mm-305 === Subject: Linear mapping problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QDbTM29795; === am hoping that someone can show me how to solve the following: 1. Find the cartesian form y = mx + c equation for the following straight lines in the z plane, z = x + jy; (a) |z - 2 + j| = |z - j + 3| (b) |z + z* + 4j(z - z*)| = 6 where z* denotes the complex congugate. === Subject: Re: Linear mapping problem > am hoping that someone can show me how to solve the following: > 1. Find the cartesian form y = mx + c equation for the following > straight lines in the z plane, z = x + jy; > (a) |z - 2 + j| = |z - j + 3| > (b) |z + z* + 4j(z - z*)| = 6 where z* denotes the complex congugate. Are you sure? I get a straight line for (a) but not one expressible in the y = m*x + c form, and get a pair of parallel lines for (b) === Subject: extraneous roots by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QGcUI11444; === does anyone at all know, how geometry and extraneous roots are rela?? or could you possibly point me in the correct direction as to where i may find out some information on them why i'm asking is that i have to do a project on the two and cant seem to get myself star thanks dennis === Subject: needs help on calculus by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QLTgl02825; === v K=Mo Vd{V/[1-(V/C[Divid e])]} (a)show that K=MoV/[1-(V[Divide ]/C)] + Mo[1-(V/C) ]- MoC (b) show that (a) can be written as K= MC- MoC, where M=Mo/[1-(V/C[Divi de])] (c)the total energy is defined as E= K + MoC Subject: Re: needs help on calculus === > v > K=Mo Vd{V/[1-(V/C[Divid e])]} (a)show that K=MoV/[1-(V[Divide ]/C)] + Mo[1-(V/C) ]- MoC > (b) show that (a) can be written as K= MC- MoC, where > M=Mo/[1-(V/C[Divid e])] > (c)the total energy is defined as E= K + MoC None of these symbols are coming through correctly. I have no idea what the problem is. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: needs help on calculus > v > K=Mo .98 Vd{V/[.941-(V/C )]} (a)show that K=MoV/[.941-(V[ CapitalODoubleDot]/C)] + Mo[.941-(V/C[Ca pitalODoubleDot])]- MoC (b) show that (a) can be written as K= MC- MoC, where > M=Mo/[.941-(V/C )] (c)the total energy is defined as E= K + MoC None of these symbols are coming through correctly. I have no idea what > the problem is. Newsgroups do not necessarily parse characters with ASCII codes higher that 127 as expec, since the characters represen by such number values vary from font to font. The character '', for example, appears to me as a radical or square root symbol, but will appear otherwise to some other viewers. Subject: Re: needs help on calculus === > v >K=Mo .98 Vd{V/[.941-(V/C )]} (a)show that K=MoV/[.941-(V[ CapitalODoubleDot]/C)] + Mo[.941-(V/C[Ca pitalODoubleDot])]- MoC >(b) show that (a) can be written as K= MC- MoC, where >M=Mo/[.941-(V/C[Copyrig ht])] >(c)the total energy is defined as E= K + MoC >>None of these symbols are coming through correctly. I have no idea what >>the problem is. > Newsgroups do not necessarily parse characters with ASCII codes higher > that 127 as expec, since the characters represen by such number > values vary from font to font. > The character '', for example, appears to me as a radical or square > root symbol, but will appear otherwise to some other viewers. Unfortunately, many people new to newsgroups want to use unicode. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Reply to How much does Fred weigh by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0QLgaW04035; === >Fred is a meat cutter who sells chicken breast for $3.88 per pound, >ground beef for $2.75 per pound, and pork chops for $3.50 per pound. >Fred wears size 36 jeans, is aobut five foot five, wears baggy >shirts and has size 12 shoes. What does Fred weigh? chicken breasts, ground beef, and pork chops....am i right? === Subject: Statistics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0REH6x21300; === Hello: Please send me the formula for this problem. Question: A machine used to regulate the amount of dye dispensed for mixing shades of paints can be set so that it discharges an average of u milliters (ml) of dye per can of paint. The amount of dye discharged is known to have a normal distribution with a standard deviation of .4 ml. If more than 6ml of dye are dischaged when making a certain shade, the shade is unacceptable. Determine the setting for u so that only 1% of the cans of paint will be unacceptable. I appreciate it. Doreen. === Subject: Re: Statistics > Hello: > Please send me the formula for this problem. > Question: > A machine used to regulate the amount of dye dispensed for mixing > shades of paints can be set so that it discharges an average of u > milliters (ml) of dye per can of paint. The amount of dye discharged > is known to have a normal distribution with a standard deviation of .4 > ml. If more than 6ml of dye are dischaged when making a certain shade, > the shade is unacceptable. Determine the setting for u so that only 1% > of the cans of paint will be unacceptable. > I appreciate it. > Doreen. Hint: For what z-score does a standardized normal distribution have an upper tail of .01? === Subject: divisibility? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0RKi8v30809; === does anybody know the rule for the divisibility of 315? or could u giv me a site to reasearch that ? === Subject: divisibility? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SFpb303421; === If you're asking about a test for divisibility BY 315, it can be broken down into three tests. Since 315 = 3x3x5x7, we need the divisibility tests for 5, 7, and 9. (And the number must pass all three tests, of course.) Divisible by 5: the number must end in 0 or 5. Divisible by 7: Double the units digit and subtract from the rest of the number. If the result is divisible by 7, the number is divisible by 7. Example: 623. Subtract twice 3 from 62: 62 - 6 = 56. Since 56 is divisible by 7, then 623 is divisible by 7. Divisible by 9: Cross out all digits whose sum is 9 or a multiple of 9. If all digits are elimina, the number is divisible by 9. Example: 16578. Cross out the 1 and 8. And the 6, 5, and 7 (since 6 + 5 + 7 = 18). Since all the digits are elimina, 16578 is divisible by 9. === Subject: Re divisibility by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SEsXg30997; === >does anybody know the rule for the divisibility of 315? or could u >giv me a site to reasearch that ? What do you mean by the divisibility of 315? It's clearly divisible by 5 since it ends in a 5: 315= 5*63. 63 is obviously divisible by 3: 63= 3*21= 3*3*7. 315= 3*3*5*7. That's about all you can say about the divisibility of 315- no rule necessary. Surely you are not asking about a rule to determine IF a number is divisible BY 315? === Subject: problem with Euclid proof - Infinite Primes by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SDMiq23293; === I am having this problem with Euclid's famous proof for Infinite Primes. the proof goes like that: 1. say that there are finitely number of primes. let us list *all* of them: p1,...,pn 2. now look at N = (p1 * p2 * ... * pn) + 1 what can wa say about N? it is either a Prime or not. If N is a prime - then it is not in our list above - and thats a contradiction to the claim that we have lis all primes. If N is not a prime, then it is divided by a prime P. P is not in our list, since N/pi {i=1,...,n} will always give a reminder => hence this again contradicts (1) that we can list all the primes in a finite list. QED. Now I have come to a proof that claims you can shorten proof the above like this: The prime numbers are the numbers p1,p2,...,pn,... of set S. Suppose finite, then p1,p2,...,pn is the complete series set, so that pn is the largest prime. Form W+1 = (p1xp2x,...,xpn) + 1. W+1 is not in the set S so it is not prime. W+1 is not divisible by any of primes in S_P so it is prime. Contradiction. Reverse supposition and primes are infinite. => my question: Is this 2nd proof correct? Does that mean that Euclid coukd have shorten his proof? Thanks for any insight & help, -Ben === Subject: Re: problem with Euclid proof - Infinite Primes > I am having this problem with Euclid's famous proof for Infinite > Primes. > the proof goes like that: > 1. say that there are finitely number of primes. > let us list *all* of them: p1,...,pn > 2. now look at N = (p1 * p2 * ... * pn) + 1 > what can wa say about N? > it is either a Prime or not. > If N is a prime - then it is not in our list above - > and thats a contradiction to the claim that we have lis all primes. > If N is not a prime, then it is divided by a prime P. > P is not in our list, since N/pi {i=1,...,n} will always give a > reminder => hence this again contradicts (1) that we can list all the > primes in a finite list. > QED. > Now I have come to a proof that claims you can shorten proof the above > like this: > The prime numbers are the numbers p1,p2,...,pn,... of set S. Suppose > finite, then p1,p2,...,pn is the complete series set, so that pn is > the > largest prime. Form W+1 = (p1xp2x,...,xpn) + 1. W+1 is not in the set > S > so it is not prime. W+1 is not divisible by any of primes in S_P so it > is prime. Contradiction. Reverse supposition and primes are infinite. > => my question: Is this 2nd proof correct? > Does that mean that Euclid coukd have shorten his proof? > Thanks for any insight & help, > -Ben It seems to me that what you claim to be Euclids proof is no it. I beleive that his proof is not one of contraction. but one of construction, more like this: For any natural number, n, let p_1,...,p_n be the first n primes and consider N = 1 + (product of p_1 through p_n). N is larger than any of the lis primes, and is not divisible by any of them, so must be divisible by (or even equal to) a prime larger than any in the list. Thus no such list can contain all the primes. === Subject: integral problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SHurV15308; === Hi everybode!!! Thanks for you attention. I have difficulties with such iintegral problem: Prove that f is periodic if f is defined on (-infty,+infty) and x+1 integral f(t)dt=0 for all x. x I'll be grateful for someone ho could help me.Great Thanks!!! Bye!!! === Subject: Re: integral problem >Hi everybode!!! >Thanks for you attention. I have difficulties with such iintegral >problem: >Prove that f is periodic if f is defined on (-infty,+infty) and > x+1 >integral f(t)dt=0 for all x. > x Think of a definite integral as the area under a curve. If we say it is 0, that must mean either the function of 0 everywhere or there is precisely as much area above the axis (>0) as below (<0). Try drawing a picture of the area that represents that integral, from say x = 3 to 4. Now draw a picture from say x = 3.2 to 4.2. That should begin to show you a pattern, which should suggest your proof. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com An expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: Re: integral problem Content-transfer-encoding: 8bit > Hi everybode!!! > Thanks for you attention. I have difficulties with such iintegral > problem: > Prove that f is periodic if f is defined on (-infty,+infty) and > x+1 > integral f(t)dt=0 for all x. > x > I'll be grateful for someone ho could help me.Great Thanks!!! > Bye!!! Let's assume f is also continuous on R. 0 = int(f(t), t = x..x+1) = int(f(t), t = 0..x + 1) + int(f(t), t = x..0) = int(f(t), t = 0..x + 1) - int(f(t), t = 0.. x). Now, use the Fundamental Theorem and take the derivative of both sides: 0 = f(x + 1) - f(x). -- Paul Sperry Columbia, SC (USA) === Subject: Re: integral problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0TDrQL15888; === >Hi everybode!!! >Thanks for you attention. I have difficulties with such iintegral >problem: >Prove that f is periodic if f is defined on (-infty,+infty) and > x+1 >integral f(t)dt=0 for all x. > x >I'll be grateful for someone ho could help me.Great Thanks!!! >Bye!!! Hi Jane, what about f(x) = 0, if x in R Z f(x) = x, if x in Z ? Maybe f is supposed to be continuous on R ? Best wishes Torsten. === Subject: Re: integral problem >Hi everybode!!! >Thanks for you attention. I have difficulties with such iintegral >problem: Prove that f is periodic if f is defined on (-infty,+infty) and > x+1 >integral f(t)dt=0 for all x. > x >I'll be grateful for someone ho could help me.Great Thanks!!! >Bye!!! > Hi Jane, > what about > f(x) = 0, if x in R Z > f(x) = x, if x in Z > ? Well, actually that looks pretty periodic to me. Period 1. > Maybe f is supposed to be continuous on R ? Well, now that I've opened my mouth I suppose I ought to say some more. A real counterexample is if f(x) = 1 for x = n + 1/n and 0 otherwise. So it looks like continuity is required. Maybe comparing int (x, 1+x, f(x)dx) to int (x + epsilon, 1 + x + epsilon, f(x)dx) (trying to get the sense here without lapsing into an actual program language) will give you some hints. Jon Miller === Subject: Re: integral problem >Hi everybode!!! >Thanks for you attention. I have difficulties with such iintegral >problem: Prove that f is periodic if f is defined on (-infty,+infty) and > x+1 >integral f(t)dt=0 for all x. > x >I'll be grateful for someone ho could help me.Great Thanks!!! >Bye!!! Hi Jane, what about f(x) = 0, if x in R Z > f(x) = x, if x in Z ? > Well, actually that looks pretty periodic to me. Period 1. Are you saying then that f(1) = f(2) when f(1) = 1 and f(2) = 2 ? Maybe f is supposed to be continuous on R ? > Well, now that I've opened my mouth I suppose I ought to say some more. > A real counterexample is if f(x) = 1 for x = n + 1/n and 0 otherwise. So it > looks like continuity is required. > Maybe comparing int (x, 1+x, f(x)dx) to int (x + epsilon, 1 + x + epsilon, > f(x)dx) (trying to get the sense here without lapsing into an actual program > language) will give you some hints. > Jon Miller === Subject: impossible or not? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SJl5U25121; === I found a simple problem... but i do not understand.. there should be an answer.. or should there be? Find a 3 digit number, which is 5 times more than the product of it's digits. Is this an impossible problem or not? If not could u give me a solution? I will be thankful for your answers. === Subject: impossible or not? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0T0bkE19204; === I found ONE solution . . . Let N = 100a + 10b + c, where {a,b,c} are digits (0 to 9). Then: 100a + 10b + c = 5abc Solving for c: c = 5abc - 100a - 10b = 5(abc - 20a - 2b) Hence, c is a multiple of 5. Since c is a digit, the only choices are 0 and 5. If c = 0, the equation becomes: 100a + 10b = 0 ... and N = 0. So, if a solution exists, c = 5. Our equation is: 100a + 10b + 5 = 5ab(5) or: 100a + 10b + 5 = 25ab Divide by 5: 20a + 2b + 1 = 5ab Solving for a: a = (2b + 1)/5(b - 4) Since a is a digit, 2b + 1 must be divisible by 5 (and by b-4). Hence, b = 2 or 7 But b = 2 produces a negative value for a. So, the only possibility is b = 7. Then a = (14 + 1)5(7 - 4) = 15/15 = 1 We have: a = 1, b = 7, and c = 5. Therefore, N = 175 . . . and 175 = 5(1)(7)(5) === Subject: Re: impossible or not? Please could you begin subject lines with Re: when posting a follow-up? That would make it easier to distinguish between new and continued threads, also between questions and answers. I tried e-mailing you this request, but it bounced. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com An expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: sum by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0T0bku19200; === grade 5 home work question ? what is the sum of the first 20 odd numbers === Subject: Re: sum >grade 5 home work question ? what is the sum of the first 20 odd >numbers Is any guidance given as to solution method? If not, try writing the first 20 odd numbers and then look at them to see any pattern. If you don't see a useful pattern (and most people other than Gauss probably don't), add them up. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com An expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: Re: sum > grade 5 home work question ? what is the sum of the first 20 odd > numbers 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 39 + 37 + 35 + 33 + 31 + 29 + 27 + 25 + 23 + 21 Add vertically first, then horizontally. === Subject: very interesting differentiation problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i0SIREt18090; === Hi everyone!!! I'm too interesting in this problem,it's seems to me that it is interesting,some difficult problem(especially for me).Here is it: Problem.Let f(x) be infinitely many differentable on R and f(x)=o(x^n) when x tends to infty.(x->infty).Prove that for any k>n (k) function f (x) have at least on zero. (k) I notice that f (x) means k-th derivative f at x. Thanks for you attention.I'll be grateful for someone who could help me. Bye!!! === Subject: Re: very interesting differentiation problem > Hi everyone!!! > I'm too interesting in this problem,it's seems to me that it is > interesting,some difficult problem(especially for me).Here is it: > Problem.Let f(x) be infinitely many differentable on R and > f(x)=o(x^n) when x tends to infty.(x->infty).Prove that for any k>n > (k) > function f (x) have at least on zero. > (k) > I notice that f (x) means k-th derivative f at x. > Thanks for you attention.I'll be grateful for someone who could help > me. > Bye!!! The derivative of an o(x^n) function is an o(x^(n-1)) function. === Subject: complex powers What interpretation/visualization can be given to the act of raising a number to a complex power? Alex -- === Subject: Re: complex powers In the following p represents Pye No doubt you already know that for n more than one, the mapping z to z^n causes the size to be raised to the nth power and the angle to be multiplied by n. The mapping z to z^(1/n) produces n solutions equally spaced in angle around z and diminished in size to the n-th root of the size of z. What about z to a complex power ? With real roots, e^[k ln(x)] = x^k, but, with complex variables ln(z) = ln(r) +i(theta) + i(2np) (ie multivalued) so, e^[k ln(z)] = [e^k ln(r)] [e^ ik(theta)] [e^ i2nkp] but any power of e^ i2p = 1 so e^[k ln(z)] = (r^k)[e^ ik(theta)] =z^k If we let k = a+ib then z^(a+ib) = (r^(a+ib))[e^ i(a+ib)(theta)] = (r^a) (e^(-b theta) (r^ib) ((e^(ia theta)) using (r^b)^i = (e^b ln r)^i, we obtain z^(a+ib) = (r^a) (e^(-b theta)) ((e^(ia theta + b ln r)) Comparing this with z = r e^ i(theta)] indicates the change of size and rotation of the result, compared with the original complex number, in terms of the size and sign of a and b. However, as expressions like these become the more complica, their usefulness in supplying visual analogies diminishes. Tristan Needham's book with its many illustrations and his idea of Amplitwist takes visual explanations of the complex as far as they can be used. I hope that some of you found this of some use . Ian Hutcheson === Subject: Re: complex powers >What interpretation/visualization can be given to the act of raising a >number to a complex power? >Alex >-- Thanks for having the decency to give an e-mail address. There is no visualization it is mathematically defined --- nor is there an interpretation. Exponentiation is precisely and mathematically defined. In the complex number system the logarithm is NOT a function but an infinitely-many-valued relation. And exponentiation is defined by: z^w = e^(w*ln(z)) === Subject: Re: complex powers >What interpretation/visualization can be given to the act of raising a >number to a complex power? > There is no visualization ??? That's a rather strange opinion, at least IMO. Perhaps you think that Needham's _Visual Complex Analysis_ didn't really deserve any awards, etc. David > it is mathematically defined --- nor > is there an interpretation. Exponentiation is precisely and > mathematically defined. > In the complex number system the logarithm is NOT a function but an > infinitely-many-valued relation. > And exponentiation is defined by: z^w = e^(w*ln(z)) === Subject: Re: complex powers > What interpretation/visualization can be given to the act of raising a > number to a complex power? Assuming real a, and z = x + iy = Re[z] + iIm[z], a^(x + iy) = exp(x*log(a) + iy*log(a)) This is a stretch of a^(Re[z]) and a rotation through an angle Im[z]*log(a). For complex a, we have a^z = |a|^z * exp(i*arg(a))^z, so exp(i*arg(a)*z) = exp(-arg(a)*Im[z] + i*arg(a)*Re[z]) - a further stretch of exp(-arg(a)*Im[z]) and a further rotation through arg(a)*Re[z]. -- === Subject: Re: JSH: Simplicity versus dirty math tricks > = >> The simple mathematical fact is that the factors of 2 on the right are >> 7/f_1 and 2/f_2 on the left. >> However, 2/f_2 is not an algebraic integer if f_2 is a non-unit factor >> of 7, so if f_2 is a non-unit factor of 7 then you're NOT in the ring >> of algebraic integers. > Can you prove this? It is not immediately obvious to me. I think James might be using: since 1= 7-2.3, if x divides 7 and 2, it must divide 1. Subject: Re: JSH: Simplicity versus dirty math tricks === >>= >The simple mathematical fact is that the factors of 2 on the right are >7/f_1 and 2/f_2 on the left. >However, 2/f_2 is not an algebraic integer if f_2 is a non-unit factor >of 7, so if f_2 is a non-unit factor of 7 then you're NOT in the ring >of algebraic integers. >>Can you prove this? It is not immediately obvious to me. > I think James might be using: > since 1= 7-2.3, > if x divides 7 and 2, it must divide 1. I doubt he is, but that gives me the seed I need for a proof. Thanks. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: JSH: Simplicity versus dirty math tricks it's probably easier for me to ignore basic algebra, than most of the other of your habitual ring of correspondents (def.: as in circle jerk .-)... be that as it may be, it still seems to revolve around what ever definition of prime or of relatively (or co-) prime, it is that you use. are you going to stick with some definition of those, perhaps one that you could paste-in from a reference that you admire? > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > where b_2(x) = a_2(x) + 1, and > a^2 - (x - 1)a + 7(x^2 + x). > So multiplying through by 7 on the right side you have > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25)x^2 + 7(30)x + 7(2) > and 7 and 2 on the left are factors of 7(2) on the right. The Daily [Yale] Voice from the Mausoleum, thus quoth: Perhaps my judgment is impaired by the fact that I am Jewish, Zionist, and generally sympathetic to Great Britain, but to me it is clear that the guy is a nut. Most people within the current establishment seem to think so as well and dismiss him as a quirky and entertaining oddity. The reality is unfortunately much more sinister. According to the Center for Responsible Politics in his current bid for the White House alone LaRouche raised over $5.5 million in small donations. According to the Federal Election Commission as late as last April his fundraising figures surpassed those of Lieberman, Dean and Graham. According to the last report from three months ago he still has more money than Kucinich, Mosley-Braun and Sharpton combined, all three of them politicians with nationwide standing. The implications of his fundraising success are astounding. He is ignored by the media, locked out of debates, and has as much chance of winning as a proverbial snowball in hell and yet he raised this truly massive amount of campaign cash. The only possible explanation is that he has a powerful grassroots organization. Just think about it: there is an ultraradical organization out there but the only reason we ever hear about it is because its leader compulsively runs for president and because a member at some point probably shoved a flier in your hand. The LaRouche experience shows how a radical organization can use the electoral system to fund its activities and spread its message. thus quoth: as you know, if you've read anyhting that I typed, there was a conspiracy of states that DID vote Gore -- and the Supremes sealed that conspiracy on March 27, 2000, by refusing to hear the appeal in LaRouche v. Fowler (Don Fowler was the DNC Chair in '96; Sentelle's 3-judge panel made the Voting Rights Act unconsitutitonal. but, hey; it's up for re-auhtorization in '07 !-) I am frankly scared of the touchscreen mentality. just like the Supremes' abrogation of the USA and Florida constitutions foments a pop-culture hatred of the electoral college on the part of some rabid Democrats. ah, so; imagine if North Dakota ... rather, imagine if Wyoming had less than one electoral college vote for president. anyway, every one of us in this debate knows about the Texas cirterium for chad -- it ain't just missing confetti! --Give the Gift of Dick Cheeny -- out of office, at last! http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: JSH: Simplicity versus dirty math tricks > I've no recently that you can't claim to be in a ring while using > elements outside of that ring. The mathematics is rather basic but > I'm seeing disagreement as if there's something to debate. > Here's an example to help take away places to hide: > 1 + 2 = 3, so you can say (1+2)/3 = 1, and some might try to claim > that my current position is like trying to say that 1/2 + 2/3 = 1 is > not in the ring of integers, as you have to sum *first* and then by > convention, since the division isn't a ring operation anyway, you can > say 3/3 = 1. > Now though, consider > 4(1+2) = 4 + 8 > as here notice that you have *two* sums on both sides wher 8 on the > right side is a factor of 4 and 2 on the left. > I've poin out that you *have* to use elements from within the ring, > for such an operation, while posters arguing with me, keep trying to > push the argument as if it's like the 1+2 = 3 example before. > Here's how they do it. > The modified Decker example is > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > where b_2(x) = a_2(x) + 1, and > a^2 - (x - 1)a + 7(x^2 + x). > So multiplying through by 7 on the right side you have > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25)x^2 + 7(30)x + 7(2) > and 7 and 2 on the left are factors of 7(2) on the right. > Notice that besides 7(2) on the right you also have 7(25)x^2 + 7(30)x, > so you have two sets of sums: the summation on the right, and the > summations on the left. > There's no simplifying possible, like with 1+2=3 to be done. > It's just a mathematical fact that the factors of 7(2) on the right > are 7 and 2 on the left. > Now then, if you factor out 7, and try to do so with factors f_1 and > f_2, such that > (5a_1(x) + 7)/f_1 (5b_2(x) + 2)/f_2 = 25x^2 + 30x + 2 At this point let us be clear about what is being claimed The claim is that (5a_1(x) + 7)/f_1 and (5b_2(x) + 2)/f_2 can be algebraic integers. The fact that 7/f1 and 2/f2 are factors of 2, and, in general, are not algebraic integers is irrelevant. An example may help to clarify. Consider functions c(x) and d(x) (x is restric to the ordinary integers) c(x) = 2x if x=0,1 mod 3; 7 if x=2 mod 3 d(x) = x if x =0,1 mod 3, (2(x^2 + 2x +1)/9 - 1) if x=2 mod 3 (note c(x) and d(x) are integers for all integers x) Then (c(x) + 2)(d(x) + 1) = 2(x^2 + 2x +1) now let, w_1(x) = 2 if x=0,1 mod 3; 1 if x = 2 mod 3 w_2(x) = 1 if x=0,1 mod 3; 2 if x = 2 mod 3 Then w_1(x) * w_2(x) = 2 for all x and (c(x) + 2)/w_1(x) is always an integer (d(x) + 1)/w_2(x) is always an integer so [(c(x) + 2)/w_1(x)] [(d(x) + 1)/w_2(x)] = (x^2 + 2x +1) is a factorization of the RHS into integers. Rewriting we have (c(x)/w_(x) + 2/w_1(x)) (d(x)/w_2(x) + 1/w_2(x)) = (x^2 + 2x +1) and 1/w_2(x) is not an integer for x=2 mod 3. - William Hughes === Subject: Re: JSH: Simplicity versus dirty math tricks > I've no recently that you can't claim to be in a ring while using > elements outside of that ring. The mathematics is rather basic but > I'm seeing disagreement as if there's something to debate. > Here's an example to help take away places to hide: > 1 + 2 = 3, so you can say (1+2)/3 = 1, and some might try to claim > that my current position is like trying to say that 1/2 + 2/3 = 1 is > not in the ring of integers, as you have to sum *first* and then by > convention, since the division isn't a ring operation anyway, you can > say 3/3 = 1. James, why don't you spend a few weeks reading a book on basic abstract algebra? It's not that you're wrong, it's that you're not even wrong. Your most basic understanding of what a ring is, is flawed. Your locutions of we are in a ring, or claiming to be in a ring, are just not the way the rest of us are thinking about rings. You claim to be frustra that you can't communicate with us. Why don't you take a few weeks off from Usenet, struggle with the basics like we all did (and do), and learn what we're all talking about. Then you can come back here and communicate better. === Subject: Re: JSH: Simplicity versus dirty math tricks Falsity alert: > Now then, if you factor out 7, and try to do so with factors f_1 and > f_2, such that f_1 and f_2 depend on x. So they should more properly be named f_1(x) and f_2(x). > (5a_1(x) + 7)/f_1 (5b_2(x) + 2)/f_2 = 25x^2 + 30x + 2 A correct rewrite: [(5a_1(x) + 7)/f_1(x)] [(5b_2(x) + 2)/f_2(x)] = 25x^2 + 30x + 2. > then necessarily you *still* have the factors of 2 on the right as > 7/f_1 and 2/f_2 on the left. When x = 0, you have factors of 2 on the right, but when x = 0, f_1(x) = 7 and f_2(x) = 1. When x = 1, you do *not* have factors of 2 on the right, neither do you have them on the left. > The simple mathematical fact is that the factors of 2 on the right are > 7/f_1 and 2/f_2 on the left. What are the factors of 2 on the right when x = 1? -- === Subject: Re: JSH: Simplicity versus dirty math tricks >I've no recently that you can't claim to be in a ring while using >elements outside of that ring. The mathematics is rather basic but >I'm seeing disagreement as if there's something to debate. >Here's an example to help take away places to hide: >1 + 2 = 3, so you can say (1+2)/3 = 1, and some might try to claim >that my current position is like trying to say that 1/2 + 2/3 = 1 is >not in the ring of integers, as you have to sum *first* and then by >convention, since the division isn't a ring operation anyway, you can >say 3/3 = 1. Yes, that *is* an accurate representation of your current position. You say that if f_2(x) is a nonunit algebraic integer divisor of 7, then (5 b_2(x) + 2)/f_2(x) cannot be in [or, as you prefer to say, pushes you out of] the ring of algebraic integers because 2/f_2(x) cannot be an algebraic integer. Even though you know perfectly well, I believe, that (5 b_2(x) + 2)/f_2(x) is itself an algebraic integer. >Now though, consider >4(1+2) = 4 + 8 >as here notice that you have *two* sums on both sides wher 8 on the >right side is a factor of 4 and 2 on the left. ??? Since when is 8 a factor of 4 or 2 ??? >I've poin out that you *have* to use elements from within the ring, >for such an operation, Wrong. The set of integers divisible by 2 is a ring. The elements are {0, +/-2, +/-4, +/-6, ... }. Note that 3 + 5 = 8 is in that ring, but neither 3 nor 5 is in it. Another example: Z[1/2]. This is a ring, not a field. The number 1/6 + 1/3 is in that ring, but neither 1/6 nor 1/3 are in it. > while posters arguing with me, keep trying to > push the argument as if it's like the 1+2 = 3 example before. You must love the word push. You use it for everything. >Here's how they do it. >The modified Decker example is >(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) >where b_2(x) = a_2(x) + 1, and >a^2 - (x - 1)a + 7(x^2 + x). >So multiplying through by 7 on the right side you have >(5a_1(x) + 7)(5b_2(x) + 2) = 7(25)x^2 + 7(30)x + 7(2) >and 7 and 2 on the left are factors of 7(2) on the right. >Notice that besides 7(2) on the right you also have 7(25)x^2 + 7(30)x, >so you have two sets of sums: the summation on the right, and the >summations on the left. >There's no simplifying possible, like with 1+2=3 to be done. Wrong again. For specific values of x, other simplifying is possible. That is, there are algebraic integers f_1(x) and f_2(x) such that f_1(x)*f_2(x) = 7, and (5 b_2(x) + 2)/f_2(x) is an algebraic integer, *even though 2/f_2(x) is NOT*. Of course also, (5 a_1(x) + 7)/f_1(x) is also an algebraic integer. In general, f_1(x) and f_2(x) are not units and are not equal to 7. Would you like to know how to define f_1(x) and f_2(x) ? [I will bet $1 that this is something that you very much do NOT want to know. Of course if I did give this, it would give you a chance to shoot holes in it. Are you sure you don't want to see the definitions?] >It's just a mathematical fact that the factors of 7(2) on the right >are 7 and 2 on the left. >Now then, if you factor out 7, and try to do so with factors f_1 and >f_2, such that >(5a_1(x) + 7)/f_1 (5b_2(x) + 2)/f_2 = 25x^2 + 30x + 2 >then necessarily you *still* have the factors of 2 on the right as >7/f_1 and 2/f_2 on the left. You are being so incredibly dense about this. Your analogy above with (1 + 2)/3 = 1 is really perfect here - it's your own analogy, and you STILL don't get it. Here is how the parallelism goes: 1. Even though 1/3 and 2/3 are not integers, 1/3 + 2/3 *is* an integer. 2. Even though 5 b_2(x)/f_2 and 2/f_2 are not algebraic integers, 5 b_2(x)/f_2 + 2/f_2 *is* an algebraic integer. See the parallelism ? There is really no conceptual difference. >So claiming that the *sum* is all that matters is just a dirty trick, >meant to fool readers into thinking that it's something like 1+2 = 3, >where you have to add first before you can divide out the 3. It is NOT a dirty trick or any other kind of trick. See the parallel above. It's just plain arithmetic! Perhaps you are hung up on the symbolic representation. Here's another parallel. Suppose you are considering r + s, and you know that s is irrational. Is it necessarily true that r + s is irrational also ? Is it OK to write a rational as a sum of irrationals? Is that 'legal' ? I think you know the answer. The fact that one piece of a whole is not of a certain class does NOT mean that the whole itself is not of that class. You are artificially focussing on a piece, 2/f_2, and saying that because this piece is not an algebraic integer, the whole expression cannot be one either. But the other piece of the expression essentially 'compensates' for the non-algebraic-integered-ness of 2/f_2. The two pieces then fit together to produce an algebraic integer. >The simple mathematical fact is that the factors of 2 on the right are >7/f_1 and 2/f_2 on the left. >However, 2/f_2 is not an algebraic integer if f_2 is a non-unit factor >of 7, so if f_2 is a non-unit factor of 7 then you're NOT in the ring >of algebraic integers. Yes. We all totally agree that 2/f_2 is not an algebraic integer. However, 5 b_2(x)/f_2 + 2/f_2 IS. >So then, if posters claim you add first, as if that matters, You're the one who is claiming it matters whether you add first and divide second, or vice versa. We on the other say *it doesn't matter*. What matters is the end result: (5 b_2(x)/f_2 + 2/f_2) *is* an algebraic integer. It doesn't matter how you got there. Just as with (1 + 2)/3: you get exactly the same thing if you add 1/3 + 2/3: the result is an integer whether you add first, then divide, or divide first, then add. Either way, it's an integer! Another point here is that as you know, 5 b_1(x) + 2 = 5 a_2(x) + 7, and, when f_1(x) and f_2(x) are defined correctly [do you want to see the definition yet?] then a_2(x)/f_2(x) and 7/f_2(x) are both algebraic integers. > consider >that they're trying to fool you with a rather basic trick that >requires that you ignore basic algebra. No one (on this side) is trying to fool ANYONE. You, however, are essentially saying that because 2/3 is not an integer, neither is (1 + 2)/3 = 1/3 + 2/3. But guess what? 1 is an integer. No, you are not literally saying this about (1 + 2)/3. But this is EXACTLY what you are saying about (5 b_2(x) + 2)/f_2. Incidentally, for the 10th or 11th time, you have launched a new thread rela to Rick Decker's quadratic example, and again you have failed to state Rick's main result: namely, that (5 a_1(1) + 7)*(5 b_2(1) + 2)/7 = (5*sqrt(-2) + sqrt(7))*(-5*sqrt(-2) + sqrt(7)), which provides a factorization in which all the coefficients and the individual factors are algebraic integers. You keep saying this is impossible. I think you are refusing to quote this result because you don't want anyone else to see it. And why not? Because it rather badly damages your claim that no such thing is possible! Your post this time is a little different. It is clear that you are very close to understanding what we are talking about: so close, that it must be true that you are *rejecting* the main idea, willfully rejecting it, rather than misunderstanding it. Of course if you went all the way and admit that you understand what we mean, you lose everything. You know that and that is what is holding you back from admitting that our claims are correct. Another hint on this. You know that if you insist on dividing 7 from both sides of your expression in the form of 7*1, with 7 dividing the first factor, and 1 dividing the second, then you get that (5 a_1(x)/7 + 1), should be an algebraic integer, and you know that for x > 0, a_1(x)/7 is NOT an algebraic integer. This fact is trying to tell you something, and you are refusing to listen. What it is trying to tell you is that 7*1 is the wrong way to decompose 7. The right way is as 7 = f_1(x)*f_2(x) [and again I will cheerfully give you the definition of those which works]. You however are misinterpreting the message. You think what it is telling you is that the definition of algebraic integers is incorrect. That of course is nonsense; a definition is just a definition; it cannot be wrong. If you really think that that is the only way to divide 7 out of both sides, and that way does not work, then what you have is not a problem with a definition; what you have is a contradiction indicating that mathematics is inconsistent. That is unfortunately the inevitable conclusion. Is that really, really what you think? That you have proved that math is inconsistent? > === Subject: Re: JSH: Simplicity versus dirty math tricks > You are being so incredibly dense about this. Your > analogy above with (1 + 2)/3 = 1 is really perfect here - > it's your own analogy, and you STILL don't get it. Here is > how the parallelism goes: > 1. Even though 1/3 and 2/3 are not integers, > 1/3 + 2/3 *is* an integer. > 2. Even though 5 b_2(x)/f_2 and 2/f_2 are not > algebraic integers, > 5 b_2(x)/f_2 + 2/f_2 *is* an algebraic integer. > See the parallelism ? There is really no conceptual difference. I think I must apologize for starting him on this track. I predict you are about to get a slew of JSH replies telling you you are lying if you claim any of this is mathematics. And, no, he won't see the parallel, as he can't seem to believe you're not allowed to divide away in non-fields. (Or division rings.) Perhaps he would like us to explicitly state there is a larger field of fractions in which we can do these calculations. >>However, 2/f_2 is not an algebraic integer if f_2 is a non-unit factor >>of 7, so if f_2 is a non-unit factor of 7 then you're NOT in the ring >>of algebraic integers. > Yes. We all totally agree that 2/f_2 is not an algebraic > integer. However, 5 b_2(x)/f_2 + 2/f_2 IS. >>So then, if posters claim you add first, as if that matters, > You're the one who is claiming it matters whether you add > first and divide second, or vice versa. We on the other say > *it doesn't matter*. What matters is the end result: > (5 b_2(x)/f_2 + 2/f_2) *is* an algebraic integer. It doesn't matter > how you got there. Just as with (1 + 2)/3: you get exactly > the same thing if you add 1/3 + 2/3: the result is an integer > whether you add first, then divide, or divide first, then add. > Either way, it's an integer! Another point here is that as you > know, 5 b_1(x) + 2 = 5 a_2(x) + 7, and, when f_1(x) and f_2(x) > are defined correctly [do you want to see the definition yet?] > then a_2(x)/f_2(x) and 7/f_2(x) are both algebraic integers. >> === Subject: Totally stumped, linear algebra 5x5 matrix determinant X-No-Archive: ?yes Hi Folks: Taking Linear Algebra this semester. I have the 2x2, 3x3 and 4x4 matrix determinants understood. However, I'm getting read to conceit defeat with the 5x5. I've spent hours so far on this one problem with no such luck. It's in my textbook. The answer they give is 138. It asks for the determinant of a 5x5 matrix: Original matrix is: 2 -1 0 4 1 3 1 -1 2 0 3 2 -2 5 1 0 0 4 -1 6 3 2 1 -1 1 I'm missing something very obvious. What it is I don't know as I'm plowing thru the properties section for determinants and nothing is jumping out at me. I was able to transform the matrix via row operations into it's diagonal form but keep coming up with large answers. One such row path I took which led me to a diagonal matrix resul in: 2 -1 0 4 1 0 -1 1 -3 -1 0 0 4 -1 6 0 0 0 7 6 0 0 0 0 46 And multiplying the diag (2)(-1)(4)(7)(46) = 2576 rather than the books answer of 138. I've tried other combinations and still end up with the same large answer rather than 138. Can someone enlighten me as to why I am not seeing the light? As usual, the textbooks skip all the steps and give you the raw answer. This is not even a homework problem as I'm going ahead and trying to learn about determinants as we cover those next week. thanks for any input, Confused === Subject: Re: Totally stumped, linear algebra 5x5 matrix determinant > Hi Folks: > Taking Linear Algebra this semester. I have the 2x2, 3x3 and 4x4 matrix > determinants understood. However, I'm getting read to conceit defeat > with the 5x5. > I've spent hours so far on this one problem with no such luck. It's in > my textbook. The answer they give is 138. > It asks for the determinant of a 5x5 matrix: > Original matrix is: > 2 -1 0 4 1 > 3 1 -1 2 0 > 3 2 -2 5 1 > 0 0 4 -1 6 > 3 2 1 -1 1 > I'm missing something very obvious. What it is I don't know as I'm > plowing thru the properties section for determinants and nothing is > jumping out at me. > I was able to transform the matrix via row operations into it's diagonal > form but keep coming up with large answers. One such row path I took > which led me to a diagonal matrix resul in: > 2 -1 0 4 1 > 0 -1 1 -3 -1 > 0 0 4 -1 6 > 0 0 0 7 6 > 0 0 0 0 46 > And multiplying the diag (2)(-1)(4)(7)(46) = 2576 rather than the > books answer of 138. I've tried other combinations and still end up > with the same large answer rather than 138. > Can someone enlighten me as to why I am not seeing the light? As usual, > the textbooks skip all the steps and give you the raw answer. This is > not even a homework problem as I'm going ahead and trying to learn about > determinants as we cover those next week. > thanks for any input, > Confused Don't know what you are doing wrong, but I agree with the book answer. === Subject: Re: Totally stumped, linear algebra 5x5 matrix determinant >Hi Folks: >Taking Linear Algebra this semester. I have the 2x2, 3x3 and 4x4 matrix >determinants understood. However, I'm getting read to conceit defeat >with the 5x5. >I've spent hours so far on this one problem with no such luck. It's in >my textbook. The answer they give is 138. >It asks for the determinant of a 5x5 matrix: >Original matrix is: >2 -1 0 4 1 >3 1 -1 2 0 >3 2 -2 5 1 >0 0 4 -1 6 >3 2 1 -1 1 >I'm missing something very obvious. What it is I don't know as I'm >plowing thru the properties section for determinants and nothing is >jumping out at me. >I was able to transform the matrix via row operations into it's diagonal >form but keep coming up with large answers. One such row path I took >which led me to a diagonal matrix resul in: >2 -1 0 4 1 >0 -1 1 -3 -1 >0 0 4 -1 6 >0 0 0 7 6 >0 0 0 0 46 I get 2 -1 0 4 1 0 2.5 -1 -4 -1.5 0 0 -0.6 4.6 1.6 0 0 0 29.6667 16.6667 0 0 0 0 -1.55056 Multiplication of diagonals yields 138. >And multiplying the diag (2)(-1)(4)(7)(46) = 2576 rather than the >books answer of 138. I've tried other combinations and still end up >with the same large answer rather than 138. >Can someone enlighten me as to why I am not seeing the light? As usual, >the textbooks skip all the steps and give you the raw answer. This is >not even a homework problem as I'm going ahead and trying to learn about >determinants as we cover those next week. To calc determinants of a matrix bigger that 3x3, don't use the method with cofactors, it's too complica (unless the matrix has a lot of zeros on a row). Always use gaussian elimination, which you call row operations. Since we don't know what you did in the row operations we can't tell you why you don't have the right answer. Maybe you could take a look at these online lectures: http://ocw.mit.edu/OcwWeb/Mathematics/18- 06Linear-AlgebraFall2002/VideoLectu res/index.htm >thanks for any input, >Confused === Subject: Re: Totally stumped, linear algebra 5x5 matrix determinant >To calc determinants of a matrix bigger that 3x3, don't use the method >with cofactors, it's too complica (unless the matrix has a lot of >zeros on a row). Always use gaussian elimination, which you call row >operations. I too, have found this last suggestion to be extremely useful advice, having struggled recently to deal with continual errors in dealing with 5 x 5 determinants and matrices. G C Subject: Re: regex === suprised no one responded. Here is what I finally used: /mailto:(.[^]*).*?gif/i Mike > I want to match mailto:null@void.com......gif in a string that contains many > similar patterns. > I used mailto:(.[^]*) to match each mailto: ... string and parsed out > the email address. > But now I can't seem to extend the match to the first 'gif'. > I have tried a number of different combinations but without the match I > need. > I know what I need. I need to do as I did in the example above except I > need to not include a group of characters instead of just one. > Mike === === Subject: Re: Explaining factorizations, relevance > Crank Information > http://www.crank.net/harris.html > http://www.crank.net/usenet.html > http://www.google.com/search?q=harris+site%3Awww.crank.net > http://www.google.com/search?q=%22james+harris%22+site% 3Ausers.pandora.be I wonder why the crank site doesn't seem to include Gregory Chaitin? He's been saying ridiculous things for decades: Mathematics is random. He's developed the most powerful form of Godel's 1st Incompleteness Theorem (he never mentions Rosser, Smullyan, et. al.) The last time get the proof of the unsolvability of the Halting Problem right. IBM has a long history of promoting flashy fakes, from E. F. Codd to Chaitin. Money talks. Nobody walks. === Subject: Re: Explaining factorizations, relevance In sci.logic, <3c65f87.0401281344.42111d88@posting.google.com>: > Mathematicians today are inheritors of a rather deserved glory from > the exploits and important work of *past* mathematicians who have done > much to advance knowledge. However, it is problematic if > mathematicians today can rely on social support garnered from the > efforts of their predecessors to hide acts of wrongdoing today. > Here I'll outline the basic mathematics underlying a find of mine of a > strange error in an esoteric branch of mathematics called algebraic > number theory. > To understand posts on this subject, it will help if you understand > the basics of factorizations. > 1. Factorizations are just ways of looking at factors of some > product. > For instance, 2(3) is a factorization of 6, which tells you that 6 is > a product of 2 and 3. > That factorization is in an integer domain, which simply means that > you have only integers, so a factorization like 5(6/5) = 6, is not > valid in integers, though it is valid in some other domain like > rational numbers that has 1/5 as a member. > 2. Algebra allows a generalization of factorizations beyond specific > number examples, for instance > (x+2)(x+3) = x^2 + 5x + 6 > which is true without regard to what domain x is in, so you can let x > be an integer, or it could be a fraction. > 3. If x is an integer, then you're guaranteed to be in the integer > domain, which logically follows from the fact that x is added to and > multiplied times other integers, as addition and multiplication are > what are called ring operations by mathematicians. > 4. Algebraic integers are roots of monic polynomials with integer > coefficients e.g. x^2 + 5x + 6, as monic just means that leading > coefficient is a factor of 1 in the integer domain. > Not surprisingly, if x is an algebraic integer, then x^2 + 5x + 6 is > an algebraic integer, and the factorization > (x+2)(x+3) = x^2 + 5x + 6 is in the algebraic integer domain. > What I've done is consider factorizations of polynomials, where the > factors are not polynomials. > 5. In an attempt at questioning an important conclusion a Rick > Decker, a professor at Hamilton College, made a post with his own > example of such a factorization, and I've used it partly because it's > a quadratic and easier to handle. > That factorization is > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > Notice that it is a factorization because you have factors > (5a_1(x) + 7) and (5a_2(x) + 7) > which multiply to give the product 7(25x^2 + 30x + 2). > Algebra allows me to make the substitution a_2(x) = b_2(x) - 1 to get > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > where the purpose is to balance out constant factors, which here are > numbers, on both sides of the equation. > Importantly, with that substitution, at x=0, a_1(0) = b_2(0) = 0, as I > emphasize that the 7 and 2 visible on the left multiply to give 7(2) > on the right. > That is, when you multiply through by 7 on the right side you have > 7(25)x^2 + 7(30)x + 7(2) > where the 7(2) is the product of the factors visible on the left in > (5a_1(x) + 7)(5b_2(x) + 2). > I have to elaborate to that level of detail because for mathematicians > to fool you about my discovery, you have to be confused on that simple > point. > Integers are rather easy to work with so now I can use them to > determine what is forced upon the variables. > Notice that multiplying out gives > 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = > 175 x^2 + 210 x + 14. > That's important because I now attempt to eliminate the factor of 7 > from both sides, and to do so while remaining in the domain of > algebraic integers. > 6. The logical conclusion is that the only way that might possibly > allow that is > (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2 > as consider another possibility, like > (5a_1(x)/sqrt(7) + sqrt(7))(5b_2(x)/sqrt(7) + 2/sqrt(7)) = > 25x^2 + 30x + 2 > which gives the factorization > sqrt(7)(2/sqrt(7)) = 2 > as multiplying out gives > 25 a_1(x) b_2(x)/7 + 10 a_1(x)/7 + 5b_2(x) + 2 = > 25 x^2 + 30 x + 2 > and the two factors are > (5a_1(x)/sqrt(7) + sqrt(7)) > and > (5b_2(x)/sqrt(7) + 2/sqrt(7)) > which internally have sqrt(7) and 2/sqrt(7) as the factors of 2. > But that second factor cannot be in the domain of algebraic integers > as 2/sqrt(7) is not an algebraic integer. > Generalizing, for any factorization of 7, with algebraic integers f_1 > and f_2, where f_1 f_2 = 7, and f_1 is assumed to be a factor of > (5a_1(x) + 7) you will have > (5b_2(x)/f_2 + 2/f_2) > as a factor, and if 2/f_2 is not an algebraic integer, then that > factor is not in algebraic integers. > 7. Logically that follows from the requirement that for a > factorization to be in a particular domain, its factors must be in > that domain as well. > Like before, with 2(3) = 6, in the domain of integers that is valid, > while 5(6/5) = 6 is not valid in the domain of integers. > So if you're trying to remain in the domain of algebraic integers then > (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2 > is the only possibility. Counterexample. x = 0. Explicit computation of your roots for a^2 - (x - 1)*a + 7*(x^2 + x) = a^2 + a + 0 gives one of the following two possibilities for root assignment. a_1(0) = 0 a_2(0) = -1 b_2(0) = 0 or a_1'(0) = -1 a_2'(0) = 0 b_2'(0) = 1 The first leads to the identity (5 * a_1(0) + 7) * (5 * b_2(0) + 2) = 7 * (25 * 0^2 + 30 * 0 + 2) (5 * 0 + 7) * (5 * 0 + 2) = 14 7 * 2 = 14 In this case, your factorization works. The second leads to the identity (5 * a_1'(0) + 7) * (5 * b_2'(0) + 2) = 7 * (25 * 0^2 + 30 * 0 + 2) (5 * (-1) + 7) * (5 * (1) + 2) = 14 2 * 7 = 14 In this case, your factorization fails. x = 1. Explicit computation of your roots for a^2 - (x - 1)*a + 7*(x^2 + x) = a^2 + 14 gives one of the following two possibilities for root assignment. a_1(1) = sqrt(-14) a_2(1) = -sqrt(-14) b_2(1) = 1-sqrt(-14) leading to the identity (5 * a_1(1) + 7) * (5 * b_2(1) + 2) = 7 * (25 * 1^2 + 30 * 1 + 2) (5 * sqrt(-14) + 7) * (5 * (1 - sqrt(-14)) + 2) = 399 In this case, your factorization fails; sqrt(-14) is not divisible by 7. Or: a_1'(1) = -sqrt(-14) a_2'(1) = sqrt(-14) b_2'(1) = 1+sqrt(-14) (5 * a_1'(1) + 7) * (5 * b_2'(1) + 2) = 7 * (25 * 1^2 + 30 * 1 + 2) (5 * (-sqrt(-14)) + 7) * (5 * (1 + sqrt(-14)) + 2) = 399 In this case, your factorization fails; -sqrt(-14) is not divisible by 7. x = 2. a^2 - (x - 1)*a + 7*(x^2 + x) = a^2 - a + 42. a_1(2) = 1/2 + 1/2 * sqrt(-167) b_2(2) = 3/2 - 1/2 * sqrt(-167) (5 * a_1(2) + 7) * (5 * b_2(2) + 2) = 7 * (25 * 2^2 + 30 * 2 + 2) (5 * (1/2 + 1/2 * sqrt(-167)) + 7) * (5 * (3/2 - 1/2 * sqrt(-167)) + 2) = 1134 Your factorization fails; 1/2 + 1/2 * sqrt(-167) is not divisible by 7. a_1'(2) = 1/2 - 1/2 * sqrt(-167) b_2'(2) = 3/2 + 1/2 * sqrt(-167) (5 * a_1(2) + 7) * (5 * b_2(2) + 2) = 7 * (25 * 2^2 + 30 * 2 + 2) (5 * (1/2 - 1/2 * sqrt(-167)) + 7) * (5 * (3/2 + 1/2 * sqrt(-167)) + 2) = 1134 Your factorization fails; 1/2 - 1/2 * sqrt(-167) is not divisible by 7. General case. x = x. a^2 - (x - 1)*a + 7*(x^2 + x) has roots a_1(x) = (x - 1) + sqrt( (x-1)^2 - 4*1*7*(x^2+x)) = (x - 1) + sqrt( x^2 - 2*x + 1 - 28*x^2 - 28*x) = (x - 1) + sqrt(-27*x^2 - 30*x + 1) b_2(x) = x - sqrt(-27*x^2 - 30*x + 1) or a_1'(x) = (x - 1) - sqrt(-27*x^2 - 30*x + 1) b_2'(x) = x + sqrt(-27*x^2 - 30*x + 1) Your factorization succeeds when either a_1(x) or a_1'(x) is a multiple of 7. Since they are conjugates if the discriminant -27*x^2 - 30*x + 1 is negative, which occurs for rational x inside of the range 15 - sqrt(252) = -0.8745... < x < 15 + sqrt(252) = 30.8745..., or if the quantity -27*x^2 - 30*x + 1 is not the square of a rational (which leads us to another realm) or 4 times the square of an algebraic integer (I'm not sure how one can solve that), your factorization fails for those cases. One can also explicitly check the equation for c_1(x) = a_1(x) / 7; c_1(x) solves the equation (c*7)^2 - (x - 1)*(c*7) + 7*(x^2 + x) = 0 = 49*c^2 - 7*(x - 1)*c + 7*(x^2 + x) and yes we can divide by 7 here: 7*c^2 - (x - 1)*c + (x^2 + x) = 0 so the factorization succeeds if this equation is reducible (x = 0 being one case thereof; the roots are 0 and 1/7) or if both x-1 and x^2+x are divisible by 7, but not for most x. So far, you're 1 for 7+.... For rational x one cannot have both x-1 and x^2+x divisible by 7: if x-1 is divisible by 7, then x = 7*y+1 for some integer y, and x^2 + x = 49*y^2 + 14*y + 1 + 7*y + 1 = 49*y^2 + 21*y + 2 which is not. If x^2 + x = x * (x + 1) is divisible by 7, then x - 1 % 7 = 5 or 6. I'm not sure one can have x-1 and x^2+x divisible by 7 for any algebraic integer x, either. > Mathematicians, however, for many years believed that if the roots of > a quadratic like > a^2 - (x - 1)a + 7(x^2 + x) > are not integers, then there's no way to figure out where factors of 7 > might go between those roots, but they believed that they did split up > in some way such that each root would have some factor of 7 in the > ring of algebraic integers. > My work proves that wrong with simple algebraic principles by using > non-polynomial factors, which is the step mathematicians did not take. Your work proves diddly-squat; you got lucky on x=0, but that's all. You are generalizing from a special case which happens to work, to a grander case which can be easily shown to not work for the vast majority of x (although not for all x, but finding such x appears to be an interesting problem in itself). > Now algebraic integers have been defined to be the roots of monic > polynomials with integer coefficients, which is actually something of > an arbitrary definition. > Mathematicians found it useful for over a hundred years without fully > understanding its limitations, like how you can be forced out of that > domain as I have shown. > It could be a small mistake, but by trying to deny and hide it, > today's mathematicians are making a HUGE deal out of it, and bringing > their own credibility into question. > They are damaging their discipline, by not telling the truth fully and > honestly. > I've shown what is mathematically correct. It's been that way from > the beginning as the correct mathematics does not change. > Today's mathematicians gain nothing by hiding from the truth. After > all, in mathematics, when you're wrong, you're just wrong. > -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Explaining factorizations, relevance In sci.logic, Uncle Al <401850F7.71DAA2F4@hate.spam.net>: > [snip] >> Here I'll outline the basic mathematics underlying a find of mine of a >> strange error in an esoteric branch of mathematics called algebraic >> number theory. > [snip] ing imbecile troll . Hey stooopid loud troll James > Harris, put up or shut up. Is a $10,000 prize no questions asked too > small to justify your submission of two little prime numbers? $20K. RSA-576 has been factored. (Of course, he could work on RSA-2048, with a $200K payoff.) > http://www.rsasecurity.com/rsalabs/challenges/factoring/ faq.html > http://www.rsasecurity.com/rsalabs/challenges/factoring/ numbers.html > http://www.crank.net/harris.html > It's not every braying jackass that gets a whole page at crank.net > James Haris in mathematics is a eunuch in a brothel, a capon in a > henhouse, a steer amidst cows; a stot, a gelding, a gelt, a havier, a > gib, a lapin, a seg, a hog, a wether... a loud stupid psychotic boor > in a science newsgroup. > -- > Uncle Al > http://www.mazepath.com/uncleal/qz.pdf > http://www.mazepath.com/uncleal/eotvos.htm > (Do something naughty to physics) -- #191, ewill3@earthlink.net It's still legal to go .sigless.