with asingle punch-card, which you would slip to somewhere in the middle === proof is attacking yourselvesIn sci.math, Arturo Magidin:>>Yeah, but really, I swear that one time while a had my back>>turned, 17 was sneakily varying a little bit. But when I>>turned around really quickly to try and catch it, it had sneakily>>changed back to 17 again.>You must have been using FORTRAN, where it cheats by giving>subroutines the actual addresses where constants are stored, instead>of copying the constant into a temporary variable, as if it was an>expression.>> My dad tells me that an old geek practical joke, back at the time>> punch-cards, was to slip a punchcard with the instruction LET 2=3>> into someone's pile. Fortran would then assign a memory location to>> the constant 2, and store the value 3 there...>Not in Fortran; perhaps in some version of BASIC. Blanks are not>significant in versions of Fortran before Fortran 90, and LET has no>special significance. The meaning of> LET 2 = 3>is to assign 3 to the integer variable named LET2.> Then maybe it was 2=3, or some particular implementation: as he> explained it, as soon as a constant was used, the compiler would> assign a memory location and assign the corresponding value to that> memory location; but that allowed you to instruct the compiler to> store a different value in that memory location later on.In Fortran it might be______2=3where '_' is actually a space character -- assuming the compilerwas dumb enough to allow '2' on the left side of an assignment.>However, some implementations of Fortran made it possible to change the>value of a constant by passing it to a subroutine that would treat it as>a variable.> Maybe that was it... but as I recall, it was supposed to happen with a> single punch-card, which you would slip to somewhere in the middle of> the other guy's program...If there was a subroutine named INCR one might______CALL INCR(2)but that's a bit of a stretch. === go .sigless.Subject: Re: Attacking a proof is attacking yourselves> I've talked about the basic truths they're attacking, about how 7and> 22 are NUMBERS and not variables, as they're constant.Yeah, but what about 17? I think I saw 17 vary a little bit one time.> Gotta> go now. Time for my shots.> That's because 17 is a more random number than 7 or 22.> - Randy> Yeah, but really, I swear that one time while a had my back> turned, 17 was sneakily varying a little bit. But when I> turned around really quickly to try and catch it, it had sneakily> changed back to 17 again. Goddam sneaky variable === constants. Driving me nuts.Subject: Re: Attacking a proof is attacking yourselves> I've talked about the basic truths they're attacking, about how 7> and> 22 are NUMBERS and not variables, as they're constant.> Yeah, but what about 17? I think I saw 17 vary a little bit onetime.> Gotta> go now. Time for my shots.That's because 17 is a more random number than 7 or 22. - Randy> Yeah, but really, I swear that one time while a had my back> turned, 17 was sneakily varying a little bit. But when I> turned around really quickly to try and catch it, it had sneakily> changed back to 17 again.> Goddam sneaky variable constants. Driving me nuts.And those constant variables are just too much. A man has tobelieve in something. === I believe I'll have a little drink.Subject: Re: Attacking a proof is attacking yourselves... > And those constant variables are just too much. A man has to > believe in something. I believe I'll have a little drink.A well-known maxim for computer programmers:constants are not constant and 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === ----http://www.crbond.comSubject: AlgebraCan anyone give me some help with this question?Let G be a finite group of order p^n, where p is prime and n>0.i) Prove that the conjugacy class ClG(g) of g in G has size p^m forsome m >=0.ii)Prove that m=0 in (i) if and only if g in Z(G)(iii) Deduce that the order of Z(G) >1iv) Use The centre Z(G) of G is defined as Z(G) = {g in G such thatgh=hg for all h in G }. Let be a group in which G/Z(G) is cyclic thisimplies thatG = Z(G) so G is abelian. to prove that any group of order p^2 === Can anyone give me some help with this question?Notation: C(g) = conjugacy class of g; | S | = number of elements in S.Since I don't know what you know don't be loath to ask.> Let G be a finite group of order p^n, where p is prime and n>0.> i) Prove that the conjugacy class ClG(g) of g in G has size p^m for> some m >=0.| C(g) | = | G | / | N(g) | where N(g) is the normalizer of g.> ii)Prove that m=0 in (i) if and only if g in Z(G)g in Z(G) <=> C(g) = {g} <=> m = 0.> (iii) Deduce that the order of Z(G) >1By (ii) We can write Z(G) = union C(g) for g in Z(G) This means p^n = | G | = | Z(G) | + | C(g_1) | + ... + | C(g_t) |. Now p divideseverything in sight except possibly | Z(G) | and hence divides | Z(G)|.> iv) Use The centre Z(G) of G is defined as Z(G) = {g in G such that> gh=hg for all h in G }. Let be a group in which G/Z(G) is cyclic this> implies that> G = Z(G) so G is abelian. to prove that any group of order p^2 is> abelian.I think you meant to write If G/Z(G) is cyclic then G is abelian.I guess we are meant to assume the quoted statements. In view of (iii),what are your choices for the order of Z(G)? If the order of Z(G) is pwhat is the order of G/Z(G)? What can you say about G/Z(G)? If theorder of Z(G) is === questionSorry to post such trivia, but I think I'm going to pluck my eyeballsout if I don't see a neat solution to this.Can anyone tell me the quickest way to solve for X in a situation likethis:x = 10^x;) Yeah I know, I must be an === idiot.Subject: Re: A quick question>Sorry to post such trivia, but I think I'm going to pluck my eyeballs>out if I don't see a neat solution to this.>Can anyone tell me the quickest way to solve for X in a situation like>this:>x = 10^x>;) Yeah I know, I must be an idiot. If you take the log (base 10) of both sides: log x = x log 10 = x log x / x =1 Try graphing that and see if it makes any sense. IMHO... MJR(This email scanned by Norton Anti-Virus and Certified VIRUS FREE!)Michael J. Reeves, AA, AScE-Mail: michaeljreeves@comcast.net------------------------------------ === ---------------------Subject: Re: A quick question> Sorry to post such trivia, but I think I'm going to pluck my eyeballs> out if I don't see a neat solution to this.> Can anyone tell me the quickest way to solve for X in a situation like> this:> x = 10^x> ;) Yeah I know, I must be an idiot.You're no idiot at all. IIRC, generally speaking the solutions to thesetypes of equations where x is both in a base *and* exponent make use offunctions that are not elementary. For example, if you had the similarequation:x^2 = 2^x...there are two obvious solutions found by simple inspection (2^2=2^2 and4^2=2^4, so x=2 and x=4 are solutions)) but there is also a third solutionthat is negative. AKAIK it is expressed in terms of the Lambert W functionwhich, again, is not an elementary function.That said, in this particular case it's not too difficult to quickly see(provided you keep your eyeballs!) that no real solution exists. Fornegative values of x where 10^x is defined, we have a negative number = apositive number since the right side can be thought of as the reciprocal of10 to the positive power and that's never negative. For x=0 we have 0=1.For positive x, we have a positive number = 10 raised to the same positivepower, but the right side is clearly greater than the left, not equal, evenfor positive values of x very close to 0. For that matter you could justgraph y=x and === === quick questionJamesSubject: Re: A quick question> Sorry to post such trivia, but I think I'm going to pluck my eyeballs> out if I don't see a neat solution to this.I'll do my best, if you'll first promise not to pluck out your eyeballs!> Can anyone tell me the quickest way to solve for X in a situation like> this:> x = 10^xFor this particular equation, it's easy to see that there is no realsolution. Just graph y = 10^x and y = x and note that the graph of theformer always lies above that of the === proofsI've put up a website called Mathematical Solutions, which I hope willfunction as a forum for promoting student proofs. I am a student atthe University of Miami, and am really interested in correspondingwith others regarding proofs on a variety of mathematical subjects.Essentially, what I'm offering is this: you go to www.mathsolve.comand submit a proof, I give you a free pop3 e-mail address with thedomain @mathsolve.com, and I publish the proof on the site. We candiscuss formatting when the time comes. As it stands, I've only gotday. The key to all this is that I really don't know what I'm doingre: math -- I just want to do it! And I want to get people to tell mewhen they think I'm wrong, and I want to explore the ways that othersdo this thing.So check it out: === www.mathsolve.comSubject: Re: Forum for promoting student proofs> I've put up a website called Mathematical Solutions, which I hope will> function as a forum for promoting student proofs. I am a student at> the University of Miami, and am really interested in corresponding> with others regarding proofs on a variety of mathematical subjects.> Essentially, what I'm offering is this: you go to www.mathsolve.com> and submit a proof, I give you a free pop3 e-mail address with the> domain @mathsolve.com, and I publish the proof on the site. We can> discuss formatting when the time comes. As it stands, I've only got> day. The key to all this is that I really don't know what I'm doing> re: math -- I just want to do it! And I want to get people to tell me> when they think I'm wrong, and I want to explore the ways that others> do this thing.> So check it out: www.mathsolve.comYou didn't include http.http://www.mathsolve.comThere I've done it. Whenever I get around to this newsgroupagain, I just might get around to opening mathsolve.What's better about your site than === newsgroups?Subject: Re: Forum for promoting student proofs> I've put up a website called Mathematical Solutions, which I hope will> function as a forum for promoting student proofs.> Essentially, what I'm offering is this: you go to www.mathsolve.com> and submit a proof, I give you a free pop3 e-mail address with the> domain @mathsolve.com, and I publish the proof on the site. We can> discuss formatting when the time comes. As it stands, I've only got> day.Your reply unreadable as it isn't plane text.There's no way to make reply.There's no separation of forums.All I can do is just post via href=mailto:...> So check it out: www.mathsolve.com> You didn't include http.> http://www.mathsolve.com> There I've done it. Whenever I get around to this newsgroup> again, I just might get around to opening mathsolve.> What's better about your site than newsgroups?For an example of a good web site forum see Ask a TopologistSome web site forums I've see are === hopeless bad.Subject: Re: Forum for promoting student proofs> I've put up a website called Mathematical Solutions, which I hope will> function as a forum for promoting student proofs.Essentially, what I'm offering is this: you go to www.mathsolve.com> and submit a proof, I give you a free pop3 e-mail address with the> domain @mathsolve.com, and I publish the proof on the site. We can> discuss formatting when the time comes. As it stands, I've only got> day.> Your reply unreadable as it isn't plane text.> There's no way to make reply.> There's no separation of forums.> All I can do is just post via href=mailto:...> So check it out: www.mathsolve.com> You didn't include http.> http://www.mathsolve.com> There I've done it. Whenever I get around to this newsgroup> again, I just might get around to opening mathsolve.> What's better about your site than newsgroups?> For an example of a good web site forum see> Ask a Topologist> Some web site forums I've see are hopeless bad.Interesting. So you suggest it becomes more of a Q&A format and lessa repository of information? I think I used the wrong word when Isaid forum... I didn't have a discussion area in mind.Why is the site === unreadable?Robhttp://www.mathsolve.comSubject: Re: Forum for promoting student proofs a website called Mathematical Solutions, which I hope will> function as a forum for promoting student proofs.> Essentially, what I'm offering is this: you go to www.mathsolve.com> and submit a proof, I give you a free pop3 e-mail address with the> domain @mathsolve.com, and I publish the proof on the site. We can> discuss formatting when the time comes. As it stands, I've only got> day.Your reply unreadable as it isn't plane text.> There's no way to make reply.> There's no separation of forums.> All I can do is just post via href=mailto:...> What's better about your site than newsgroups?> For an example of a good web site forum see> Ask a Topologist> Some web site forums I've see are hopeless bad.> Interesting. So you suggest it becomes more of a Q&A format and less> a repository of information? I think I used the wrong word when I> said forum... I didn't have a discussion area in mind.If it's proofs you want, I've 1/2 meg ofof plain text proofs in class note stylemostly on topology and abstract algebra.Interested? What a huge mammoth undertaking,an encyclopedia of proofs and problem solutions.Yes, 'forum' didn't describe your thoughts.Other familar forms are Q&A and FAQ.Perhaps you wish FAQ. I suggest you consider how http://en.wikipedia.organ user authored encyclopedia, is designed.Facilities you'll need or eventually wantAs we now live in a virtual computerized Tower of BabelDeclaration of text type for entries plain text extended plain text html text and special characters TeX characters TeX, ps. pdf, etc filesWriters' guide lines including notation usage.Present proof or solution to editor. Editor decides what subject it goes into checks for cerical and mathematical correctnessPeer review of proofs and solutions to problem.Search facilityClassification system Analysis series integration differentiation ...> Why is the site unreadable?The site isn't unreadable. It's simple and easy to fathom.What's unreadable is the proof you entered into your site.I think it's numberous hyper-text special characters showing up like$#AR56F or something such, that make it unreadable.> === http://www.mathsolve.comSubject: Re: Forum for promoting student proofs> I've put up a website called Mathematical Solutions, which I hopewill> function as a forum for promoting student proofs.> Essentially, what I'm offering is this: you go to www.mathsolve.com> and submit a proof, I give you a free pop3 e-mail address with the> domain @mathsolve.com, and I publish the proof on the site. We can> discuss formatting when the time comes. As it stands, I've only got> day.Your reply unreadable as it isn't plane text.> There's no way to make reply.> There's no separation of forums.> All I can do is just post via href=mailto:...> So check it out: www.mathsolve.com> You didn't include http.> http://www.mathsolve.com> There I've done it. Whenever I get around to this newsgroup> again, I just might get around to opening mathsolve.What's better about your site than newsgroups?For an example of a good web site forum see> Ask a Topologist> Some web site forums I've see are hopeless bad.> Interesting. So you suggest it becomes more of a Q&A format and less> a repository of information? I think I used the wrong word when I> said forum... I didn't have a discussion area in mind.> Why is the site unreadable?> Rob> http://www.mathsolve.comMy $.02: Someone made and advertised a website, then someone complainedthat the site is not better than a newsgroup and additionally complainedthat the proper syntax was not used to describe the link to facilitatesoftware properly redirecting to the site when clicked upon. More or less,that's about what happened. So while we're in complaint mode, let mepresent my own. Leave the guy alone. He is entitled to make his site andinvite others to visit. We are by no means obligated to do so, with orwithout some software allowing us to just click the address he provided andthe page automatically appear on our screen.And nothing, BTW, deems a website to be inappropriate for the simple reasonthat something else may exist that may be better.And finally, to be fair I must also point out that everyone is entitled totheir opinion, even if the opinion seems very rude and could be presentedmuch more effectively if it was done in a respectful and polite manner. Inshort, everyone has a right to have eyebrows raised by others as a result ofthe rude manner in === which they state their opinion.Subject: Re: Forum for promoting student proofs> I've put up a website called Mathematical Solutions, which I hope will> function as a forum for promoting student proofs.Essentially, what I'm offering is this: you go to www.mathsolve.com> and submit a proof, I give you a free pop3 e-mail address with the> domain @mathsolve.com, and I publish the proof on the site. We can> discuss formatting when the time comes. As it stands, I've only got> day.> Your reply unreadable as it isn't plane text. ^^^^^You can't read text which isn't topologically equivalent of Iraqis want to live in a peaceful, free world.And we will === find these people and we will bring them to justice.Subject: Re: Enthalpy> how do u intergrated > 1> ----- dp G= Cv/Cp> p ^1/G No answer to this has shown up at my news server, so perhaps peopleare puzzled by your notation. It appears you want to integrate p^(-1/G) which is just a power of p, so you get(p^(1 - 1/G))/(1 - 1/G) + a constant.*However*, this assumes that the ratio G of specific heats is constant(which is often quite a good approximation). Are you integrating volume with respect to pressure for a gas inadiabatic equilibrium? If so, the title enthalpy is a bit puzzling, butmy recollection of thermodynamics is _very_ rusty. :-( Ken === Pledger.Subject: elementary double integralDomain:y <= x <= 2y0 <= x <= 2Now i want this as a double integral where dx is inner and dy as outer.And apparantly it yields:[0,1] dy [y,2y] f(x,y) dx + [1,2] dy [y,2] f(x,y) dxI dont understand why, why is the integral splitted in two parts ?i dont know ascii notation for integrals so [0,1] means integral symbolwith === upper and lower value.Subject: Re: elementary double integral>Domain:>y <= x <= 2y>0 <= x <= 2>Now i want this as a double integral where dx is inner and dy as outer.>And apparantly it yields:>[0,1] dy [y,2y] f(x,y) dx + [1,2] dy [y,2] f(x,y) dx>I dont understand why, why is the integral splitted in two parts ?>i dont know ascii notation for integrals so [0,1] means integral symbol>with upper and lower === value.Subject: Re: elementary double 2y> 0 <= x <= 2I think you meant 0 <= y <= 2.> Now i want this as a double integral where dx is inner and dy as outer.> And apparantly it yields:> [0,1] dy [y,2y] f(x,y) dx + [1,2] dy [y,2] f(x,y) dx> I dont understand why, why is the integral splitted in two parts ?Nor do I. What was the rest of the problem?> i dont know ascii notation for integrals so [0,1] means integral symbol> with upper and lower value.What you have is int ( int ( f(x,y), x = y..2*y), y = 0..2) =( int ( f(x,y), x = y..2*y), y = 0..1) + ( int ( f(x,y), x = y..2*y), y = 1..2).Which is surely true. _Why_ you want to rewrite that === === elementary double integral----- Original Message -----Subject: Re: elementary double integral> Domain:> y <= x <= 2y> 0 <= x <= 2> I think you meant 0 <= y <= 2.y <= x <= 2y0 <= x <= 2Let me clarify my question. First, we note that the domain can be rewrittenx/2 <= y <= x0 <= x <= 2Which yields the integral [0,2] dx [x/2,x] f(x,y) dyThe question is now, REVERSE the integration order of the integral. i DONTunderstand how to do that. The answer in my textbook reads: [0,1] dy [y,2y]f(x,y) dx + [1,2] dy [y,2] f(x,y) dxmy problem is, i dont understand WHY the reverse order integral can berewritten that === way.Subject: Re: elementary double === Message -----> Subject: Re: elementary double integral> Domain:> y <= x <= 2y> 0 <= x <= 2> I think you meant 0 <= y <= 2.> y <= x <= 2y> 0 <= x <= 2OK. It looked odd to me and typos do happen. ( I also misread youranswer. )> Let me clarify my question. First, we note that the domain can be rewritten> x/2 <= y <= x> 0 <= x <= 2> Which yields the integral [0,2] dx [x/2,x] f(x,y) dy> The question is now, REVERSE the integration order of the integral. i DONT> understand how to do that. The answer in my textbook reads: [0,1] dy [y,2y]> f(x,y) dx + [1,2] dy [y,2] f(x,y) dx> my problem is, i dont understand WHY the reverse order integral can be> rewritten that way.I'll try again. If you graph the region, you'll see a triangle with two slanty sidesand one side parallel to the y-axis.The total variation of y's for that region is 0 <= y <= 2. But note for0 <= y <= 1, the x's vary from one slanty side to the other. That is,y <= x <= 2y. However, for 1 <= y <= 2, the x's vary from the y = xside to the x = 2 side. That is y <= x <= 2. Hence the limits on yourdxdy integrals. The key point is that, for the dydx integral, for any x between 0 and2, the y's always vary from x/2 to x. However, for the dxdy integraland y between 0 and 2, sometimes the x's vary from y to 2y andsometimes the x's vary from y to 2. So you need two integrals.Instead of one region given by x/2 <= y <= x and 0 <= x <= 2, you have_two_ regions. One is given by y <= x <= 2y and 0 <= y <= 1 and theother region is given by y <= x <= 2 and 1 <= y <= 2. > -- > Stan Brown, Oak Road Systems, Cortland County, New York, USA> http://OakRoadSystems.com> Address munging may or may not useful answers you get.> http://www.cs.tut.fi/~jkorpela/usenet/laws.html Is this what you mean by not posting upside down? to answer yourquestions about what topics are covered in intermediate Algebra. they arelike Polynomials the operation of them and terms, factors, and coefficients,etc... operations with polynomials, factoring Polynomials, SimplifyingRadical and Rational Expressions, Solving Quadratic Equations and RationalEquations, Graphs of Quadratic Functions, Graphs of a Region Enclosed byQuadratic Functions and a Straight Line and sub-topics under these topics. You may be right I may not have learned Elementary algebra effetelyand that is what I told the instructor. He says it should al come togetherfor me. However I think I will need some help like with this problem onsubstitution. I don't think I understand it here is an example of a problemmaybe you can help with how they(the textbook) came up the answer. this is under the heading solving a system by graphing andsubstitution.y = x + 2x + y = 4it says first write in slope interceptI believe it is (1,3) then3 = 1 + 2x = 1 and y = 3 in x + y = 41 + 3 = 4then they have a graph I have not figure out === how to post a graphSubject: Stepped Out Key Core Error ArgumentThe following proof steps through a rather basic argument which is keyin proving an over one hundred year old error, a previously unexpectedconsequence of the definition of the ring of algebraic integers.Note that ultimately the proof relies on 22 NOT having 7 as a factor,and constant terms like 7 and 22, being constant, not variablesdependent on x, which may seem like odd things to emphasize, but I'vefaced posters who've gotten away with challenging those truths becausepeople seem unaware that's what they're doing.1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x isin the ring of algebraic integers, notice that P(x) has a constantterm that is 1078.2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3where you should note that using v = -1 + 49x, givesP(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3where the *same* polynomial has been put in a form which allows afactorization into non-polynomial factors so that I haveP(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).3. Now let x=0, soP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's at x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) anda_2(0) to equal 0, which leaves a_3(0) with a value of 3.4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then IhaveP(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).5. Now P(x) has a factor of 49 asP(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22which means that(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)has a factor of 49.6. However, the constant term of P(x)/49 is 22, which is verified byagain setting x=0, which gives P(0)/49 = 22.But for two of the factors of P(x), the constant terms is 7, which isNOT a factor of 22. Therefore, *none* of the constant terms ofP(x)/49 as they multiply to give 22 can have 7 as a factor.(By saying that 7 is NOT a factor of 22, I'm making a choice as towhere the proof is going. Since I've been talking about algebraicintegers, where 7 is NOT a factor of 22, it's natural to go with achoice where 7 is NOT a factor of 22.)Given that the constant terms are independent of x's value, it must bethe case that dividing P(x) by 49 divides the two constant terms equalto 7, by 7.7. But to divide 7 from those constant terms requires dividingthrough two of the factors, so(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22from reverse use of the distributive property, which gives constantterms that don't have 7 as a factor, as required.Notice that it's a rather short and direct argument, where if youaccept that 22 does not have 7 as a factor, it's obvious enough whatthe constant terms of the factors must be as you go from 7, 7 and 22,necessarily to 1, 1, and 22, when you divide P(x) by 49.James === Harrishttp://mathforprofit.blogspot.com/Subject: Re: Stepped Out Key Core Error ArgumentThe following proof steps through a rather basic argument which is keyin proving an over one hundred year old error, a previously unexpectedconsequence of the definition of the ring of algebraic integers.Note that ultimately the proof relies on 22 NOT having 7 as a factor,and constant terms like 7 and 22, being constant, not variablesdependent on x, which may seem like odd things to emphasize, but I'vefaced posters who've gotten away with challenging those truths becausepeople seem unaware that's what they're doing.1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x isin the ring of algebraic integers, notice that P(x) has a constantterm that is 1078.2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3where you should note that using v = -1 + 49x, givesP(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3where the *same* polynomial has been put in a form which allows afactorization into non-polynomial factors so that I haveP(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7)where the a's are roots ofa^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).3. Now let x=0, soP(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22)as the cubic defining the a's at x=0 isa^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) anda_2(0) to equal 0, which leaves a_3(0) with a value of 3.4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then IhaveP(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7)P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22).5. Now P(x) has a factor of 49 asP(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22which means that(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22)has a factor of 49.6. However, the constant term of P(x)/49 is 22, which is verified byagain setting x=0, which gives P(0)/49 = 22.But for two of the factors of P(x), the constant terms is 7, which isNOT a factor of 22. Therefore, *none* of the constant terms ofP(x)/49 as they multiply to give 22 can have 7 as a factor.(By saying that 7 is NOT a factor of 22, I'm making a choice as towhere the proof is going. Since I've been talking about algebraicintegers, where 7 is NOT a factor of 22, it's natural to go with achoice where 7 is NOT a factor of 22.)Given that the constant terms are independent of x's value, it must bethe case that dividing P(x) by 49 divides the two constant terms equalto 7, by 7.7. But to divide 7 from those constant terms requires dividingthrough two of the factors, so(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22from reverse use of the distributive property, which gives constantterms that don't have 7 as a factor, as required.Notice that it's a rather short and direct argument, where if youaccept that 22 does not have 7 as a factor, it's obvious enough whatthe constant terms of the factors must be as you go from 7, 7 and 22,necessarily to 1, 1, and 22, when you divide P(x) by 49.James === Harrishttp://mathforprofit.blogspot.com/Subject: INVERSE Function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9UE5Xs05197;Inverse === Functionf(x)= 1/3x-2Please help!!!Subject: Re: INVERSE Function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9VIOXC25048;>Inverse Function>f(x)= 1/3x-2>Please help!!! Rule 1: write the problem carefully!1/3x-2 could be interpreted as (1/3)x- 2 or 1/(3x)- 2 or 1/(3x-2). You probably meant the last since that's the most complicated. Rule 2: If y= f(x), then x= f^(-1)(y) (f^(-1) mean inverse function) If y= 1/(3x-2) is the given function, === swap x and y to getx= 1/(3y-2) and solve for y.Subject: Reply to Inverse Function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9V2Kqk22264;They just want you to solve for x in terms of f(x), i.e.x = ?By the way, the inverse function you're surely the most familiar withis SquareRoot(x) - its just the inverse of that other function you'reprobably very familiar with, === namely f(x) = x^2.Subject: i need a math project idea! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9UE3kG04997;What about estimating how many smarties (or M and M's, or any otherflattish lolly that's to your taste) it would take to cover thesurface of the === Earth ....Subject: HELP! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9V2LRP22423;I'm returning to college and need help with this problem:You can't use the same letter twice, if each different sequence ofletters constitutes a different word in the language, what is themaximum number of six letter words theat this language can employ. the letters are a, b, c, d, e, and === f.Subject: Re: HELP!> I'm returning to college and need help with this problem:> You can't use the same letter twice, if each different sequence of> letters constitutes a different word in the language, what is the> maximum number of six letter words theat this language can employ.> the letters are a, b, c, d, e, and f.Start with a simpler problem. Suppose there were one letter words. How manycould there be. 6. (a, b, ... ) Right? Suppose there were 2 letter words. Howmany could there be. You would have 6 choices for the first letter, and (sinceno repeats are allowed) 5 choices for the second. Right? === Continue the logic tolarger words.BillSubject: Group Theory by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9V2KrU22293;Hey guys,need some help with this question please!Prove that the follwing groups are Abelian. what are their orders? Arethey cyclic(i) {z e C : z^12 = 1} e = Element, C = complex numbers(ii) {z e R : z^12 = === 1} Any help would really be appreciated!Subject: Re: Group Theory> Hey guys,> need some help with this question please!> Prove that the following groups are Abelian.Are (R{0}, *) and (C{0}, *) Abelian, and why would this be relevant?> what are their orders? Are> they cyclic> (i) {z e C : z^12 = 1} e = Element, C = complex numbersIf w = exp((2*pi*i)/12), what is w^12?Now consider positive integer powers of w. Are these in group (i)?Is w^n a periodic function of n? If so, what is its period?What elements of (i), if any, cannot be expressed as a power of w?> (ii) {z e R : z^12 = 1}Which elements of (i) are to live in a peaceful, free world.And we will find these === people and we will bring them to justice.Subject: Re: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9VEMOY06951;Sorry forgot to mention binary operations acting on both the groupsare === multiplicationSubject: Maximal Ideals by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA13UpE31944;i'm having some trouble with the following question:> Prove that the ideal (x + y^2, y + x^2 + 2xy^2 + y^4) in C[x,y] isa maximal ideal.my initial thought was to use hilbert's nullestellensatz, but i'm notsure how to factor the second polynomial (having the y term makes itpretty tough!). if anyone has any suggestions, i'd === appreciate it.jaySubject: re: Can anyone Try this Problem? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA13Umr31921;Hello!Lora Bush wants to buy a photocopier. The salesperson has thefollowing information on 3 models. if all 3 are used, a specific jobcan be done in 50 minutes. If copier A operates for 20 minutes andcopier B for 50 mininutes, one-half of the job is finished. If copierB operates for 30 minutes and copier C for 80 minutes, three-fifths ofthe job is done. Which is the fastest copier, and how long does it takes for thiscopier to finish the whole job?Torsten is absolutely correct.I'll baby-step through the problem. (Don't be insulted.)Let copier A complete the job in a minutes, copier B in b minutes,and copier C in c minutes.In one minute, A can complete 1/a of the job, B can complete 1/b ofthe job,and C can complete 1/c of the job.Together, in one minute, they can complete: 1/a + 1/b + 1/c = 1/50 (ofthe job).That's Equation #1.In 20 minutes, A can complete 20/a of the job.In 50 minutes, B can complete 50/b of the job.Together, they complete: 20/a + 50/b = 1/2 (of the job).That's Equation #2.In 30 minutes, B can complete 30/b of the job.In 80 minutes, C can complete 80/c of the job.Together, they complete: 30/b + 80/c = 3/5 (of the job).That's Equation #3.We have a system of three equations in three variables (a,b,c).#1: 1/a + 1/b + 1/c = 1/50#2: 20/a + 50/b = 1/2 ---> 2/a + 5/b = 1/20#3: 30/a + 80/c = 3/5 ---> 3/a + 8/c = 3/50And we can solve this system WITHOUT clearing the denominators.Multiply #1 by 8, and subtract #3:8/a + 8/b + 8/c = 8/503/a + 8/c = 3/50-----------------------#4: 5/a + 8/b = 5/50 = 1/10Now, solve the 2-by-2 system:#2: 2/a + 5/b = 1/20#4: 5/a + 8/c = 1/10Multiply #2 by 8, multiply #4 by 5, and subtract:16/a + 40/b = 8/20 = 4/1025/a + 40/a = 5/10We get: 9/a = 1/10 ---> a = 90 minutes.Substitute this into #2: 2/90 + 5/b = 1/20 ---> b = 180 minutes.And into #3: 3/90 + 8/c = 3/50 ---> c = 300 minutes.Therefore, copier A is fastest with === a time of 90 minutes.Subject: Re: Can anyone Try this Problem? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9VEpXK08955;>Lora Bush wants to buy a photocopier. The salesperson has the>following information on 3 models. if all 3 are used, a specific job>can be done in 50 minutes. If copier A operates for 20 minutes and>copier B for 50 mininutes, one-half of the job is finished. If copier>B operates for 30 minutes and copier C for 80 minutes, three-fifths>of>the job is done. >Which is the pastest copier, and how long does it takes for this>copier to finish the whole job?the solution x_i of the linear system50*x_1 + 50*x_2 + 50*x_3 = 1 20*x_1 + 50*x_2 = 1/2 30*x_2 + 80*x_3 = 3/5will the fraction of the job photocopier i is able to copy in one minute, and so 1/x_i is the time it takes for photocopier i to do the whole job.Best === wishesTorsten.Subject: Integrations..need some help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA1Jo8q26880;I do not knwo how to go about integrating the following expressionswith respect to x..Can anyone give me some === ideas???e^[sqrt(3x+9)]and1/(sinx-2)RonSubject: Integration help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA248G425689;1) INT e^[sqrt(3x+9)] dxLet u = sqrt(3x + 9), then u^2 = 3x + 9And x = (u^2 - 9)/3 ---> dx = (2/3)u duSubstitute: (2/3)INT u e^u duThis can be integrated by parts: (2/3)(u - 1)e^u + CAnswer: (2/3)[sqrt{3x + 9) - 1]e^[sqrt{3x + 9)] + C2) INT dx/(sin x - 2)I suggest the substitution: z = tan(x/2)The following statements come from a long series of steps.So if you've never seen it, you'll have to take my word for it.sin x = 2z/(1 + z^2), dx = 2 dz/(1 + z^2)Substituting, the integral simplifies to: - INT dz/(z^2 - z + 1)We can complete-the-square in the denominatorand get: - INT dz/[(z - 1/2)^2 + 3/4]Now, we can let: u = z - 1/2and === the integral is of the arctangent form.Good luck!Subject: integration by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA248E325665;why not try using the heinemann books - they have one for each unitand are === aceSubject: Re: integration>why not try using the heinemann books - they have one for each unit>and are aceWhat question Cortland County, New York, USA http://OakRoadSystems.comAddress munging may or may not reduce you get. === http://www.cs.tut.fi/~jkorpela/usenet/laws.htmlSubject: 1!+2!+3!+...+n!=? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id === hA2JcXp20885;1!+2!+3!+...+n!=? in easy formSubject: Re: 1!+2!+3!+...+n!=?> 1!+2!+3!+...+n!=? in easy formI doubt it. If an easy form were known, it should have been given at. === DavidSubject: Re: 1!+2!+3!+...+n!=? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA3DAb322085;I don't know about your problem, but it's close to another sum resultthat is interesting. Namely,1*1!+ 2*2!+ 3*3!+...+ n*n! = (n+1)! -1The sum on the left is the largest integer that you can write in thestandard Cantor representation b1*1!+b2*2!+b3*3!+...+bn*n! = M whereall the bi are integers and 0<=bi<=i -- the largest number withouthaving to use the next bigger factorial (n+1)!. For any integer M <(n+1)! it is easy to write M in the form above(uniquely). In fact, you can use a top-down recursive algorithm whichcompares M to multiples of n!, i.e. looks at floor[M/n!]=bn, or youcan use a bottom up algorithm where you divide M successively by2,3,4,5... and pick off the bi as remainders to get === M=q1*2+b1=[q2*3+b2]*2+b1=q3*4!+b3*3!+b2*2!+b1*1! etc. Subject: GR and QT by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA4FV8t01405;Why can't the theory of general relativity and quantum theory becombined to give a single thory? And what is this string === theorystuff?Hmm....Subject: Reply to GR QT by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA65F6803077;Well, General relitivity relies on the geometry of space-time to beflat...smooth without the presence of matter. Quantum mechanics onthe other hand, creates this quantum foam that distorts space-time onvery small scales. When the two theories are put together they justdont jive mathematically speaking. You get inconsistent mathematicalresults such as 5=6 wich is purely impossible. Now for string theory.That smoothes out the quantum foam and the equations don't yeildanomalies. The downside is that these strings are so tiny (less thanwhat is called planck's length) that it is impossible to verify thereexistance experimentally. It should be exciting to see === what itdevelopes into though.Subject: re: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA63B3D26321;hi jimmy,did you try long division or synthetic division? the first one comesout with no remainer. With the second one, try arranging the divisorin terms of decreasing powers of u, then divide. As for usingmatrices, I don't know where that comes from...this is === pretty simpleusing plain algebra.Subject: Math manipulatives by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA65SdS03822; I am currently teaching 5th grade right now, and want to make thesubject matter more interesting. I am looking for websites that giveexamples of manipulatives that can be used for different types ofproblems (i.e. addition with regrouping). I would love === to see thedifferent manipulatives that can be used.Subject: Quick Math Help: PercentagesIts been a while since I'v been in a math course and I'm trying to write aresearch paper. How do I figure out what the percentage of a number is if Ihave the total number?For instance, if I know that in the year 2000, there were 125 homicides inMassachusetts, and the population of Massachusetts at that time was6,349,097. How do I figure out the percentage of people that were murdered?In other words, how do I figure out what the percentage that 125 is out of6, 349,097. I tried dividing the two numbers but I got a number x 10(-5power). Any idea how I translate this to an k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: === http://www.thinkspot.net/k12math/charter.htmlSubject: Re: Quick Math Help: PercentagesYou need to first think of this as a fraction, which you have:125 / 6 349 097Then you need to convert it to a decimal form, which you have:1.968783... e-5This is scientific notation to avoid writing tons of zeroes... but convertedto a standard decimal, it's:0.00001968783...This is still your fraction... so now you need to pretend you had only 100people and figure the same fraction of those 100 people:0.00001968783... x 100Which gives you:0.001968783... of 100 or0.001968783... %Hope this helps. And hope I didn't make a mistake along the way and thenpost it. heh hehBJ MacNevin> Its been a while since I'v been in a math course and I'm trying to write a> research paper. How do I figure out what the percentage of a number is ifI> have the total number?> For instance, if I know that in the year 2000, there were 125 homicides in> Massachusetts, and the population of Massachusetts at that time was> 6,349,097. How do I figure out the percentage of people that weremurdered?> In other words, how do I figure out what the percentage that 125 is out of> 6, 349,097. I tried dividing the two numbers but I got a number x 10(-5> power). Any idea how I translate this to an k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: === http://www.thinkspot.net/k12math/charter.htmlSubject: JSH: Clearing up confusionIt seems to me that there's some confusion about constant terms withpolynomials and factors of polynomials, so it might help for me topost in an attempt to clear up the confusion.Let's say you have the polynomials x+1 and x+7, and you multiply toget(x+1)(x+7) = x^2 + 8x + 7so the constant term is 7, but of course, the polynomial doesn'tnecessarily have 7 as a factor.Also, dividing back through by x+7 gives you x+1, so the constant termafter the division is 1, whereas before it was 7.Ok, yeah, it's kid stuff, but remember, I'm facing people who seem tobe confused on these issues, which is why I'm posting.But now consider having 7 itself as a factor.Notice that it's different from x+7 as one is a polynomial while 7 isa constant.Can you ever divide 7 off as a variable?Some of you seem intent on arguing as long as you can that you can.However, division of constants by constants is a constant operation,and is NOT variable dependent.No matter how much you wish it is, it's not, and refusing to acceptthat it's not merely makes you a crank who refuses to believe inmathematical logic.So you may wonder about relevancy.Well I have an expression where I divide off 49, a constant factor ofa polynomial P(x), which changes you from having 7, 7 and 22 asconstant terms of factors of that polynomial to having 1, 1, and 22 asthe constant terms of the factors of P(x)/49, so it should bestraightforward.However, because the result has an impact on how mathematicians thinkof themselves and their discipline, several posters have kept uparguments, which basically require that 7 can divide off as a factorof x.It's time for that to stop, or for those people to get the cranklabels they deserve.To see the full argument that those cranks are trying to dispute go tothe blog:http://mathforprofit.blogspot.com/I hope you'll check it out, and help me because it's good formathematics to check the crank assertions.James === Harrishttp://mathforprofit.blogspot.com/Subject: Re: JSH: Clearing up confusion>It seems to me that there's some confusion about constant terms with>polynomials and factors of polynomials, so it might help for me to>post in an attempt to clear up the confusion.>Let's say you have the polynomials x+1 and x+7, and you multiply to>get>(x+1)(x+7) = x^2 + 8x + 7>so the constant term is 7, but of course, the polynomial doesn't>necessarily have 7 as a factor.>Also, dividing back through by x+7 gives you x+1, so the constant term>after the division is 1, whereas before it was 7.>Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to>be confused on these issues, which is why I'm posting.>But now consider having 7 itself as a factor.>Notice that it's different from x+7 as one is a polynomial while 7 is>a constant.>Can you ever divide 7 off as a variable? You have put your finger right on where the disagreementoriginates. You are thinking of factoring 7 out of a polynomial. Forexample, 7 factors out of 14*x + 7, but it does not factorout of 13*x + 7. No one disagrees with that. That is not the problem. The kind of polynomials you are considering look like 5*a_1 + 7. If a_1 is a constant, then 7 either will or will notfactor out depending on what that constant is. Forexample, if a_1 = 14 and never changes from that, then theexpression is always divisible by 7. But a_1 is not a constant. It varies. It is dependenton the variable x in your polynomial P(x). You are not dealing with things like 14*x + 7. Your coefficient a_1 is not a constant like 14. It varies. You know that. For x = 0, a_1 = 0. So for that value of x, clearly 7*can* be factored out of the whole thing. In general you don't know what values a_1 has for othervalues of x. Writing a_1 as a function of x, say, a_1(x),you know that a_1(0) = 0, but you probably cannot easilyfigure out what a_1(1) is, or a_1(17), etc.. So the question is not whether 5*a_1(x) + 7 is divisibleby 7 AS A FUNCTION. It is whether it is divisible by 7FOR EXPLICIT VALUES OF x. Its divisibility by 7 is*dependent* on x, in general. The constant term, on which you have put so muchstress, is not what's important. [*No one* here hasargued that the constant term is not constant.] Itsdivisibility by 7 is almost irrelevant. What isabsolutely essential to your argument is divisibilityof the *whole expression* 5*a_1(x) + 7 for specificvalues of x *other than x = 0*. Which means, of course,that you must prove divisibility of a_1(x) by 7 forvalues of x other than 0. Knowing that 5*a_1 + 7 is divisible by 7 for x = 0doesn't tell you whether it is divisible by 7 for x = 1.In fact we know that, for x = 1, a_1 is neither coprimeto 7 nor divisible by 7. This is not rocket science. This is simple. This iselementary. A grade-school student could understand thepoint I am making. You for some reason are not gettingit. All of us are trying to tell you this same thing. You are not getting it. There is really only oneexplanation. You are wearing blinders. Nora B.>Some of you seem intent on arguing as long as you can that you can.>However, division of constants by constants is a constant operation,>and is NOT variable dependent.>No matter how much you wish it is, it's not, and refusing to accept>that it's not merely makes you a crank who refuses to believe in>mathematical logic.>So you may wonder about relevancy.>Well I have an expression where I divide off 49, a constant factor of>a polynomial P(x), which changes you from having 7, 7 and 22 as>constant terms of factors of that polynomial to having 1, 1, and 22 as>the constant terms of the factors of P(x)/49, so it should be>straightforward.>However, because the result has an impact on how mathematicians think>of themselves and their discipline, several posters have kept up>arguments, which basically require that 7 can divide off as a factor>of x.>It's time for that to stop, or for those people to get the crank>labels they deserve.>To see the full argument that those cranks are trying to dispute go to>the blog:>http:// mathforprofit.blogspot.com/I hope you'll check it out, and help me because it's good for>mathematics to check the crank assertions.>James Harris>http:// === mathforprofit.blogspot.com/[...]> This is not rocket science. This is simple. This is>elementary. A grade-school student could understand the>point I am making. You for some reason are not getting>it. All of us are trying to tell you this same thing. >You are not getting it. There is really only one>explanation.> You are wearing blinders.> Nora B.Another explanation has often occurred to me:He often goes on about how we ignore the Truth, decidingwhat to believe on the basis of what we're told to believe,etc etc ad nauseam. The funny thing about that is thathe's often demonstrated that that _is_ his approach:he doesn't believe something when someone showshim a simple proof, but he believes it when he hearsthat Dirichlet proved it (or another example: the onlyreason he believes that the algebraic integers forma ring is that that's what he's been told - he's certainlynever read and understood a proof of that fact.)So it could be that he thinks that things actually_are_ decided on that sort of basis - if so he canwin fame, glory and babes just by being persistent,doesn't matter whether he's === actually right.************************Subject: Re: JSH: Clearing up confusion>It seems to me that there's some confusion about constant terms with>polynomials and factors of polynomials, so it might help for me to>post in an attempt to clear up the confusion.>Let's say you have the polynomials x+1 and x+7, and you multiply to>get>(x+1)(x+7) = x^2 + 8x + 7>so the constant term is 7, but of course, the polynomial doesn't>necessarily have 7 as a factor.>Also, dividing back through by x+7 gives you x+1, so the constant term>after the division is 1, whereas before it was 7.>Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to>be confused on these issues, which is why I'm posting.>But now consider having 7 itself as a factor.>Notice that it's different from x+7 as one is a polynomial while 7 is>a constant.>Can you ever divide 7 off as a variable?> You have put your finger right on where the disagreement> originates.> You are thinking of factoring 7 out of a polynomial. For> example, 7 factors out of 14*x + 7, but it does not factor> out of 13*x + 7.> No one disagrees with that. That is not the problem.> The kind of polynomials you are considering look like> 5*a_1 + 7.> If a_1 is a constant, then 7 either will or will not> factor out depending on what that constant is. For> example, if a_1 = 14 and never changes from that, then the> expression is always divisible by 7.> But a_1 is not a constant. It varies. It is dependent> on the variable x in your polynomial P(x). You are not > dealing with things like 14*x + 7. Your coefficient a_1 is not > a constant like 14. It varies. You know that.Notice readers that Nora Baron is being inconsistent as 14x *does*vary, and I've been writing a_1(x) by request to show that it's afunction of x, but notice how Nora Baron, one of the posters whoused to get so upset at my just writing a_1 when it's a function of x,switched to just writing a_1 and now calls it a coefficient.Did you catch the sleight-of-hand before I pointed it out?The reality is that Nora Baron is a crank poster who got away withfooling you for a long time until I found a simple enough presentationof the argument to catch the poster in the act. And now I can moreeasily point out to you how the poster is doing it.And think about it, given what I've shown, how could Nora Baron NOTbe deliberately trying to === mislead you?James HarrisSubject: Re: JSH: Clearing up confusion>It seems to me that there's some confusion about constant terms with>polynomials and factors of polynomials, so it might help for me to>post in an attempt to clear up the confusion.>Let's say you have the polynomials x+1 and x+7, and you multiply to>get>(x+1)(x+7) = x^2 + 8x + 7>so the constant term is 7, but of course, the polynomial doesn't>necessarily have 7 as a factor.>Also, dividing back through by x+7 gives you x+1, so the constant term>after the division is 1, whereas before it was 7.>Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to>be confused on these issues, which is why I'm posting.>But now consider having 7 itself as a factor.>Notice that it's different from x+7 as one is a polynomial while 7 is>a constant.>Can you ever divide 7 off as a variable?> You have put your finger right on where the disagreement> originates.> You are thinking of factoring 7 out of a polynomial. For> example, 7 factors out of 14*x + 7, but it does not factor> out of 13*x + 7.> No one disagrees with that. That is not the problem.> The kind of polynomials you are considering look like> 5*a_1 + 7.> If a_1 is a constant, then 7 either will or will not> factor out depending on what that constant is. For> example, if a_1 = 14 and never changes from that, then the> expression is always divisible by 7.> But a_1 is not a constant. It varies. It is dependent> on the variable x in your polynomial P(x). You are not > dealing with things like 14*x + 7. Your coefficient a_1 is not > a constant like 14. It varies. You know that.>Notice readers ... yes ... notice the appeal to the grandstand ...>that Nora Baron is being inconsistent as 14x *does*>vary, and I've been writing a_1(x) by request to show that it's a>function of x, but notice how Nora Baron, one of the posters who>used to get so upset at my just writing a_1 when it's a function of x,>switched to just writing a_1 and now calls it a coefficient.14*x + 7 was intended to parallel the polynomial youstarted with in this post, x^2 + 8*x + 7 = (x + 1)*(x + 7). You have not been consistent in writing a_1 or a_1(x). Mypreference is to write a_1(x), so let's stick with that. However when I write 14*x + 7, it is 14 which isanalogous to a_1(x), and x is analogous to 5 in yourexpression for the factor 5*a_1(x) + 7. I think that wouldbe clear to most people who have been keeping up on this.>Did you catch the sleight-of-hand before I pointed it out? None intended. Others seemed to understand what I was saying with no trouble.>The reality is that Nora Baron is a crank poster who got away with>fooling you for a long time until I found a simple enough presentation>of the argument to catch the poster in the act. And now I can more>easily point out to you how the poster is doing it. Since you are clearly playing to an audience, why notexplain why you did not reply to the real substance of mypost ? You deleted it. You did not answer anything, essentially.I am repeating it below to allow for a less evasive response -------------------------------------------------------------- --------------- For x = 0, a_1 = 0. So for that value of x, clearly 7can be factored out of the whole thing. In general you don't know what values a_1 has for othervalues of x. Writing a_1 as a function of x, say, a_1(x),you know that a_1(0) = 0, but you probably cannot easilyfigure out what a_1(1) is, or a_1(17), etc.. So the question is not whether 5*a_1(x) + 7 is divisibleby 7 AS A FUNCTION. It is whether it is divisible by 7FOR EXPLICIT VALUES OF x. Its divisibility by 7 is*dependent* on x, in general. The constant term, on which you have put so muchstress, is not what's important. [*No one* here hasargued that the constant term is not constant.] Itsdivisibility by 7 is almost irrelevant. What isabsolutely essential to your argument is divisibilityof the *whole expression* 5*a_1(x) + 7 for specificvalues of x *other than x = 0*. Knowing that 5*a_1 + 7 is divisible by 7 for x = 0doesn't tell you whether it is divisible by 7 for x = 1.Or if it does, you certainly have not offered a proof ofit. This is not rocket science. This is simple. This iselementary. A grade-school student could understand thepoint I am making. You for some reason are not gettingit. All of us are trying to tell you this same thing. You are not getting it. There is really only oneexplanation. You are wearing blinders. Nora B.------------------------------------------------------------ --------------->Some of you seem intent on arguing as long as you can that you can.>However, division of constants by constants is a constant operation,>and is NOT variable dependent. Of course. No disagreement. However divisibility of 5*a_1(x) + 7depends on more than just the constant term, 7. It also depends on a_1(x). You need to know divisibility whenx = 1, x = 5, x = 173, etc, etc..>No matter how much you wish it is, it's not, and refusing to accept>that it's not merely makes you a crank who refuses to believe in>mathematical logic.>So you may wonder about relevancy.>Well I have an expression where I divide off 49, a constant factor of>a polynomial P(x), which changes you from having 7, 7 and 22 as>constant terms of factors of that polynomial to having 1, 1, and 22 as>the constant terms of the factors of P(x)/49, so it should be>straightforward.>However, because the result has an impact on how mathematicians think>of themselves and their discipline, several posters have kept up>arguments, which basically require that 7 can divide off as a factor>of x.you think divisibility of 5*a_1(x) + 7 is ENTIRELY determinedby that last term, 7. Of course it is not. The variablepart, a_1(x), must also be considered. Why is this so hardfor you to understand? Dik Winter has pointed out that there are factorizationsof 7*7 = 49 of the form 49 = r * s * t, where r, s, and tare algebraic integers, and where (5*a_1 + 7)/r, (5*a_2 + 7)/s, and (5*b_3 + 22)/tare also all algebraic integers. This factorization does notresult in any conflict regarding the constant terms, because, since r*s*t = 49, (7/r)*(7/s)*(22/t) = 49*22/(r*s*t)= 22. Here r, s, and t are NOT constants; they are dependenton x. When x = 0, r = 7, s = 7, and t = 1. There areinfinitely many such factorizations into nonunits r, s, and t, one for each value of x. For x <> 0 they aredifficult to calculate. They nevertheless exist. I really think your error in this is related to yourhabit of thinking in terms of simple examples involving the integers or polynomials with integer coefficients, where prime factorization is true and important. It is not true in the algebraic integers. There, numbers (like 49) can be factored in infinitely many ways with counterintuitive results.>It's time for that to stop, or for those people to get the crank>labels they deserve. If labels are that important to you, feel free to use any that you want. The essential thing here is the math, isn't it? One other question, related to the issues regarding 5*a_1(x) + 7.The third term in your factorization is 5*b_3(x) + 22,where the constant term is 22. Of course this constant term is coprime to 7. If each of two numbers is coprime to 7, that does*not* imply that their sum is coprime to 7. Example: 20 + 22. You know that b_3(0) = 0, so when x = 0, b_3(x) is notcoprime to 7. But again, when x <> 0, b_3(x) is hard tocalculate, and it will be difficult to tell whether it iscoprime to 7, divisible by 7, or whatever. Knowing thatb_3(0) = 0 tells you nothing about, for example, b_3(1). Can you tell us why you think b_3(1) is *not* coprime to 7 ? A further problem. Algebraic numbers are trickier than onemight think. For example, assume A is not coprime to 7,while B is coprime to 7 (like, 22). What about A + B ? Is A + B necessarily (1) coprime to 7 or (2) not coprime to 7 or (3) could be either Which do you think is right? A further reminder. Assuming your usual factorization, P(x) = (5*a_1 + 7)*(5*a_2 + 7)*(5*b3 + 22),there are at least two different proofs, one based onelementary Galois theory and one based on elementaryalgebraic number theory, which show that both a_1 and a_2 are*not coprime* to 7 and both are *not divisible* by 7. You have seenthese proofs and have not found errors in them. They implymore than that your proof is wrong. They imply that yourcentral claim is wrong. Therefore your proof is not onlywrong: it cannot be fixed. If your argument were valid, this would imply that there isindeed a problem in core mathematics - it would imply adirect contradiction and *inconsistency* of mathematics. We have pointed to an exact place in your argument wherethere is a huge gap. In your numbered-steps version of yourargument, it is step #6. The contradiction noted above plusthis unexplained gap lead to one conclusion: you think youhave a proof, but you are simply wrong. Your central claim isfalse. You cannot fill the gap. Arturo described it well. He said you appear to assign'almost magical powers' to the constant term of yourpolynomial and its factors. No one has argued, as you haveclaimed, that the constant terms are not constant. You assignmuch more power and significance to the constant terms thanis reasonable, and without explanation. We know you have made an error and we know where you madeit. We have pointed out exactly where you have a gap. Youcontinue to say and perhaps believe that no one has foundan error in your stepped argument. That is false.The one thing we don't know is what you are thinkingwhen you leap across that gap. Only you can explain that. So far you have not done so. You simply keep repeating thefact that 49 can factor out of 7*7*22 in only one way - atrue fact in the ordinary integers, but false in thealgebraic integers. Once you understand this, everythingelse unravels. Why is it so hard for you to see?>To see the full argument that those cranks are trying to dispute go to>the blog:>http:// mathforprofit.blogspot.com/I hope you'll check it out, and help me because it's good for>mathematics to check the crank assertions.>James Harris>http:// mathforprofit.blogspot.com/And think about it, given what I've shown, how could Nora Baron NOT>be deliberately trying to mislead you? Indeed, a good question. My reputation would not be worthmuch if I were caught doing that. And certainly I am givinga lot of opportunities to be caught. I would be a fool totry to mislead people here. Thousands of people can see whatI have done, and if you think every single one of them wouldagree to hush up my trying to mislead people, you should joina conspiracy chat group right away. My advice to your loyal readers on this is, just judge what is said on its own merits and decide for yourself without anyguff from either me or JSH about whether you are beingmisled.Nora B.>James === HarrisSubject: Re: JSH: Clearing up confusion>It seems to me that there's some confusion about constant terms with>polynomials and factors of polynomials, so it might help for me to>post in an attempt to clear up the confusion.> >Let's say you have the polynomials x+1 and x+7, and you multiply to>get>(x+1)(x+7) = x^2 + 8x + 7>so the constant term is 7, but of course, the polynomial doesn't>necessarily have 7 as a factor.>Also, dividing back through by x+7 gives you x+1, so the constant term>after the division is 1, whereas before it was 7.>Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to>be confused on these issues, which is why I'm posting.>But now consider having 7 itself as a factor.>Notice that it's different from x+7 as one is a polynomial while 7 is>a constant.>Can you ever divide 7 off as a variable?> You have put your finger right on where the disagreement> originates.> You are thinking of factoring 7 out of a polynomial. For> example, 7 factors out of 14*x + 7, but it does not factor> out of 13*x + 7.> No one disagrees with that. That is not the problem.> The kind of polynomials you are considering look like> 5*a_1 + 7.> If a_1 is a constant, then 7 either will or will not> factor out depending on what that constant is. For> example, if a_1 = 14 and never changes from that, then the> expression is always divisible by 7.> But a_1 is not a constant. It varies. It is dependent> on the variable x in your polynomial P(x). You are not> dealing with things like 14*x + 7. Your coefficient a_1 is not> a constant like 14. It varies. You know that.> Notice readers that Nora Baron is being inconsistent as 14x *does*> vary, and I've been writing a_1(x) by request to show that it's a> function of x, but notice how Nora Baron, one of the posters who> used to get so upset at my just writing a_1 when it's a function of x,> switched to just writing a_1 and now calls it a coefficient.Nora Baron's reference here is to James' factor in his previous posts: g_1(m) = a_1(m)*x + u*fwhere his a_1(m) *is* a coefficient of x.Re the 2nd objection, it's obvious that Nora Baron switched to just writinga_1 above because she was giving an example in which a1 is a constant givenby 14. One could hardly say if a_1(x) is a constant. Also she neversaid that 14*x doesn't vary with x.> Did you catch the sleight-of-hand before I pointed it out?There was no sleight-of-hand. Only Nora Baron's usual argument on themerits of the math. I only wish James would stop obfuscating and speak tothe math as well.KeithK[Smoke screen Snipped]> James === HarrisSubject: Re: JSH: Clearing up confusion>It seems to me that there's some confusion about constant terms with>polynomials and factors of polynomials, so it might help for me to>post in an attempt to clear up the confusion.>Let's say you have the polynomials x+1 and x+7, and you multiply to>get>(x+1)(x+7) = x^2 + 8x + 7>so the constant term is 7, but of course, the polynomial doesn't>necessarily have 7 as a factor.>Also, dividing back through by x+7 gives you x+1, so the constant term>after the division is 1, whereas before it was 7.>Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to>be confused on these issues, which is why I'm posting.>But now consider having 7 itself as a factor.>Notice that it's different from x+7 as one is a polynomial while 7 is>a constant.>Can you ever divide 7 off as a variable?> You have put your finger right on where the disagreement> originates. You are thinking of factoring 7 out of a polynomial. For> example, 7 factors out of 14*x + 7, but it does not factor> out of 13*x + 7. No one disagrees with that. That is not the problem. The kind of polynomials you are considering look like 5*a_1 + 7. If a_1 is a constant, then 7 either will or will not> factor out depending on what that constant is. For> example, if a_1 = 14 and never changes from that, then the> expression is always divisible by 7. But a_1 is not a constant. It varies. It is dependent> on the variable x in your polynomial P(x). You are not> dealing with things like 14*x + 7. Your coefficient a_1 is not> a constant like 14. It varies. You know that.> Notice readers that Nora Baron is being inconsistent as 14x *does*> vary, and I've been writing a_1(x) by request to show that it's a> function of x, but notice how Nora Baron, one of the posters who> used to get so upset at my just writing a_1 when it's a function of x,> switched to just writing a_1 and now calls it a coefficient.> Nora Baron's reference here is to James' factor in his previous posts:> g_1(m) = a_1(m)*x + u*f> where his a_1(m) *is* a coefficient of x.How do you know?Did you ask that === poster? Are you two in communication?James HarrisSubject: Re: JSH: Clearing up confusion>It seems to me that there's some confusion about constant termswith>polynomials and factors of polynomials, so it might help for me to>post in an attempt to clear up the confusion.>Let's say you have the polynomials x+1 and x+7, and you multiply to>get>(x+1)(x+7) = x^2 + 8x + 7>so the constant term is 7, but of course, the polynomial doesn't>necessarily have 7 as a factor.>Also, dividing back through by x+7 gives you x+1, so the constantterm>after the division is 1, whereas before it was 7.>Ok, yeah, it's kid stuff, but remember, I'm facing people who seemto>be confused on these issues, which is why I'm posting.>But now consider having 7 itself as a factor.>Notice that it's different from x+7 as one is a polynomial while 7is>a constant.>Can you ever divide 7 off as a variable?> You have put your finger right on where the disagreement> originates.> You are thinking of factoring 7 out of a polynomial. For> example, 7 factors out of 14*x + 7, but it does not factor> out of 13*x + 7.> No one disagrees with that. That is not the problem.> The kind of polynomials you are considering look like> 5*a_1 + 7.> If a_1 is a constant, then 7 either will or will not> factor out depending on what that constant is. For> example, if a_1 = 14 and never changes from that, then the> expression is always divisible by 7.> But a_1 is not a constant. It varies. It is dependent> on the variable x in your polynomial P(x). You are not> dealing with things like 14*x + 7. Your coefficient a_1 is not> a constant like 14. It varies. You know that.Notice readers that Nora Baron is being inconsistent as 14x *does*> vary, and I've been writing a_1(x) by request to show that it's a> function of x, but notice how Nora Baron, one of the posters who> used to get so upset at my just writing a_1 when it's a function of x,> switched to just writing a_1 and now calls it a coefficient.> Nora Baron's reference here is to James' factor in his previous posts:> g_1(m) = a_1(m)*x + u*f> where his a_1(m) *is* a coefficient of x.> How do you know?> Did you ask that poster? Are you two in communication?> James HarrisHow did he know a_1(m) is a coefficient? It is multiplied by the x term.That's a coefficient; Basic === Algebra.Subject: Re: JSH: Clearing up confusionit's about time that you'd cleared that up, monsieur Harris! > How did he know a_1(m) is a coefficient? It is multiplied by the x term.> That's a coefficient; Basic Algebra.--ils duces === d'Enron!http://larouchepub.com/radio/index.htmlSubject: Re: JSH: Clearing up confusion>And think about it, given what I've shown, how could Nora Baron NOT>be deliberately trying to mislead you?Oh, that's easy, but I'll leave it as a proof for === the reader.DougSubject: Re: JSH: Clearing up confusion > Let's say you have the polynomials x+1 and x+7, and you multiply to > get > (x+1)(x+7) = x^2 + 8x + 7 > so the constant term is 7, but of course, the polynomial doesn't > necessarily have 7 as a factor.It works *if* the factors are polynomials over the same ring. In yourcase they are *not* polynomials over the same ring, they are not evenpolynomials. Giving a polynomial factorisation works as a red herring. > Also, dividing back through by x+7 gives you x+1, so the constant term > after the division is 1, whereas before it was 7.Yup, indeed, you divided a function with constant term 7 by a functionwith constant term 7 to get a function with constant term 1. > Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to > be confused on these issues, which is why I'm posting.Yup it is. So *you* are allowed to divide by functions with constantterm 7, but *I* am not allowed to do such? > But now consider having 7 itself as a factor. > Notice that it's different from x+7 as one is a polynomial while 7 is > a constant. > Can you ever divide 7 off as a variable?Nope, and nobody does so. But how the factors of 7 distribute may dependon some variable. Consider x^2 + 3x + 2in the integers. It factors as (x + 1)(x + 2) and for all x it is divisibleby 2. So we can say: (x^2 + 3x + 2)/2 = (x+1)/w1(x) . (x+2)/w2(x)where w1 and w2 are defined as follows: w1(x) = gcd(x+1, 2) w2(x) = gcd(x+2, 2).You may check that it does *not* work as (x^2 + 3x + 2)/2 = (x + 1)(x/2 + 1)in the integers. Which is similar to what you require *your* polynomialto do in the algebraic integers.You may check that w1(x).w2(x) = 2, a constant. You may also checkthat the constant terms of both factors with w1 and w2 are 1.... > Well I have an expression where I divide off 49, a constant factor of > a polynomial P(x), which changes you from having 7, 7 and 22 as > constant terms of factors of that polynomial to having 1, 1, and 22 as > the constant terms of the factors of P(x)/49, so it should be > straightforward.Yup, it is. There are three functions w1(x), w2(x) and w3(x) that multiplytogether to get the constant 49. They have *constant terms* 7, 7 and 1.So dividing the factors by w1, w2 and w3 will give you 1, 1 and 22 asthe constant terms of P(x)/49 = P(x)/w1(x)w2(x)w3(x). Do you have aproblem with this? Note that I actually gave instructions how toconstruct those three functions (and it is not dissimilar to theconstruction in the little example above). [ You may note that myinstructions do not work when x = 7y + 5, because there is a slightcomplication in that case... Your polynomial is divisible by 343 forthose values, so the gcd's have to be adjusted to cater for the additional7.] > However, because the result has an impact on how mathematicians think > of themselves and their discipline, several posters have kept up > arguments, which basically require that 7 can divide off as a factor > of x.No, that has never been said. This must be a result of your bad reading.7 can divide of as a *function* of x is closer to reality. But thathappens when you use gcd-like amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn === amsterdam, nederland; http://www.cwi.nl/~dik/Subject: Re: JSH: Clearing up confusion> It seems to me that there's some confusion about constant terms with> polynomials and factors of polynomials, so it might help for me to> post in an attempt to clear up the confusion.> Let's say you have the polynomials x+1 and x+7, and you multiply to> get> (x+1)(x+7) = x^2 + 8x + 7Ok so far.> so the constant term is 7, but of course, the polynomial doesn't> necessarily have 7 as a factor.That's true, the polynomial does not necessarily have 7 as a factor. But if you EVALUATE the polynomial at various values of x, the result may well have 7 as a factor. For example if f(x) = x^2+8x+7, then f(0) does have 7 as a factor.> Also, dividing back through by x+7 gives you x+1, so the constant term> after the division is 1, whereas before it was 7.> Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to> be confused on these issues, which is why I'm posting.We're all in agreement so far.> But now consider having 7 itself as a factor.7 itself as a factor of what?> Notice that it's different from x+7 as one is a polynomial while 7 is> a constant.x+7 is a 1-degree poly, and 7 is a 0-degree poly. > Can you ever divide 7 off as a variable?> Some of you seem intent on arguing as long as you can that you can.> However, division of constants by constants is a constant operation,> and is NOT variable dependent.> No matter how much you wish it is, it's not, and refusing to accept> that it's not merely makes you a crank who refuses to believe in> mathematical logic.What are you talking about?> So you may wonder about relevancy.> Well I have an expression where I divide off 49, a constant factor of> a polynomial P(x), which changes you from having 7, 7 and 22 as> constant terms of factors of that polynomial to having 1, 1, and 22 as> the constant terms of the factors of P(x)/49, so it should be> straightforward.Now you are babbling. You and I were in total agreement right up until this paragraph, which has no context and makes no sense. Why don't you keep on going with your x+7 example and say what === it is you're talking about?Subject: Re: JSH: Clearing up confusion> It seems to me that there's some confusion about constant terms with> polynomials and factors of polynomials, so it might help for me to> post in an attempt to clear up the confusion.> Let's say you have the polynomials x+1 and x+7, and you multiply to> get> (x+1)(x+7) = x^2 + 8x + 7 > Ok so far.Good.> so the constant term is 7, but of course, the polynomial doesn't> necessarily have 7 as a factor.> That's true, the polynomial does not necessarily have 7 as a factor. > But if you EVALUATE the polynomial at various values of x, the result > may well have 7 as a factor. For example if f(x) = x^2+8x+7, then f(0) > does have 7 as a factor.That's not important for this discussion. > Also, dividing back through by x+7 gives you x+1, so the constant term> after the division is 1, whereas before it was 7.> Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to> be confused on these issues, which is why I'm posting.> We're all in agreement so far.Good. > But now consider having 7 itself as a factor.> 7 itself as a factor of what?Of a polynomial.> Notice that it's different from x+7 as one is a polynomial while 7 is> a constant.> x+7 is a 1-degree poly, and 7 is a 0-degree poly. Yeah, it's constant. My point is that 7 doesn't divide off as avariable, as it's a constant. It's still kid stuff but it needs to bementioned given the posts I've been seeing from certain people. > Can you ever divide 7 off as a variable?> Some of you seem intent on arguing as long as you can that you can.> However, division of constants by constants is a constant operation,> and is NOT variable dependent.> No matter how much you wish it is, it's not, and refusing to accept> that it's not merely makes you a crank who refuses to believe in> mathematical logic.> What are you talking about?You know, like if you have P(x) = 7S(x), like 7(x+1) = 7x + 7, youcan't have it where 7 divides off as some variable, as it's just 7.> So you may wonder about relevancy.> Well I have an expression where I divide off 49, a constant factor of> a polynomial P(x), which changes you from having 7, 7 and 22 as> constant terms of factors of that polynomial to having 1, 1, and 22 as> the constant terms of the factors of P(x)/49, so it should be> straightforward.> Now you are babbling. You and I were in total agreement right up until > this paragraph, which has no context and makes no sense. Why don't you > keep on going with your x+7 example and say what it is you're talking > about?My point is that the x+7 example isn't what the situation is, but someof you seem to think that it is. The situation with my work is likewith 7 itself--a constant--and NOT like x+7, a constant plus avariable.So the constant terms in the expressions I use go from being 7, 7 and22 to being 1, 1 and 22, and the only way to go from 7 to 1 is todivide by 7.It's like if you have 7 divided by *something* such that the result is1, you have7/x = 1, so x=7.Get it?The posters who are arguing with me trying to dispute my proof arearguing against that x being 7 because people are letting them getaway with the error.Some of you see w_1(x) or whatever and seem to believe that *maybe*someway, if you have some really weird function, like 7/w_1(x), thatw_1 does not have to equal 7, if 7/w_1(x) = 1, but that's nonsense.Basically those of you who are fighting my proof are === cranks.James HarrisSubject: Re: JSH: Clearing up confusion... > Some of you see w_1(x) or whatever and seem to believe that *maybe* > someway, if you have some really weird function, like 7/w_1(x), that > w_1 does not have to equal 7, if 7/w_1(x) = 1, but that's nonsense. > Basically those of you who are fighting my proof are cranks.Basically you are misstating the objection. 7/w_1(x) = 1 is onlynecessary if x = 0. So if w_1(0) = 7, there is no problem.Like w_1(x) = gcd((5 amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn === amsterdam, nederland; http://www.cwi.nl/~dik/Subject: Re: JSH: Clearing up confusion> ...> Some of you see w_1(x) or whatever and seem to believe that *maybe*> someway, if you have some really weird function, like 7/w_1(x), that> w_1 does not have to equal 7, if 7/w_1(x) = 1, but that's nonsense.> Basically those of you who are fighting my proof are cranks.> Basically you are misstating the objection. 7/w_1(x) = 1 is only> necessary if x = 0. So if w_1(0) = 7, there is no problem.> Like w_1(x) = gcd((5 a1(x) + 7), 49).Ok, so Dik Winter wants to have some w_1(x) that divides as a functionof x.So divide 5a_1(x) + 7 by w_1(x), and you get some function I'll callg(x).But at x=0, g(0) = 1, so letting h(x) = g(x) + 1, I have my new factorafter 49 is dividing off of P(x).But I say that the factor is 5a_1(x)/7 + 1, and notice that then youhave 5a_1(x)/7 + 1 = h(x) + 1, so h(x) = 5a_1(x)/7.That's why Dik Winter keeps writing his expressions out as ratiosJames === HarrisSubject: Re: JSH: Clearing up confusion > ... > Some of you see w_1(x) or whatever and seem to believe that *maybe* > someway, if you have some really weird function, like 7/w_1(x), that > w_1 does not have to equal 7, if 7/w_1(x) = 1, but that's nonsense. > Basically those of you who are fighting my proof are cranks. > Basically you are misstating the objection. 7/w_1(x) = 1 is only > necessary if x = 0. So if w_1(0) = 7, there is no problem. > Like w_1(x) = gcd((5 a1(x) + 7), 49). > Ok, so Dik Winter wants to have some w_1(x) that divides as a function > of x. > So divide 5a_1(x) + 7 by w_1(x), and you get some function I'll call > g(x). > But at x=0, g(0) = 1, so letting h(x) = g(x) + 1, I have my new factor > after 49 is dividing off of P(x).You mean h(x) = g(x) - 1, a bit precision might help. > But I say that the factor is 5a_1(x)/7 + 1, and notice that then you > have > 5a_1(x)/7 + 1 = h(x) + 1, so h(x) = 5a_1(x)/7. > That's why Dik Winter keeps writing his expressions out as ratiosDivide 5a1(x) + 7 by w1(x), we get: g(x) = 5a1(x)/w1(x) + 7/w1(x).Set h(x) = g(x) - 1, so h(x) = 5a1(x)/w1(x) + 7/w1(x) - 1.We get h(0) = 0, so the constant term of h(x) = 0.Furthermore, with the definition above, h(x) is an algebraic integer.When (as you wish) you divide by 7, you do not necessarily get analgebraic integer. This all, regardless how a1(x) is defined, andwith the 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, === 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/Subject: Re: JSH: Clearing up confusion> ...> Some of you see w_1(x) or whatever and seem to believe that *maybe*> someway, if you have some really weird function, like 7/w_1(x), that> w_1 does not have to equal 7, if 7/w_1(x) = 1, but that's nonsense.> Basically those of you who are fighting my proof are cranks.> Basically you are misstating the objection. 7/w_1(x) = 1 is only> necessary if x = 0. So if w_1(0) = 7, there is no problem.> Like w_1(x) = gcd((5 a1(x) + 7), 49).> Ok, so Dik Winter wants to have some w_1(x) that divides as a function> of x.> So divide 5a_1(x) + 7 by w_1(x), and you get some function I'll call> g(x).> But at x=0, g(0) = 1, so letting h(x) = g(x) + 1, I have my new factor> after 49 is dividing off of P(x).> You mean h(x) = g(x) - 1, a bit precision might help.Oh yeah, easy mistake, but yours Dik Winter is MUCH more embarrassing.After all the factorization is(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where you want to divide through by w_1(x) w_2(x) w_3(x) = 49, sodoing that gives(5 a_1(x)/w_1(x) + 7/w_1(x))(5 a_2(x)/w_2(x) + 7/w_2(x))(5b_3(x)/w_3(x) + 22/w_3(x)) = 300125 x^3 - 18375 x^2 - 360 x + 22and notice that necessarily, if you multiply everything out andsimplify, you have(7/w_1(x))(7/w_2(x))(22/w_3(x)) = 22 which leaves you === out of luck Dik Winter.James HarrisSubject: Re: JSH: Clearing up confusion... > Oh yeah, easy mistake, but yours Dik Winter is MUCH more embarrassing. > After all the factorization is > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where you want to divide through by w_1(x) w_2(x) w_3(x) = 49, so > doing that gives > (5 a_1(x)/w_1(x) + 7/w_1(x))(5 a_2(x)/w_2(x) + 7/w_2(x))(5 > b_3(x)/w_3(x) + 22/w_3(x)) = 300125 x^3 - 18375 x^2 - 360 x + 22 > and notice that necessarily, if you multiply everything out and > simplify, you have > (7/w_1(x))(7/w_2(x))(22/w_3(x)) = 22 > which leaves you out of luck Dik Winter.Yes, the product of those three terms is 22. What is the problem?You home: bovenover 215, 1025 jn amsterdam, nederland; === http://www.cwi.nl/~dik/Subject: Re: JSH: Clearing up confusion>Oh yeah, easy mistake, but yours Dik Winter is MUCH more embarrassing.Much more embarrassing than starting a new thread every time youcan't talk your way out of a === corner?DougSubject: Re: JSH: Clearing up confusion> ...> Some of you see w_1(x) or whatever and seem to believe that*maybe*> someway, if you have some really weird function, like 7/w_1(x),that> w_1 does not have to equal 7, if 7/w_1(x) = 1, but that'snonsense.> Basically those of you who are fighting my proof are cranks.> Basically you are misstating the objection. 7/w_1(x) = 1 is only> necessary if x = 0. So if w_1(0) = 7, there is no problem.> Like w_1(x) = gcd((5 a1(x) + 7), 49).> Ok, so Dik Winter wants to have some w_1(x) that divides as afunction> of x.> So divide 5a_1(x) + 7 by w_1(x), and you get some function I'll call> g(x).> But at x=0, g(0) = 1, so letting h(x) = g(x) + 1, I have my newfactor> after 49 is dividing off of P(x).> You mean h(x) = g(x) - 1, a bit precision might help.> Oh yeah, easy mistake, but yours Dik Winter is MUCH more embarrassing.> After all the factorization is> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) => 49(300125 x^3 - 18375 x^2 - 360 x + 22)> where you want to divide through by w_1(x) w_2(x) w_3(x) = 49, so> doing that gives> (5 a_1(x)/w_1(x) + 7/w_1(x))(5 a_2(x)/w_2(x) + 7/w_2(x))(5> b_3(x)/w_3(x) + 22/w_3(x)) = 300125 x^3 - 18375 x^2 - 360 x + 22> and notice that necessarily, if you multiply everything out and> simplify, you have> (7/w_1(x))(7/w_2(x))(22/w_3(x)) = 22> which leaves you out of luck Dik Winter.> James HarrisWhy are you so hateful toward people who make mistakes? Everyone makesmistakes, at least he admits when he is wrong. By the way, your use === ofswearing makes you look very immature.Subject: Re: JSH: Clearing up confusion... > Oh yeah, easy mistake, but yours Dik Winter is MUCH more embarrassing. > After all the factorization is > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where you want to divide through by w_1(x) w_2(x) w_3(x) = 49, so > doing that gives > (5 a_1(x)/w_1(x) + 7/w_1(x))(5 a_2(x)/w_2(x) + 7/w_2(x))(5 > b_3(x)/w_3(x) + 22/w_3(x)) = 300125 x^3 - 18375 x^2 - 360 x + 22 > and notice that necessarily, if you multiply everything out and > simplify, you have > (7/w_1(x))(7/w_2(x))(22/w_3(x)) = 22 > which leaves you out of luck Dik Winter. > Why are you so hateful toward people who make mistakes? Everyone makes > mistakes, at least he admits when he is wrong. By the way, your use of > swearing makes you look very immature.The strangest about this is that there is no mistake. The product indeedcomes to 22. When you do the multiplication James suggest you come at22, just what was needed. So I really do not understand what James isdoing nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, === nederland; http://www.cwi.nl/~dik/Subject: Re: JSH: Clearing up confusion> It seems to me that there's some confusion about constant terms with> polynomials and factors of polynomials, so it might help for me to> post in an attempt to clear up the confusion.> Let's say you have the polynomials x+1 and x+7, and you multiply to> get> (x+1)(x+7) = x^2 + 8x + 7> so the constant term is 7, but of course, the polynomial doesn't> necessarily have 7 as a factor.> Also, dividing back through by x+7 gives you x+1, so the constant term> after the division is 1, whereas before it was 7.> Ok, yeah, it's kid stuff, but remember, I'm facing people who seem to> be confused on these issues, which is why I'm posting.> But now consider having 7 itself as a factor.> Notice that it's different from x+7 as one is a polynomial while 7 is> a constant.> Can you ever divide 7 off as a variable?> Some of you seem intent on arguing as long as you can that you can.> However, division of constants by constants is a constant operation,> and is NOT variable dependent.> No matter how much you wish it is, it's not, and refusing to accept> that it's not merely makes you a crank who refuses to believe in> mathematical logic.> So you may wonder about relevancy.> Well I have an expression where I divide off 49, a constant factor of> a polynomial P(x), which changes you from having 7, 7 and 22 as> constant terms of factors of that polynomial to having 1, 1, and 22 as> the constant terms of the factors of P(x)/49, so it should be> straightforward.Unfortunately, you have still not proven that the individual factorsdistribute themselves this way for values of 'x' other than x = 0.Certainly P(0) suppresses every contribution from the polynomial exceptthe constant term. But P(1) consists of the *sum* of all coefficients ofthe polynomial and without performing the calculations, you can't justassert that the partition you claim actually occurs. In fact, P(1)/49 =281412 so your discussion of 1, 1, 22 isn't even relevant.