mm-1040 === Subject: Re: terms in Math by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, >Can anyone tell me the URL with pages devoted to math terms. One such is http://mathworld.wolfram.com/ === Subject: Re: coverings of a Mobius band by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iBHMLmT09372; >Are the even degree covers of a Mobius band annulus(es)? Are the odd degree covers of a Mobius band, Mobius bands again? What is the universal cover of a Mobius band? Yes, yes, and R x [-1, 1]. One can imagine a strip of paper with n half-twists before gluing the ends as the total space of the degree n cover of the Mobius band. The strip of paper is I x [-1, 1], and the gluing map for taping the ends is the homeomorphism (-1)^n * - : [-1, 1] --> [-1, 1], so we get an annulus for n even and a Mobius strip for n odd. For the universal cover, imagine an infinite strip with infinitely many half-twists. Todd Trimble === Subject: Re: The Possible Cure For AIDS in part: > There are no As, Gs, 4s, Os, >>The G is gimmel the 3 rd letter of the Hebrew alphabet. >>The vowels (mplicit in old Hebrew but made explicit by the Masorites) >>include an ah sound and a oh sound. >AFAIK the Torah codes are never done with the vowels. Thats quite true. Including vowels makes it harder to find matches. But if someone wanted to find English words in the Tanakh, the standard convention is, of course, Aleph for A and Ayin for O. As for the number 404, two letters would represent that in standard Hebrew numeric notation: Tau and Daleth. But Hebrew has a word for silver, although *ancient* Hebrew hardly had a word for oxygen, even if it certainly has one today. Actually, we should be asking what letters of the Hebrew alphabet are supposed to stand for and ! John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: The Possible Cure For AIDS > But Hebrew has a word for silver, although *ancient* Hebrew hardly had a Kessef. > word for oxygen, even if it certainly has one today. Khamtzan which means sour-er. This corresponds to the German Saurstuff. > Actually, we should be asking what letters of the Hebrew alphabet are > supposed to stand for and ! There is a typefont used to crossreference passages in the Talmud to the Shulkhan Arooch. It looks like superscripts because of the way it is printed. This was originated about 1500 of the common era. Bob Kolker === Subject: Re: The Possible Cure For AIDS in part: >There are no As, Gs, 4s, Os, etc. in Hebrew. Did you transliterate >these according to some arbitrary scheme, or did you search for Hebrew >letter strings? Not that it matters, of course. But while were talking about a cure for AIDS... Some years back, there were news stories about how, for AIDS research, modified mice were created with human immune systems. This was done by taking cells from human fetuses. Naturally, this was controversial because of the abortion issue. I noted that nobody cares about mouse fetuses. If you can fix a mouse so that it _can_ get AIDS by giving it a human beings immune system, it would seem you could fix a human so that he cant get AIDS by giving him a mouses immune system. There were probably very good reasons why this couldnt _really_ be done, but I had thought it worth mentioning. John Savard http://home.ecn.ab.ca/~jsavard/index.html === Subject: Re: Beals conjecture > => A large prize is offered by banker Andrew Beal for a solution to > => the Beal Conjecture: the equation x^p + y^q = z^r has no solutions > => for p, q, r > 2 and coprime integers x, y, z. > = => Sorry if this has been discussed here - my only justification > => is that I just recently discovered this group but I am curious to > => know if Wiles proof of FLT also covers the Beal Conjecture > => Or have any counter-examples been found? Wiles proof covered the case p=q=r and relaxed the coprime requirement for x, y, & z. IE: There is no integer solution when p=q=r whatever the coprimality of x, y, & z. tom -- We have discovered a therapy ( NOT a cure ) for the common cold. Play tuba for an hour. === Subject: Mathforge.net :: Near-numbers, autonomic computing, John Derbyshire, and Church-Turing posting-account=Jy65eAwAAADVuRECxXfbsbHI5uTpc8FY The Latest Math News from Mathforge.net http://mathforge.net ****An invitation to additive prime number theory**** A.V. Kumchev and D.I. Tolev have compiled a short document entitled An Invitation to Additive Prime Number Theory[~60pp, pdf]. The document serves as an introductory guide to graduate-level students on... http://mathforge.net/index.jsp?page=seeReplies&messageNum=981 ****Are there encoded messages in the Bible?**** researchers both supporting and denying the statistical evidence for Ôhidden messages found in the Old Testament tell their tales. Some... http://mathforge.net/index.jsp?page=seeReplies&messageNum=974 ****Near-numbers: the new Ôlimit**** Theres a very interesting paper by Frank J Swenton of Middlebury College called Limits and the System of Near-Numbers[19pp, pdf]. What seemed at first glance (at the title and abstract) like an uninspired... http://mathforge.net/index.jsp?page=seeReplies&messageNum=970 ****Introduction to autonomic computing**** While not itself mathematical in nature, the concept is built around ideas garnered from areas of artificial intelligence research and it is easy to see that autonomic computing could have applications... http://mathforge.net/index.jsp?page=seeReplies&messageNum=969 ****Derbyshires Diary**** John Derbyshire, author of Prime Obsession, has a mathematical problem accompanying each of his Diary entries located in his Web Journalism folder. If you sift through enough of the partisan propaganda... http://mathforge.net/index.jsp?page=seeReplies&messageNum=968 ****Maple 9.5 Released**** MapleSoft has released version 9.5 of their popular symbolic and numeric computational software suite Maple. New Features include added packages (optimization, logic, and root finding), OpenMaple access... http://mathforge.net/index.jsp?page=seeReplies&messageNum=967 ****Mathematica 5.1 released**** The Mathematica version has jumped a tenth of a point, and Wolfram has added scores of new features to the new release, including Web Services support, a benchmarking package, string manipulation functions,... http://mathforge.net/index.jsp?page=seeReplies&messageNum=966 ****Quantum computers and the Church-Turing Thesis**** The original Church-Turing Thesis states that every function which would naturally be regarded as computable can be computed by a Turing Machine and Petrus H. Potgeiter mentions in his paper Zeno Machines... http://mathforge.net/index.jsp?page=seeReplies&messageNum=963 ****Sobering U.S. Student Math Scores**** In a disheartening follow-up to the Putnam story below, news (Seattle Times) outlets (New York Times)everywhere are reporting the horrid state of U.S. student math skills. The Organisation for Economic... http://mathforge.net/index.jsp?page=seeReplies&messageNum=959 ÔMathforge ran a story about the Putnam Competition... http://mathforge.net/index.jsp?page=seeReplies&messageNum=958 === Subject: What kind of matrix can map positive element vectors to positive element vectors? Hi all, Suppose I have a matrix A, and a positive element vector x, then y=A*x, I want y to be also positive element vector... What can I say about A? I want all such kinds of A? What kinds of A can let me have both directions: x positive elements <=> y=A*x positive elements? === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > Suppose I have a matrix A, and a positive element vector x, > then y=A*x, > I want y to be also positive element vector... > What can I say about A? I want all such kinds of A? > What kinds of A can let me have both directions: > x positive elements <=> y=A*x positive elements? Take a simple case and try to learn from it. The unit vectors for x will pull out the columns of A. Sure seems to say that A must at least have all of its elements positive. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? > Hi all, > > Suppose I have a matrix A, and a positive element vector x, > > then y=A*x, > > I want y to be also positive element vector... > > What can I say about A? I want all such kinds of A? > > What kinds of A can let me have both directions: > > x positive elements <=> y=A*x positive elements? > > > > Take a simple case and try to learn from it. > The unit vectors for x will pull out the columns of A. > Sure seems to say that A must at least have all of its > elements positive. Necessary and sufficient. === Subject: Re: What kind of matrix can map positive element vectors to positive element vectors? posting-account=ZeRDXwsAAACLpj2mpKc97NFPxBaFxAzp >> Hi all, >> Suppose I have a matrix A, and a positive element vector x, >> then y=A*x, >> I want y to be also positive element vector... >> What can I say about A? I want all such kinds of A? >> What kinds of A can let me have both directions: >> x positive elements <=> y=A*x positive elements? >> Take a simple case and try to learn from it. >> The unit vectors for x will pull out the columns of A. >> Sure seems to say that A must at least have all of its >> elements positive. >Necessary and sufficient. Neither necessary (unless your positive means >= 0) nor sufficient. Note that lucy wanted <=>. Ill assume A is a square matrix. Its easy to see A must be nonsingular, else given vector x > 0 (i.e. all x_i > 0) with Ax > 0, you could add to x a suitable multiple of a vector w with Aw = 0 to get a vector v not > 0 with Av = Ax > 0. A must map the nonnegative cone C = {x in R^n: all x_i >= 0} onto itself. Note that the extreme rays of C pass through the standard unit vectors e(j) (with e(j)_i = 1 if i=j, 0 otherwise). That is, these are the only members w of C such that if w = t x + (1-t) y with x,y in C and 0 < t < 1, then x and y are scalar multiples of w. Now A must map extreme rays to extreme rays, and from this its easy to see that A must be of the form A = D P where P is a permutation matrix and D a diagonal matrix with positive elements on the diagonal. Conversely, everything of this form has the desired property. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: How to determine parameter integer values such that quadratic has integer solutions? posting-account=uJhfTw0AAACZ85X1hg4ZQuYw9kXQVPPG Hi all, In particular, Im interested in finding the positive integer values of g such that the following quadratic in v has integer roots: v^2 - 7v + (12 - 12g) Evidently this requires that 1 + 48g be a perfect square. I can have Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, 20, 35, ... This sequence is more conveniently summarized as something-mod-something-else, but I forget exactly what the two somethings are. How to figure this out? cdj === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? > In particular, Im interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) v = (7 +- sqr(49 - 48 + 48g)/2 = (7 +- sqr(1 + 48g)/2 sqr(1 + 48g) must be odd number, 2n + 1 1 + 48g = 4n^2 + 4n + 1 12g = n^2 + n = n(n + 1) Case 12 | n. Done Case 4 | n, not 3 | n. 3 | n+1; n = 3k - 1 Case 3 | n, not 2 | n. 4 | n+1; n = 4k - 1 Case 2 | n, not 4 | n. Not possible Three family parametrization of solutions. n = 12k n = 3k - 1 provided 4 | n n = 4k - 1 provided 3 | n Expect some duplicates. > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? posting-account=uJhfTw0AAACZ85X1hg4ZQuYw9kXQVPPG The square root of 1+48g is odd... for Ôs sake Im blind.... === Subject: Re: How to determine parameter integer values such that quadratic has integer solutions? cdj: Heres an outline: We know that 1+48g is a perfect square. Since g is an integer, 48g is even, and 1+48g is odd. So, the square root of 1+48g is odd, and we can write that square root as 2n+1 for some integer n. Thus, (2n+1)^2 = 1+48g Expanding and simplifying, 4n^2+4n+1 = 1+48g 4n^2+4n = 48g n^2+n = 12g We need a little number theory here: Consider the above equation (mod 12): (n^2+n) (mod 12) = 0 Trying the 11 possibilities, we find that n (mod 12) must be 0, 3, 8, or 11, that is, we must be able to write n as 12k or 12k+3 or 12k+8 or 12k+11 for some integer k. (The above condition is equivalent to (n^2+n) = 0 (mod 3) AND (n^2+n) = 0 (mod 4), so with some work, we really need only to consider 3+4 = 7 cases.) Substituting for the first possibility (n = 12k, that is, n is a multiple of 12), n^2+n = (12k)^2+(12k) = 144k^2+12k = 12(12k^2+k) But, n^2+n = 12g, so this case gives: 12(k^2+k) = 12g, that is, g = 12k^2+k The second possibility (n = 12k+3, that is, n is 3 more than a multiple of 12) similarly gives n^2+n = (12k+3)^2+(12k+3) = 144k^2+84k+12 = 12(12k^2+7k+1) 12(12k^2+7k+1) = 12g, so, g = 12k^2+7k+1 The other possibilities (n = 12k+8, n=12k+11) give after similar calculations, g = 12k^2+17k+6, and g = 12k^2+23k+11 If k > 0, then all four possibilities clearly give nonnegative values for g. (For k = 0, we generate the first four values of your sequence: 0, 1, 6, 11.) Suppose we want to generate a list of such values: We can exhaust all the possibilities for g by evaluating the above four formulas for g for each k = 0,1,2,3,... (When writing your list, because you need positive integer values for g, youll need to throw away the first element of your list, 12(0)^2+(0)=0.) Considering the values of g generated by a particular k, we have (clearly): 12k^2+k < 12k^2+7k+1 < 12k^2+17k+6 < 12k^2+23k+11. So, the smallest value of g generated by k+1 is: 12(k+1)^2+(k+1) = 12k^2+25k+13 > 12k^2+23k+11 That is, the smallest value for g generated by k+1 is larger than the largest value generated by k. So, we can write down a complete ordered list (a sequence that gives the possible solutions for g in increasing order) by writing down the four possibilities for g generated by 0 (in increasing order), then the four generated by 1, then those for 2, and so on. I dont know how to write the sequence more compactly than by writing down the above rule. Travis > Hi all, > In particular, Im interested in finding the positive integer values of > g such that the following quadratic in v has integer roots: > v^2 - 7v + (12 - 12g) > Evidently this requires that 1 + 48g be a perfect square. I can have > Mathematica spit out such values, which begin with: 0, 1, 6, 11, 13, > 20, 35, ... > This sequence is more conveniently summarized as > something-mod-something-else, but I forget exactly what the two > somethings are. How to figure this out? > cdj === Subject: Integral posting-account=DH5daAwAAADTxT0WC17CRoCKZYkI-swj Hi. What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral >Hi. >What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? http://integrals.wolfram.com/ says: RootSum[-1+#1+#1^2+#1^5 &, (Log[x-#1] / (1+2#1+5#1^4)) &] Thomas === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? As soon as you come up with a closed form of factorization of x^5 + x^2 + x - 1 (one linear factor and two quadratic factors), I will be able to tell you more. === Subject: Re: Integral Mike4ty, Ugly, I think. Factor the polynomial (warning, its irreducible over the integers), then apply the method of partial fractions. Travis > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? === Subject: Re: Integral > Hi. > What is the integral of 1/(x^5 + x^2 + x - 1) dx in closed form? The hard part is solving the quintic. There is a real root near 0.568544, a complex conjugate pair near 0.91612 +/- 0.57771 i, and another pair near 0.622848 +/- 1.03222 i. Once you have the factors, its an easy partial fractions decomposition, provided you dont mind approximate answers. Mathematica 5.0 for Mac OS X -- Terminal graphics initialized -- In[1]:= Integrate[1./(x^5+x^2+x-1),x] Out[1]= 1. (-0.22894 ArcTan[0.484391 (-1.2457 + 2. x)] - > 0.189874 ArcTan[0.865487 (1.83224 + 2. x)] + > 0.361679 Log[0.586544 - 1. x] - 2 > 0.0641575 Log[1.45342 - 1.2457 x + x ] - 2 > 0.116682 Log[1.17302 + 1.83224 x + x ]) -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Galois group = A_4 The following question has been bugging me : I am trying to come up with a quartic polynomial whose Galois group is A_4. I know I can go about the business of finding a discriminant that is the square of a rational number, and the resolvent cubic is irreducible. But I was wondering if there was a more illuminating way to geometrically come up with a quartic polynomial. In particular, we know A_4 has no transpositions and no 4-cycles. So, what can I say about the quartic? Wouldnt it be true that since there are no transpositions, the quartic cannot have exactly 2 real roots? What can I deduce about the fact that the Galois group has no 4-cycles? Tony === Subject: Re: How to find the extremum of the Absolute value of a function=?big5?Q?=EF=BC=9F?= My method is to differentiate |Z| w.r.t y, and then subtitute y1 and y2 into the above equation. Finally I got a simultaneous equations and solving for x1 and x2. But where confused me is that the derivative of an absolute value seems not exist, so my known method didnt work due to the absolute value. is there any other method to solve a extremum problem of an obsolute value? Á¡ (Randy Poe)ÁnÛ¤Èæ2 50´ÁG > Suppose I have a complex-valued function Z, and Z=3DZ(x1,x2,y) > where x1,x2 and y are three real variables. I wanna |Z| has local > minimums at two given points y=3Dy1 and y=3Dy2, where x1 and x2 should be > adjusted to met this demand. > i.e. > Q=EF=BC=9AHow to find x1 and x2 such that |Z| has local minimum at two > given points > y1 and y2 ? > To find the minimimum for y=3Dy1, define a new function: > W1(x1,x2) =3D Z(x1, x2, y1) > and use your favorite minimization technique. Similarly for y2. > - Randy -- Á¡ Origin: ´.beÛjË.b3.bc®[UDo ubleDot].90[Degre e]Tøü  === Subject: swjpam format=ßowed; fyi, The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer exists due to budgetary contraints. === Subject: Re: swjpam Discussion, linux) > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. I will never doubt the hammer again. Golly. -- Conservative, n: A statesman who is enamored of existing evils, as distinguished from the Liberal who wishes to replace them with others. -- Ambrose Bierce === Subject: Re: swjpam posting-account=UtgH7gwAAACpBhTelVPOXNP7RAfbtQrK > fyi, > The Southwest Journal of Pure and Applied Mathematics (swjpam) no longer > exists due to budgetary contraints. Is that just the excuse? Is the real reason editorial incompetence? Is this the first blow of The Hammer? === Subject: do you have any smart way of finding which number is bigger ? Using the fatest way: compare: 0.9^10 vs. 2*(0.9^19)+0.9^20 how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? lucy escribi.97: > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? 2*(0.9^19)+0.9^20 = 0.9^10(2*0.9^9 + 0.9^10) = = 0.9^10(20*9^9 + 9^10)/10^10 = 0.9^10* 9^9*29/10^10 But 9^2 = 81 > 80, then 9^8 > 8^4*10^4 = 4096*10^4 ==> 9^9 > 36*10^7 ==> 29*9^9 > 36*29*10^7 = 1044*10^7 > 10^10 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: do you have any smart way of finding which number is bigger ? posting-account=Sny74g0AAADy66iGh6ZMdSlIFta_KAXh Err, the other thing you need to notice along this line is the 2* and the addition adding up to 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? posting-account=Sny74g0AAADy66iGh6ZMdSlIFta_KAXh Err, the other thing you need to notice along this line is the 2* and the addition adding up to over 5x Of course 0.9^20 < 0.9^10 === Subject: Re: do you have any smart way of finding which number is bigger ? posting-account=Sny74g0AAADy66iGh6ZMdSlIFta_KAXh A: 0.9^10 B: 2*(0.9^19)+0.9^20 Notice both pieces of B are positive. b1 > 0 b2>0 So if I can see one side of + larger than A then B > A 0.9^20 > 0.9^10 20 > 10 ergo B > A 5 seconds === Subject: Re: do you have any smart way of finding which number is bigger ? lucy, Heres a way that doesnt use much explicit calculation. (It admittedly uses the fact that e < 2.9, which can easily be derived analytically, anyway). The Taylor series for log about 1 is: log(1+x) = x - (1/2)x^2 + (1/3)x^3 - ... Setting x = 1/9, we have: log(10/9) = log(1 + 1/9) = (1/9) - (1/2)(1/9)^2 + (1/3)(1/9)^3 - ... < 1/9 9 * log (10/9) < 1 9 log 10 - 9 log 9 < 1 Adding log 10, 10 log 10 - 9 log 9 < 1 + log 10 = log (e * 10) < log (29) [Here is where I invoke e < 2.9.] Subtracting log 9, 10 log 10 - 10 log 9 < log 29 - log 9 10 log (10/9) < log (29/9) = log (2 * (10/9) + 1) Exponentiating, (10/9)^10 < 2*(10/9) + 1 Multiplying both sides by (9/10)^20 (9/10)^10 < 2*(9/10)^19 + (9/10)^20 I suspect there is a much more elegant way. Maybe something with the expression 10 log 10 - 9 log 9 < log (29)? Travis > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? === Subject: Re: do you have any smart way of finding which number is bigger ? > Using the fatest way: > compare: > 0.9^10 vs. 2*(0.9^19)+0.9^20 > how long does it take you to figure out which number is larger? I dont know if its smart or fastest, but you can factor. 0.9^10 >?< 2 * (0.9^19) + 0.9^20 0.9^10 >?< 2 * (0.9^20)/0.9 + 0.9^20 0.9^10 >?< 0.9^20 * (2/0.9 + 1) 0.9^10 / 0.9^20 >?< 2/0.9 + 1 0.9^-10 >?< 2.2222... + 1 2.8679... < 3.2222... -- john === Subject: Re: do you have any smart way of finding which number is bigger ? ETAtAhUAvgUxhHwS5c+oabk25UmVuHI06JUCFAza+ gLCc85dGRk3KQeD5aQy6jyx 0.9^10 ? 2*(0.9^19)+0.9^20 1 ? 2^(0.9^9) + 0.9^10 10^10 ? 20*9^9 + 9^10 = 29*9^9 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. Therefore ? is <. --OL === Subject: Re: do you have any smart way of finding which number is bigger ? > 0.9^10 ? 2*(0.9^19)+0.9^20 > 1 ? 2^(0.9^9) + 0.9^10 > 10^10 ? 20*9^9 + 9^10 = 29*9^9 I followed you up to this point; you are now comparing 10^10 with 29 * 9^9 I dont understand the next line. > 9^9 = 729^3 > 720^2*700 = 518400*700 > 3.5e8, and 3.5*29 >100. > Therefore ? is <. -- john === Subject: Re: do you have any smart way of finding which number is bigger ? posting-account=qrPIWAwAAABKr36mTyR-AQd_YQJSbfcG .9^10 = 0.3486784401 2*(0.9^19) + 0.9^20 = 0.39174699812516770581 The second one is larger time to do the problem -- (2 seconds for cut and paste maybe?) === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, > 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark But he assumes .999... = 1 in his equation before it is proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >> Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). >> He assumes 9(1) = 9 (going from the second to last line that you >> quoted, to the last line). >> One reason this proof is deficient is because of the assumption >> that 10(.9999999...) = 9.9999999... (which is true, but needs to be >> proven). >> --Mark > But he assumes .999... = 1 in his equation before it is proven. Would you point out where he makes this assumption? I repeat the entire proof, expanded a bit, with line numbers added for your convenience: [1] 9=9 [2] 9=9.9999999...-.9999999... [3] 9(1)=10(.9999999...)-.9999999 [4] Let x = .9999999... and substitute in [3] [5] 9(1) = 10x - x [6] 9(1) = 9(x) [7] Therefore x=1 In which line is the assumption .99999... = 1 used? --Mark === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Mark Nudelman : > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) > Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). > He assumes 9(1) = 9 (going from the second to last line that you > quoted, to the last line). > One reason this proof is deficient is because of the assumption > that 10(.9999999...) = 9.9999999... (which is true, but needs to be > proven). > --Mark >> But he assumes .999... = 1 in his equation before it is proven. > Would you point out where he makes this assumption? I repeat the entire > proof, expanded a bit, with line numbers added for your convenience: > [1] 9=9 > [2] 9=9.9999999...-.9999999... > [3] 9(1)=10(.9999999...)-.9999999 > [4] Let x = .9999999... and substitute in [3] > [5] 9(1) = 10x - x > [6] 9(1) = 9(x) > [7] Therefore x=1 > In which line is the assumption .99999... = 1 used? [3]. The possibility of an infinite borrow generates headaches. > --Mark -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 > 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ jesus christ! do you know anything about mathematical induction?????????????? let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = 9*(1.11111111...) = 9.999999.. then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - 1)/(1 - 1/10))| = |1/10^n| = 1/10^n < e for all n >= N > -log(e) that means, the difference between the infinitely repeating decimal with period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work for any number, not just 9) if you dont believe that x_n = 9.9999999999999999999 then thats your fault, you need to learn some simple math.... just try to find me a number sticktly between .999999999999..... and 1! you can do this for all x if you want... x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) if x is terminating or repeating in its tail, then the sum has a simple solution and its easy to calculate the answer. if you put x = 1, the {x} = 0 x = .99999...... then sum is just over 9/10^k which is easily to compute again, the only thing that you can have any sorta problem with is how .9999999999 could be reprsented by the sum, but that is your problem... as any halfwit knows that. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) >Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). >One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >--Mark >> But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your fault, >you need to learn some simple math.... just try to find me a number sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. Hey .999... IS NOT A REAL NUMBER (PERIOD). See math link below: http://mathworld.wolfram.com/HyperrealNumber.html .999... is of the form of a hyper-real number because there is a space between the real numbers between .999... and 1. .999... | | 1 ^ | See space A Hyperreal number is of the form Where n is a real number, x < n x = .999... n = 1 .999... < 1 THEREFORE, .999... =/= 1 Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: re:PROOF that 0.99999... = 1 >> heres a much simpler proof: >> statement: .9999999...=1 >> since 9x=10x-x, >> 9=9 >> 9=9.9999999...-.9999999... >> 9(1)=10(.9999999...)-.9999999 9(1) =/= 9(.999...) >>Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >>assumes 9(1) = 9 (going from the second to last line that you quoted, to >>the >>last line). >>One reason this proof is deficient is because of the assumption that >>10(.9999999...) = 9.9999999... (which is true, but needs to be proven). >>--Mark > But he assumes .999... = 1 in his equation before it is proven. > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ >>jesus christ! >>do you know anything about mathematical induction?????????????? >>let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >>9*(1.11111111...) = 9.999999.. >>then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >>1)/(1 - 1/10))| >>= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >>that means, the difference between the infinitely repeating decimal with >>period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >>work >>for any number, not just 9) >>if you dont believe that x_n = 9.9999999999999999999 then thats your >>fault, >>you need to learn some simple math.... just try to find me a number >>sticktly >>between .999999999999..... and 1! >>you can do this for all x if you want... >>x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >>if x is terminating or repeating in its tail, then the sum has a simple >>solution and its easy to calculate the answer. >>if you put x = 1, the {x} = 0 >>x = .99999...... >>then sum is just over 9/10^k which is easily to compute >>again, the only thing that you can have any sorta problem with is how >>.9999999999 could be reprsented by the sum, but that is your problem... as >>any halfwit knows that. > Hey .999... IS NOT A REAL NUMBER (PERIOD). > See math link below: > http://mathworld.wolfram.com/HyperrealNumber.html > .999... is of the form of a hyper-real number because there is a space > between > the real numbers between .999... and 1. > .999... | | 1 > ^ > | > See space > A Hyperreal number is of the form > Where n is a real number, > x < n > x = .999... > n = 1 > .999... < 1 > THEREFORE, > .999... =/= 1 > Smarts Alt. Physics News Group > http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 > S. Enterprize (Science Journal) > http://smart1234.s-enterprize.com/ your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! === Subject: Re: PROOF that 0.99999... = 1 In sci.math, Jon Slaughter <10sbusfei7k2lee@corp.supernews.com>: [snipped for sanity] >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ > your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! FSVO genius. Most of us use an alternate word with one less letter. :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: re:PROOF that 0.99999... = 1 > heres a much simpler proof: > statement: .9999999...=1 > since 9x=10x-x, 9=9 > 9=9.9999999...-.9999999... > 9(1)=10(.9999999...)-.9999999 >> 9(1) =/= 9(.999...) Huh? Nowhere in this proof does he assume that 9(1) = 9(.999...). He >assumes 9(1) = 9 (going from the second to last line that you quoted, to >the >last line). One reason this proof is deficient is because of the assumption that >10(.9999999...) = 9.9999999... (which is true, but needs to be proven). --Mark > But he assumes .999... = 1 in his equation before it is proven. >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >jesus christ! >do you know anything about mathematical induction?????????????? >let x_n = 9*sum((1/10)^k,k=0..n) = 9*(1 + 1/10 + 1/100 + .. 1/10^n) = >9*(1.11111111...) = 9.999999.. >then |10 - x_n| = |10 - 9*sum((1/10^k,k=0..n))| = |10 - ((1/10)^(k+1) - >1)/(1 - 1/10))| >= |1/10^n| = 1/10^n < e for all n >= N > -log(e) >that means, the difference between the infinitely repeating decimal with >period one is the same as 10, i.e. 9.9999999...... = 10 (ofcourse, this >work >for any number, not just 9) >if you dont believe that x_n = 9.9999999999999999999 then thats your >fault, >you need to learn some simple math.... just try to find me a number >sticktly >between .999999999999..... and 1! >you can do this for all x if you want... >x = [x] + {x} = ßoor(x) + sum((ßoor((n-x)*10^k) mod 10)/10^k) >if x is terminating or repeating in its tail, then the sum has a simple >solution and its easy to calculate the answer. >if you put x = 1, the {x} = 0 >x = .99999...... >then sum is just over 9/10^k which is easily to compute >again, the only thing that you can have any sorta problem with is how >.9999999999 could be reprsented by the sum, but that is your problem... as >any halfwit knows that. >> Hey .999... IS NOT A REAL NUMBER (PERIOD). >> See math link below: >> http://mathworld.wolfram.com/HyperrealNumber.html >> .999... is of the form of a hyper-real number because there is a space >> between >> the real numbers between .999... and 1. >> .999... | | 1 >> ^ >> | >> See space >> A Hyperreal number is of the form >> Where n is a real number, >> x < n >> x = .999... >> n = 1 >> .999... < 1 >> THEREFORE, >> .999... =/= 1 >> Smarts Alt. Physics News Group >> http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 >> S. Enterprize (Science Journal) >> http://smart1234.s-enterprize.com/ >your a freaken genius!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! youre not your Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > .999... | | 1 > ^ > | > See space Pure scribble. > A Hyperreal number is of the form You would not know a hyperreal if it bit you. You have not the foggiest notion of how the real number system R is extended to *R. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> .999... | | 1 >> ^ >> | >> See space >Pure scribble. >> A Hyperreal number is of the form >You would not know a hyperreal if it bit you. You have not the foggiest >notion of how the real number system R is extended to *R. >Bob Kolker Hey, I thought you said I didnt know what it was. You are wrong again, and again, again. http://mathworld.wolfram.com/HyperrealNumber.html Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 In sci.math, S. Enterprize Company .999... | | 1 > ^ > | > See space >>Pure scribble. > > A Hyperreal number is of the form >>You would not know a hyperreal if it bit you. You have not the foggiest >>notion of how the real number system R is extended to *R. >>Bob Kolker > Hey, I thought you said I didnt know what it was. You are wrong again, and > again, again. > http://mathworld.wolfram.com/HyperrealNumber.html Like that tells him *anything*. Heres a few Qs for you. [1] If d is such that 0 < d < 1/n for all n in N, what is d^2? d^3? sqrt(d)? [2] Why is 5/5 != 9/9? 5/5 = 1, of course; 0.2 * 5 = 1. 9/9, by contrast, is 0.111... * 9 = 0.999... = 1 - d. In base 12, 1/9 = 0.14(12) but 1/5 = .24972497...(12) ; therefore in this case 9/9 = 1 but 5/5 = 1-d. Does it matter what base one uses for arithmetic? [3] Explain how one computes D_10[.999..., w-1], where w (omega) is the first transfinite ordinal, and D_10[r,n] is rs nth digit to the right of the decimal point, if n is an integer, then evaluate D_10[(.999... + 9)/10, w-1] and D_10[.999... * 10 - 9, w-1]. (n can be negative but thats not all that important here.) [.sigsnip] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 > jesus christ! > do you know anything about mathematical induction?????????????? Enterprise does not even know what end comes out of. He is a total mathematical incompetent. He makes JSH look intelligent by comparison. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> jesus christ! >> do you know anything about mathematical induction?????????????? >Enterprise does not even know what end comes out of. He is a total >mathematical incompetent. He makes JSH look intelligent by comparison. >Bob Kolker Whats a hyper-real number? Do you even know anything about math? Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? No. But I do know how the hyperrals are constructed. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 In sci.math, robert j. kolker : >> Whats a hyper-real number? Do you even know anything about math? > No. But I do know how the hyperrals are constructed. > Bob Kolker http://mathworld.wolfram.com/HyperrealNumber.html is extremely bare-bones (is there one hyperreal? more than one? arithmetic operations? proofs?) but at least its a start. A reference link http://members.tripod.com/PhilipApps/line.html looks to be little more than my attempts at d-math, though there might be more than one d -- or H, its dual. No doubt one could claim at least three theories: [1] An infinite hierarchy of d < 1/n for all n in N: 0 < ... < d^4 < d^3 < d^2 < d < 1, with a more or less standard algebra (e.g., (1-d)^3 = 1 - 3d + 3d^2 - d^3). [2] d^k = d for some k in N. [3] Some other esoteric condition. I suppose one might even notate this as R[d] -- a standard polynomial group over R, with a slightly weird ordering. And again, I must complain that S. Enterprize is being extremely sloppy here. (Not that Im all that neat, but hopefully my notations clear at least.) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are You dont even know what a hyper-real number is??? And you are name calling people here like you know everything?????? Why not admit you ARE WRONG! constructed. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > You dont even know what a hyper-real number is??? And you are name calling > people here like you know everything?????? Why not admit you ARE WRONG! Quick. Define an ultra-filter. No, dont look it up. Bob Kolker === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. >Bob Kolker Oh this is so hard to understand, I might need to take an asprin for a headache. Ill define it with an example. Suppose you have alot of people here making noise here on this NG and they dont know what they are talking about with .999..., and then comes along an ultrafilter F_Smart1234 with the correct information. What we do is apply ultrafilter F_Smart1234 to the whole set S of noise on the NG, and then just the pure correct answer is shown. The ultrafilter is then said to be a success and has worked very well, and is therefore proven. Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 > Whats a hyper-real number? Do you even know anything about math? >>No. But I do know how the hyperrals are > > You dont even know what a hyper-real number is??? And you are name >>calling > people here like you know everything?????? Why not admit you ARE WRONG! >>Quick. Define an ultra-filter. No, dont look it up. >>Bob Kolker > Oh this is so hard to understand, I might need to take an asprin for a >headache. > Ill define it with an example. Suppose you have alot of people here >making >noise here on this NG and they dont know what they are talking about with >.999..., and then comes along an ultrafilter F_Smart1234 with the correct >information. What we do is apply ultrafilter F_Smart1234 to the whole set S >noise on the NG, and then just the pure correct answer is shown. > The ultrafilter is then said to be a success and has worked very well, and >is therefore proven. Your turn. Perform a ANOVA statistical test between .999... and 1. And of course go into details explaining what the ANOVA test is. hurry hurry dont look... Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: PROOF that 0.99999... = 1 >> Whats a hyper-real number? Do you even know anything about math? >No. But I do know how the hyperrals are >> You dont even know what a hyper-real number is??? And you are name >calling >> people here like you know everything?????? Why not admit you ARE WRONG! >Quick. Define an ultra-filter. No, dont look it up. Oh, but I do have the right to refresh my memory. I even gave you time to do this and you still dont know what a hyper-real number is. >Bob Kolker Smarts Alt. Physics News Group http://pub39.bravenet.com/forum/show.php?usernum=3320272813& cpv=1 S. Enterprize (Science Journal) http://smart1234.s-enterprize.com/ === Subject: Re: A Quantum Poem for Xmas