mm-485 === Subject: Re: calculus help (e/d limit proofs) Felix, > But if you want to see a different good exposition, try Foundations > of Mathematical Analysis by Fikhtengoltz Are you sure that you have the name/ author correct? I can only find Foundations of Mathematical Analysis by Johnsonbaugh & Pfaffenberger, http://www.amazon.com/exec/obidos/tg/detail/-/0486421740/qid= 1065159257/sr=8 -1/ref=sr_8_1/102-6922965-6261706?v=glance&s=books&n=507846 reasonably sized math library: http://madcat.library.wisc.edu/index.html === Subject: Re: calculus help (e/d limit proofs) All: sleepless brain ;) WWW: I just reread the problem, and you may be interested to note that Spivak specifically does not include the condition N>0. Felix: === Subject: Re: calculus help (e/d limit proofs) > WWW: > I just reread the problem, and you may be interested to note that Spivak > specifically does not include the condition N>0. Right, there's no need to put that in. === Subject: Re: Primitive Element Theorem > Incidentally, I think the title of the thread is somewhat misleading -- > I don't think the term primitive element is normally used in this sense. > Usually an extension k(w) is said to be _simple_, I think, > so the result should probably be called the Simple Extension Theorem. Agree that an extension k(w) is called simple, but still I have always known the theorem we're talking about as the theorem of the primitive element. > I'm not sure if you are just trying to prove the result, > or are looking for an alternative proof. > The latter seems to me an entirely worthy enterprise, > as the standard proof has always struck me as rather unsatisfying, > since it doesn't really give any idea _why_ the result holds. There is a very slick proof using Galois Theory. First you show that if F has characteristic zero and K is a finite extension of F then there are only finitely many intermediate fields, that is, fields E containing F and contained in K. Once you have that result you look at all the fields F(u + av) with a in F. Since these are intermediate fields between F and K = F(u, v), and since F is infinite, two of them must be the same: F(u + av) = F(u + bv) = E, say, with a not equal to b. But now e contains both u + av and u + bv, hence also their difference, (a - b)v, hence also v, hence also u, hence E = F(u, v). It's in the proof of the existence of only finitely many intermediate fields that Galois Theory comes in. Let L be a normal closure of K over F. L is a finite extension of F, the Galois group of L over F is finite, it has only finitley many subgroups, the fields intermediate between F and L are in one-to-one correspondence with these subgroups, so there are only finitely many of them, a fortiori for fields between F and K. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Primitive Element Theorem === Subject: Re: Primitive Element Theorem Primitive Element Theorem: If u,v are algebraic over F, a field of characteristic zero, then there's some algebraic w with F(u,v) = F(w). >First you show that if F has characteristic zero and K is a finite >extension of F then there are only finitely many intermediate fields, >that is, fields E containing F and contained in K. Once you have >that result you look at all the fields F(u + av) with a in F. Since >these are intermediate fields between F and K = F(u,v), and since F >is infinite, two of them must be the same: F(u + av) = F(u + bv) = E, >say, with a not equal to b. But now E contains both u + av and u + >bv, hence also their difference, (a - b)v, hence also v, hence also >u, hence E = F(u,v). >It's in the proof of the existence of only finitely many intermediate >fields that Galois Theory comes in. Let L be a normal closure of K >over F. L is a finite extension of F, the Galois group of L over F >is finite, it has only finitley many subgroups, the fields >intermediate between F and L are in one-to-one correspondence with >these subgroups, so there are only finitely many of them, a fortiori >for fields between F and K. Galois theory, I'll remain content with what I've learned for now. ---- === Subject: Re: Non-Polynomial Time Algorithms |But in GENERALIZED CHESS and GENERALIZED CHECKERS there can be games |with exponentially many moves. These games are known to be |EXPTIME-complete and so not in PSPACE. method produces a problem in EXPTIME but not in PSPACE? Keith Ramsay === Subject: Re: Non-Polynomial Time Algorithms > But in GENERALIZED CHESS and GENERALIZED CHECKERS there can be games > with exponentially many moves. These games are known to be > EXPTIME-complete and so not in PSPACE. Oops. My mistake. It's not known whether PSPACE is different from EXPTIME. I should have said so likely not in PSPACE. > method produces a problem in EXPTIME but not in PSPACE? You can't diagonalize over PSPACE in EXPTIME -- or at least not using the naive method. An EXPTIME algorithm has running time O(2^(n^j)) where j is constant and n is the problem size. But there are PSPACE algorithms requiring space O(n^k) for arbitrarily large k and so taking O(2^(n^k)) to simulate. So any given EXPTIME algorithm will find some PSPACE algorithms it can't simulate. (You can diagonalise over PSPACE in EXPSPACE, or over P in EXPTIME. The time and space hierarchy theorems generalize these results.) -- Gareth Rees === Subject: Terminology question What do you call a statement, after a proof of a theorem, that says: This theorem does NOT apply if such-and-such assumption is removed, as we'll show here? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ Last year he disrespected me - and then he showed lack of respect. - Anthony Mason === Subject: Re: Terminology question > What do you call a statement, after a proof of a theorem, that says: > This theorem does NOT apply if such-and-such assumption is removed, > as we'll show here? Ralph. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Codense sets in R are 0-dimensional? I've been thinking about topological dimension a bit. I've come to the conclusion that codense sets in R are 0-dimensional. Proof: Codense sets have empty interiors. This means that codense sets in R cannot contain intervals, otherwise they would have at least one interior point. This also means, according to Hurewicz and Wallman, that codense sets in R are 0-dimensional. Does this apply in the other direction too? Are 0-dimensional sets in R codense? What about other spaces, for example R^2? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ You can pick your friends, you can pick your nose, but you can't pick your relatives. - MAD Magazine === Subject: Re: Codense sets in R are 0-dimensional? >I've been thinking about topological dimension a bit. I've come to the >conclusion that codense sets in R are 0-dimensional. >Proof: Codense sets have empty interiors. This means that codense sets >in R cannot contain intervals, otherwise they would have at least one >interior point. This also means, according to Hurewicz and Wallman, >that codense sets in R are 0-dimensional. >Does this apply in the other direction too? Are 0-dimensional sets in R >codense? >What about other spaces, for example R^2? Consider the x-axis in the plane ... In general: in R^n a set has empty interior if and only if its topological dimension is less than n. KP -- the Netherlands === Subject: Re: Codense sets in R are 0-dimensional? K. P. Hart scribbled the following: > In general: in R^n a set has empty interior if and only if its > topological dimension is less than n. So this confirms my theorem for the case R^1? But it does not follow that a codense set in R^n in general has to be *0-dimensional*? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ This isn't right. This isn't even wrong. - Wolfgang Pauli === Subject: Re: Codense sets in R are 0-dimensional? > K. P. Hart scribbled the following: >> In general: in R^n a set has empty interior if and only if its >> topological dimension is less than n. > So this confirms my theorem for the case R^1? But it does not follow > that a codense set in R^n in general has to be *0-dimensional*? Let X be Q^n, the set of all rational points in R^n, and let Y = R^n - Q^n, its complement. Both X and Y are dense in R^n, so both are codense. But R contains subspaces homeomorphic to R^{n-1}, e.g., {pi} x R^{n-1}. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Codense sets in R are 0-dimensional? Robin Chapman scribbled the following: >> K. P. Hart scribbled the following: > In general: in R^n a set has empty interior if and only if its > topological dimension is less than n. >> So this confirms my theorem for the case R^1? But it does not follow >> that a codense set in R^n in general has to be *0-dimensional*? > Let X be Q^n, the set of all rational points in R^n, and let > Y = R^n - Q^n, its complement. Both X and Y are dense in R^n, so > both are codense. But R contains subspaces homeomorphic to > R^{n-1}, e.g., {pi} x R^{n-1}. I do not understand the last sentence: But R contains subscapes homeomorphic to R^{n-1}, eg. {pi} x R^{n-1}. Are you sure you meant *R* contains...? Don't you mean Y contains... as {pi} x R^{n-1} is a subspace of Y? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ I am lying. - Anon === Subject: Re: Codense sets in R are 0-dimensional? > Robin Chapman scribbled the > following: > K. P. Hart scribbled the following: >> In general: in R^n a set has empty interior if and only if its >> topological dimension is less than n. > > So this confirms my theorem for the case R^1? But it does not follow > that a codense set in R^n in general has to be *0-dimensional*? >> Let X be Q^n, the set of all rational points in R^n, and let >> Y = R^n - Q^n, its complement. Both X and Y are dense in R^n, so >> both are codense. But R contains subspaces homeomorphic to >> R^{n-1}, e.g., {pi} x R^{n-1}. > I do not understand the last sentence: But R contains subscapes > homeomorphic to R^{n-1}, eg. {pi} x R^{n-1}. Are you sure you meant > *R* contains...? Don't you mean Y contains... as {pi} x R^{n-1} is > a subspace of Y? What do you think ? :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Codense sets in R are 0-dimensional? Robin Chapman scribbled the following: >> Robin Chapman scribbled the >> following: >> K. P. Hart scribbled the following: > In general: in R^n a set has empty interior if and only if its > topological dimension is less than n. >> >> So this confirms my theorem for the case R^1? But it does not follow >> that a codense set in R^n in general has to be *0-dimensional*? > Let X be Q^n, the set of all rational points in R^n, and let > Y = R^n - Q^n, its complement. Both X and Y are dense in R^n, so > both are codense. But R contains subspaces homeomorphic to > R^{n-1}, e.g., {pi} x R^{n-1}. >> I do not understand the last sentence: But R contains subscapes >> homeomorphic to R^{n-1}, eg. {pi} x R^{n-1}. Are you sure you meant >> *R* contains...? Don't you mean Y contains... as {pi} x R^{n-1} is >> a subspace of Y? > What do you think ? :-) I think I am correct. {pi} x R^{n-1} is a subspace of R^n-Q^n, and it is homeomorphic to R^{n-1} which is not necessarily 0-dimensional, {pi} x R^{n-1} is not necessarily 0-dimensional either, although as a subset of a codense set it is codense. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ The question of copying music from the Internet is like a two-barreled sword. - Finnish rap artist Ezkimo === Subject: Re: Codense sets in R are 0-dimensional? > Robin Chapman scribbled the following: >> >> So this confirms my theorem for the case R^1? But it does not follow >> that a codense set in R^n in general has to be *0-dimensional*? >> > Let X be Q^n, the set of all rational points in R^n, and let > Y = R^n - Q^n, its complement. Both X and Y are dense in R^n, so > both are codense. But R contains subspaces homeomorphic to > R^{n-1}, e.g., {pi} x R^{n-1}. > I think I am correct. {pi} x R^{n-1} is a subspace of R^n-Q^n, and it > is homeomorphic to R^{n-1} which is not necessarily 0-dimensional, > {pi} x R^{n-1} is not necessarily 0-dimensional either, although as a > subset of a codense set it is codense. {0}xR is codense subset R^2 and {0}xR isn't zero dimensional. === Subject: Re: Codense sets in R are 0-dimensional? William Elliot scribbled the following: >> Robin Chapman scribbled the following: > So this confirms my theorem for the case R^1? But it does not follow > that a codense set in R^n in general has to be *0-dimensional*? > >> Let X be Q^n, the set of all rational points in R^n, and let >> Y = R^n - Q^n, its complement. Both X and Y are dense in R^n, so >> both are codense. But R contains subspaces homeomorphic to >> R^{n-1}, e.g., {pi} x R^{n-1}. >> I think I am correct. {pi} x R^{n-1} is a subspace of R^n-Q^n, and it >> is homeomorphic to R^{n-1} which is not necessarily 0-dimensional, >> {pi} x R^{n-1} is not necessarily 0-dimensional either, although as a >> subset of a codense set it is codense. > {0}xR is codense subset R^2 and {0}xR isn't zero dimensional. Which would prove what I have been thinking for the past few hours, that in the space R, a set is 0-dimensional iff it is codense, but in the space R^n where n>1, this does not apply any more. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ No, Maggie, not Aztec, Olmec! Ol-mec! - Lisa Simpson === Subject: Re: Codense sets in R are 0-dimensional? > I think I am correct. {pi} x R^{n-1} is a subspace of R^n-Q^n, and it > is homeomorphic to R^{n-1} which is not necessarily 0-dimensional, > {pi} x R^{n-1} is not necessarily 0-dimensional either, although as a > subset of a codense set it is codense. What dimension do you think that {pi} x R^{n-1} is? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: Codense sets in R are 0-dimensional? Robin Chapman scribbled the following: >> I think I am correct. {pi} x R^{n-1} is a subspace of R^n-Q^n, and it >> is homeomorphic to R^{n-1} which is not necessarily 0-dimensional, >> {pi} x R^{n-1} is not necessarily 0-dimensional either, although as a >> subset of a codense set it is codense. > What dimension do you think that {pi} x R^{n-1} is? Um, the same dimension as R^{n-1}, as they're homeomorphic, and dimension is a topological invariant? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ I wish someone we knew would die so we could leave them flowers. - A 6-year-old girl, upon seeing flowers in a cemetery === Subject: Re: Codense sets in R are 0-dimensional? > I've been thinking about topological dimension a bit. I've come to the > conclusion that codense sets in R are 0-dimensional. What's a codense set? > Proof: Codense sets have empty interiors. This means that codense sets > in R cannot contain intervals, otherwise they would have at least one > interior point. This also means, according to Hurewicz and Wallman, > that codense sets in R are 0-dimensional. > Does this apply in the other direction too? Are 0-dimensional sets in R > codense? > What about other spaces, for example R^2? === Subject: Re: Codense sets in R are 0-dimensional? William Elliot scribbled the following: >> I've been thinking about topological dimension a bit. I've come to the >> conclusion that codense sets in R are 0-dimensional. > What's a codense set? One whose complement is dense. >> Proof: Codense sets have empty interiors. This means that codense sets >> in R cannot contain intervals, otherwise they would have at least one >> interior point. This also means, according to Hurewicz and Wallman, >> that codense sets in R are 0-dimensional. >> Does this apply in the other direction too? Are 0-dimensional sets in R >> codense? >> What about other spaces, for example R^2? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ I said 'play as you've never played before', not 'play as IF you've never played before'! - Andy Capp === Subject: Re: Codense sets in R are 0-dimensional? === Subject: Re: Codense sets in R are 0-dimensional? William Elliot scribbled the following: > codense sets in R are 0-dimensional. >> What's a codense set? >One whose complement is dense. D codense subset S when SD dense subset S D codense subset S iff cl SD = S iff int D = nulset Thus codense sets are exactly the sets with empty interiors. >> Proof: Codense sets have empty interiors. This means that codense sets >> in R cannot contain intervals, otherwise they would have at least one >> interior point. This also means, according to Hurewicz and Wallman, >> that codense sets in R are 0-dimensional. >> Does this apply in the other direction too? Are 0-dimensional sets in R >> codense? Assume A is zero dimensional subset R and int A /= nulset. Thus some open interval U with U subset A Now U is subspace of zero dimensional space and zero dimensional spaces are hereditary Thus U is zero dimensional which cannot be as U is interval of R >> What about other spaces, for example R^2? Does not the above proof suffice for any space with an open base having no zero dimensional sets? How is it that an empty interior subset of R is zero dimensional? ---- === Hello to all, What is your favorite popularization - book - review - author - internet site ? why ? A. === > Hello to all, > What is your favorite popularization > - book > - review > - author > - internet site ? book > why ? I hate commentators, you can't hear the sound, so reviews and such are out. Though internet site is close second, one day I'll have a 17 inch screen wireless laptop and combine the 2. Herc === Subject: Re: Second Order PDE (Linear) I was a little late in sending this but here goes: the solution of the PDE that you specified. I also posted it on Sci.Math under Verse and Solution to PDE Under truth there often lies mysteries that wise eyes would love to discover but until the truth is well known mystery and truth will be alone in their knowledge of each other The following discussion will be non-rigorous but hopefully informative and interesting: Let P= P(x, y) and let the Linear Partial Differential Operator L(P) be defined by L(P) = P_xx + P_yy +(x+2y)P_x +(x^2 + y)P_y + x^3 P (eq.1a) Now consider the equation L(P) = 0 (eq.1b) with the auxiliary conditions, P(0, y) = F(y) (eq.2a) P_x(0, y) = G(y) (eq.2b) For ease of computation, we will set F(y) = 1 and G(y) = 0. Therefore our auxiliary conditions become P(0, y) = 1 (eq.3a) P_x(0, y) = 0 (eq.3b) We propose the following infinite series (in x and y) solution which satisfies (1b) and the auxiliary conditions (3a) and (3b) P(x, y) = 1 - 1/20x^5 + 1/168x^7 +1/60yx^6 - 1/1728x^9 - 1/280yx^8 - 1/210y^2x^7 + 1/2700x^10 - 29/332640x^12 - 1/3960yx^11 - 1/378000x^18 + . . . (eq.4) Note: the numbers in parenthesis are equation #'s The reader will note that the right side of (4) is truncated after the18th degree term in x . Of course any (finite representation of an) infinite series must necessarily be truncated at some point. Eq. (4) is the result generated after the 2nd operation of an infinite series of operations that generate the terms of P(x, y) . More on the details of this procedure will be presented later. But presently we will show that (4) formally satisfies eqs. (1b) and the auxili- ary conditions eqs. (3a & 3b), when terms of degree > or = 7 are neglected . Then we will dis- cuss convergence (and uniqueness). We now calculate, starting from (4), all the partial derivatives that appear in (1a). We have P_x = - 1/4x^4 + 1/24x^6 + 1/10yx^5 - 1/192x^8 - 1/35yx^7 - 1/30y^2x^6 + 1/270x^9 - 29/27720x^11 - 1/360yx^10 - 1/25200x^14 + . . . (5) P_xx = -x^3 + 1/4x^5 + 1/2yx^4 - 1/24x^7 - 1/5yx^6 - 1/5y^2x^5 + 1/30x^8 - 29/2520x^10 - 1/36yx^9 - 1/1800x^13 + . . . (6) P_y = 1/60x^6 - 1/280x^8 - 1/105yx^7 - 1/3960x^11 + . . . (7) P_yy = -1/105x^7 (8) Now if we substitute eqs. (5) thru (8) into (1a) then after performing the indicated operations and simplifying, the reader may do the elementary algebra perhaps using software such as Ma- thematica, eq. (1b) becomes L(P) = -1/105x^7 - 1/192x^9 - 143/3360yx^8 - 1/10y^2x^7 - 1/15y^3x^6 - 41/7560x^10 -5/378yx^9 - 1081/665280x^12 - 241/27720yx^11 -13/1260y^2x^10 -13/29700x^13 - 211/1663200x^15 - 23/69300yx^14 -1/378000x^18 +. . . (9) The right side of (9) represents the truncation error . It will be seen that all terms of de- gree (in x and y) of 6 or less cancel out. If we NEGLECT TERMS OF DEGREE 7 AND HIGHER then (9) may be written as follows: L(P) = 0 (neglecting terms of the 7th degree and higher) (10) If a ( higher) accuracy of order k is desired, non-rigorously measured by the cancellation of all terms of degree < or = k, for an arbitrary integer k >0, then (by conjecture) this can be accomplished by calculating more terms via the sequence of operations referred to previously. Eq. (4) was obtained after only two such operations. Referring to the expressions that were simplified in order to obtain (10), the terms of lowest order (or degree in x and y) to appear were of the third order. All terms of order three thru six canceled out. However, it should be noted that the terms of order 4 and 6 canceled out trivially since no terms of order 4 or 6 appeared originally. Moreover, all of the terms of order 7 canceled out except for x^7 which has a coefficient with an abso- lute value of 1/105. In addition, the only term of order 8 canceled out. If the reader will refer to the statement of the Cauchy- Kovalevshy Theorem it will be seen that eqs. (1a), (1b), (3a) and (3b) satisfy the hypothesis of that theorem, therefore this set of equations has a unique solution. Consequently we will pose the question: if the first few terms of the solution of the PDE, that is stated in (1b), that satisfies the conditions (3a) and (3b), does not coincide with the first few terms on the right of (4), then what are those first few terms? A Dream Of A Scholar Oh thou sands of time thousands of you my hands can hold. But there are billions of you on the beach. Oh thou stars that shine in the night sky like gold like the sand you too are within my reach. Abstract Algebra Rapping Is there a one to one mapping between the stars of a galaxy and grains of sand on a beach? Notwithstandingly, an ordered pair contains just one of each. A specific statement of the conjecture referred to above will be presented later. Your comments are welcomed and please feel free to point out any typos and/or errors. The verse(s) are courtesy of yours truly. === Subject: Re: Talk to a mathematician (Op-Ed) >> >> >> .... After all, it's a >> simpler world where I'm just a nut, instead of a major discoverer who >> found an over hundred year old error in core mathematics, right? >> >> Yes, right. Or at least, nearly so. I'd replace nut with silly >> person who could learn some [not all that difficult] mathematics, but >> chooses not to. >> >> Which defies the fact that I *explained* my mathematical argument >> point-by-point in person to an actual math professor. That professor >> did NOT find any error, but instead claimed my work was out of his >> area. >> >> However, given that information, you choose to claim that I am a >> silly person who needs to learn mathematics, which is an odd >> response. >> Many people here have pointed out your errors and asked you questions. >> You do not respond. Because one person does not find an error that does >> not mean those who do can be ignored. Silly seems ok to me, but if >> you find it terribly offensive I'll withdraw it and apologize. >The one person is a distinguished math professor at Vanderbilt >University, my alma mater. And it follows that if (note the if!) he disagrees with everyone here then he's right and everyone here is wrong? Leads to one of the many questions that you've never answered: When replying to your critics, the ones with advanced degrees in particular, you often state correctly that the fact that someone has a PhD doesn't prove he's right. Credentials are not how we decide mathematical validity, fine. But then _why_ are credentials relevant when someone _agrees_ with you? >The assertion that others have found an error in my work is stated >without proof. Right. Without proof, unless one's newsreader has an Up button, or one has access to Google - then proof is easy to find. What _is_ being stated without proof is that McKenzie found no errors! You _say_ that you answered all his objections. But nobody here actually believes that, because it happens so often that you also claim you've answered the objection of someone here, when in fact it's clear to anyone who can read that you simply have not done so. Did he _agree_ that all his objections had been answered? >What seems ok to you is to insinuate things you cannot prove. >I see you as a proven liar, as an error in my proof of the problem in >core mathematics can NOT be found. >As you're lying about an error of such magnitude you are standing >against mathematics as a discipline, proving yourself to be an enemy >of it. >> Do you think that you _don't_ need to learn mathematics? You may be the >> only poster here who thinks that he does not need to learn some >> mathematics. Silly seems more and more reasonable. >What you call reasonable is suspect since you've proven yourself to be >an enemy of mathematics. >Many of you seem to be lost on the primary point here. >It's not about me. Trying personal attacks against me is what is >silly. >I have not only found an error in core mathematics, which is over a >hundred years old, I've proven it with very concise algebra, and >successfully explained it to a distinguished math professor. >Some of you seem hellbent on attacking me for those achievements. >And that is, I strongly suggest, what is silly. >James Harris === Subject: Re: Talk to a mathematician (Op-Ed) > [Gabriele Rossetti] has left a vast body of writings... in which > he has attempted to prove the truth of his unorthodox interpre- > tation of medieval literature. They present a formidable > record of unsystematic research in which we see an enthusiast > plunging farther and farther and farther from the logic of facts > and good sense until truth is lost in the dreadful nightmare > of an idee fixe. There is no real evolution of the Theory > although it grows and expands until it embraces ever wider > horizons. The numerous inaccuracies of deduction, mis-statements > of historical fact, and self-contradictions...have caused critics > to turn away from them in disgust... [...] It is impossible to > read far... without realizing that we have to deal with a work of > faith and imagination rather than of reasoning. There is an > appearance of reason, for the author is set on proving by logic > the truth of what he already believes by intuition. The truth > is plain to him and he cannot comprehend why others do not > immediately accept it, but as they desire demonstration he has > multiplied his proofs. It is the redundancy and confusion of a > prophet expounding by a familiar method the truth revealed to his > own simple soul in a flash of inspiration... In such work as > this... it is idle to look for the calm reasoning of a scholar; > we do not find it, and there is little or no advantage in > attacking the obvious inconsistencies and absurdities that abound. > -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in > _The Shakespearan Ciphers Examined_, by William F. > Friedman and Elizebeth S. Friedman As you say, my good knight! There ought to be laws to protect the body of acquired knowledge. Take one of our good pupils, for example: modest and diligent, from his earliest grammar classes he's kept a little notebook full of phrases. After hanging on the lips of his teachers for twenty years, he's managed to build up an intellectual stock-in-trade; doesn't it belong to him as if it were a house or money? (P. Claudel, _Le Soulier du Satin_; cited in _Language and Symbolic Power_, p. 43) === Subject: Re: Fundamental Reason for High Achievements of Jews -Ju === >Subject: Re: Fundamental Reason for High Achievements of Jews >-Ju > >> Google searrch: brainwashed author:tom author:potter >> >> 387 hits >> >> As also can be seen, >> I actively expose the brainwashed people, >> >> 388 > >As can be seen, >I know how to stimulate brainwashed people, >and get them to expose their conditioning. > > There's a word for that: troll. > > >I can make brainwashed people >come to me like moths to a light. > > Come to you? You're the idiot posting this crap all over sci.math. > > >389 > > Yeah, that's something to be proud of. It would be different if you were a > clever troll with witty banter like James Harris, but the ad nauseam > posting of > the same tiresome boiler-plate responses just makes people dismiss you as a > pathetic fool. What's the matter old man, can't get it up anymore? > > Ask yo momma. > Holy ! > Tom Potter drops his rubber stamp, dusts off his keyboard, > AND ACTUALLY TYPES A REPLY! > Touched a nerve, huh? > Not the kind of repartee that makes a good flame war, but you were probably > woozy from the adrenaline rush. But that's short-lived, so I expect you'll be > back to the usual prattle about boiler plates and attacking the messenger. So > you may as well just give up. Better men than you have tilted with The > Mensanator > only to have their hopes dashed and their fears realized. Note that your beloved Mensanator crawled into the gutter with his remark: What's the matter old man, can't get it up anymore? and I simply responded to him at the same level. As you can see from a Google search of my posts, I generally try to communicate at the level of the people I interact with, so they understand. When I encounter immature, low class people, I try to communicate at their level, as they are unable to engage in mature. rational, intelligent, moral discussions. When in Rome, do as the Romans. BTW, if you want to see how your hero fared in his dichotomies with me, take a look at these threads. I see that I have him so bent out of shape, that he just bushwhacked me in a thread that I have not participated in. That's your hero. -- Tom Potter === Subject: Re: Fundamental Reason for High Achievements of Jews -Ju === >Subject: Re: Fundamental Reason for High Achievements of Jews >-Ju === >>Subject: Re: Fundamental Reason for High Achievements of Jews >>-Ju >> > Google searrch: brainwashed author:tom author:potter > > 387 hits > > As also can be seen, > I actively expose the brainwashed people, > > 388 >> >>As can be seen, >>I know how to stimulate brainwashed people, >>and get them to expose their conditioning. >> >> There's a word for that: troll. >> >> >>I can make brainwashed people >>come to me like moths to a light. >> >> Come to you? You're the idiot posting this crap all over sci.math. >> >> >>389 >> >> Yeah, that's something to be proud of. It would be different if you >were a >> clever troll with witty banter like James Harris, but the ad nauseam >> posting of >> the same tiresome boiler-plate responses just makes people dismiss you >as a >> pathetic fool. What's the matter old man, can't get it up anymore? >> >> Ask yo momma. >> Holy ! >> Tom Potter drops his rubber stamp, dusts off his keyboard, >> AND ACTUALLY TYPES A REPLY! >> Touched a nerve, huh? >> Not the kind of repartee that makes a good flame war, but you were probably >> woozy from the adrenaline rush. But that's short-lived, so I expect you'll >> back to the usual prattle about boiler plates and attacking the messenger. >> you may as well just give up. Better men than you have tilted with The >> Mensanator >> only to have their hopes dashed and their fears realized. >Note that your beloved Mensanator >crawled into the gutter with his remark: >What's the matter old man, can't get it up anymore? >and I simply responded to him at the same level. >As you can see from a Google search of my posts, >I generally try to communicate at the level >of the people I interact with, so they understand. >When I encounter immature, low class people, >I try to communicate at their level, >as they are unable to engage in >mature. rational, intelligent, moral discussions. Well, that's an unusual tactic. Most of the rational, moral people I know won't descend into the gutter to duke it out with a troll. They either killfile or ignore them. But instead of passively ignoring them, this is a way of actively ignoring them? Something like a print version of the echo game? >When in Rome, do as the Romans. >BTW, if you want to see how your hero fared in his >dichotomies with me, take a look at these threads. Mensanator%26safe%3Dimages%26ie%3DUTF-8%26oe%3DUTF-8%26as_ uauthors%3Dtom%2 520potter%26lr%3D%26hl%3Den en&lr=&ie=UTF-8&oe=UTF-8&selm=b9nrc7%24li147%241%40ID-188019. news.dfncis.d e&rnum=4 Okay, you can be rational when you choose to. Who knew all this rubber-stamping is a sublime ploy on your part? When you post like a troll, the casual lurker might mistake you for a troll. >I see that I have him so bent out of shape, >that he just bushwhacked I assume this is a metaphor about attacking from a place of concealment? How does one hide in the Usenet? >me in a thread that I have not participated in. Well, you obviously were lurking in that thread or you wouldn't have seen it. Does that still count as bushwhacking? >That's your hero. >Tom Potter === Subject: Re: Fundamental Reason for High Achievements of Jews -Ju === >Subject: Re: Fundamental Reason for High Achievements of Jews >-Ju === >>Subject: Re: Fundamental Reason for High Achievements of Jews >>-Ju >> > Google searrch: brainwashed author:tom author:potter > > 387 hits > > As also can be seen, > I actively expose the brainwashed people, > > 388 >> >>As can be seen, >>I know how to stimulate brainwashed people, >>and get them to expose their conditioning. >> >> There's a word for that: troll. >> >> >>I can make brainwashed people >>come to me like moths to a light. >> >> Come to you? You're the idiot posting this crap all over sci.math. >> >> >>389 >> >> Yeah, that's something to be proud of. It would be different if you > were a >> clever troll with witty banter like James Harris, but the ad nauseam > posting of >> the same tiresome boiler-plate responses just makes people dismiss you > as a >> pathetic fool. What's the matter old man, can't get it up anymore? >> >> Ask yo momma. >> >> Holy ! >> >> Tom Potter drops his rubber stamp, dusts off his keyboard, >> AND ACTUALLY TYPES A REPLY! >> >> Touched a nerve, huh? >> >> Not the kind of repartee that makes a good flame war, but you were probably >> woozy from the adrenaline rush. But that's short-lived, so I expect you'll > be >> back to the usual prattle about boiler plates and attacking the messenger. > So >> you may as well just give up. Better men than you have tilted with The >> Mensanator >> only to have their hopes dashed and their fears realized. >Note that your beloved Mensanator >crawled into the gutter with his remark: >What's the matter old man, can't get it up anymore? >and I simply responded to him at the same level. >As you can see from a Google search of my posts, >I generally try to communicate at the level >of the people I interact with, so they understand. >When I encounter immature, low class people, >I try to communicate at their level, >as they are unable to engage in >mature. rational, intelligent, moral discussions. > Well, that's an unusual tactic. Most of the rational, moral people > I know won't descend into the gutter to duke it out with a troll. > They either killfile or ignore them. But instead of passively ignoring > them, this is a way of actively ignoring them? Something like > a print version of the echo game? >When in Rome, do as the Romans. >BTW, if you want to see how your hero fared in his >dichotomies with me, take a look at these threads. > Mensanator%26safe%3Dimages%26ie%3DUTF-8%26oe%3DUTF-8%26as_ uauthors%3Dtom%2 > 520potter%26lr%3D%26hl%3Den > en&lr=&ie=UTF-8&oe=UTF-8&selm=b9nrc7%24li147%241%40ID-188019. news.dfncis.d > e&rnum=4 > Okay, you can be rational when you choose to. Who knew all this > rubber-stamping is a sublime ploy on your part? When you post like > a troll, the casual lurker might mistake you for a troll. >I see that I have him so bent out of shape, >that he just bushwhacked > I assume this is a metaphor about attacking from a place of > concealment? How does one hide in the Usenet? >me in a thread that I have not participated in. > Well, you obviously were lurking in that thread or you wouldn't have > seen it. Does that still count as bushwhacking? I don't lurk in threads, but I routinely search for threads that refer to me, so I can respond if necessary. Bushwhacking is a coward's way of retaliating, from what they think is a safe position. I have noticed, that when someone gets their ass kicked in a thread, that they go away and lick their wounds, and bushwhack in an unrelated post to get back at the person who wounded them. -- Tom Potter === Subject: Re: Fundamental Reason for High Achievements of Jews -Ju >>I see that I have him so bent out of shape, >>that he just bushwhacked >> I assume this is a metaphor about attacking from a place of >> concealment? How does one hide in the Usenet? >>me in a thread that I have not participated in. >> Well, you obviously were lurking in that thread or you wouldn't have >> seen it. Does that still count as bushwhacking? >I don't lurk in threads, >but I routinely search for threads that refer to me, >so I can respond if necessary. That sounds paranoid. But it goes to show they have to get up pretty early to put one over on you, huh? >Bushwhacking is a coward's way of retaliating, >from what they think is a safe position. Aren't you taking this too seriously? >I have noticed, that when someone gets their ass kicked >in a thread, that they go away and lick their wounds, That's the trouble with getting in the last word, you never know for sure whether your antagonist read your final reply. >and bushwhack in an unrelated post to get back at >the person who wounded them. Look on the bright side. You at least have the satisfaction of knowing they went away hurt and didn't just abandon the thread. It's important to know that you've won, right? === Subject: Re: Change all f=ma; to f=wa/g >snippety snip< > Mathematics should NEVER be used to simplify anything! LOL. very good :-) Andy === Subject: Re: Subfield of algebraics? >> > > Let phi be a bijection from R onto the power set P(N). Let > A_1, A_2, ... be disjoint infinite subsets of N with union equal to N. > Let f_j be a bijection from N onto A_j. For S a subset of N define > f(S) = {f(x) : x in S}, as usual. > > Then > > (S1, S2, ...) |-> union(f_j(S_j)) > > defines a bijection from P(N)^N to P(N), and hence > > (x1, x2, ...) |-> phi^(-1)(union(f_j(phi(x_j))) > > defines a bijection from R^N to R. > >> >> >>I don't quite understand this yet. >> Well of course you don't - how could you? >>I notice you define A_j above but >>haven't used it in the mapping descriptions. >> Not true. I define A_j, then I say f_j is a bijection >> from N onto A_j, and then f_j is used in the >> description of the mapping. >>{S_1, S_2, ...} = P(N) =/= {A_1 U A_2 U ...} = N >> Huh? Both equalities there are obviously false. >>You say that f_j is a bijection from N to A_j, but in the above you >>have the variable of f_j being an element S_j. >> Huh? Nothing above implies that f_j is an element of S_j. >>While each A_j is an >>S_k there are many more S's, subsets of N, than A's, disjoint infinite >>subsets of N the union of which is N, in terms of that {S_1, ...} is a >>proper superset of {A_1, ...}..] >> >>R^N, here again talking about unit intervals of reals for >>convenience's sake, is composed of elements like (0, 0, 0, ...) and >>(1, 1, 1, ...) and (0, 1, e/pi, 2-e, sqrt(2)/2, ...). That reminds me >>of the sequence representing a real where the base, instead of being >>2, is infinite and even similar to aleph_one. >> >>Please make your example clearer thus that it is easier to understand. >> out any details, without any informal abuses of notation. Any >> competent mathematician will have no problem understanding it. >> But don't feel bad, we can never understand the things you >> write either, so it all sort of balances out. >> ************************ >> David C. Ullrich >About the equalities, it's not {A_1 U A_2 U ... } it should be A_1 U >A_2 U ... = N, that I understand. About S_1, etcetera, where you say >S is a subset of N, I thought you meant that S_1, etcetera, were >subsets of N. What are S1, S2, etcetera, with no underscores >(subscript), and what then is S_j? >That's a pretty strong curse, there. >About the variable of f_j being A_j, that is to say, f_j(A_j) is a >bijective mapping from A_j to N, where A_j is one of infinitely many >disjoint infinite subsets of N, then you have f_j(S_j), that would >seem to imply S_j=A_j. Yet, this is where the definition of S_j is >unclear. >Is S_j any element of P(N) or is it one of infinitely many disjoint >infinite subsets of N, or neither? >You say {S1, S2, ...} mapping bijectively to union (f_j(S_j)) is the >same thing as mapping P(N) to P(N)^N. The expression union (f_j(A_j)) >evaluates to N. As S is a subset of N f_j maps elements of N to >elements of N, union f_j(A_j) is N. >You have phi and inverse phi, there, phi is a bijection from R to >P(N), its inverse describes a mapping from P(N) to R, phi^(-1)(phi(R)) >= R. >The images of f_j are subsets of N, that's what can be said about them >as described. The union of f_j(A_j) is N, I am thinking that union >(f_jS_j)) is N. Where you have x being an element of S, that would >only imply f_j(A_j)=N and union (f_j(A_j)) = N. >You have the sequences (S1, S2, ...), and (x1, x2, ...). Where x is >an element of N, (x_1, x_2, ...) is a coordinate of N^N. >Here I think you write |-> to mean bijection, where that is often >written <->. Also (S1, S2, ...) and (x1, x2, ...) are in parentheses >implying that they are sequences. Please correct the >misunderstanding. >Perhaps I'm particularly thick or ignorant of some your notations, >please explain further. Please explain what are S1, S2, (S1, S2, >...), x1, x2, (x1, x2, ...), f_j(S_j), and phi(x_j). none of your complaints make any sense. If you want to read mathematics you need to learn how first. I mean I'd be happy to explain the things you ask me to explain, but the explanations would not look any different from what I don't see anything unclear about any of it, or any way to place. >Ross ************************ David C. Ullrich === Subject: Re: JSH: Ok, so now maybe you know >Some of you may accept Jim Ferry's comments at face value No. The only person on the planet who did not immediately realize he was joking is you. I appreciate the comment in that other thread about how it was curious to see someone posting in support of your work - made my day. >hile others >may look for some escape from my discoveries. >But it's mathematics. >Mathematics is not an invention of human minds but a fundamental >constant which defines and shapes all of reality. >It defines and controls all of you as well, so you do not wish to >properly comprehend just how bad of an idea it is to try and defy the >math. >When I talk about the math, I'm talking about the idea construct by >which a human being can get some sense of that which is infinite, as >it shows traces of its activity, which stands outside of time. >Because of that, there is no limitation from time or space, no way to >stop it, no way to control it, kill it, or in any way for any of you >to affect it at all. >However, it is consistent. >Think you understand Goedel if you will, but I *do* understand his >work, and I understand what many of you do not, which is that it >*proves* the finiteness of our world and your inherent (as well as my) >limitations. >We are mortal. But we are not alone. >So when I tell you that it is important to accept the math, I'm not >talking mysticism, or giving vague threats or maybe's as it's not >about dying--you're going to die anyway--it's about living a better >life till you inevitably die. >And make no mistake, it's *provable* that immortality, as many of you >unfortunately conceive it, for us is impossible, and you will all >eventually die, as will I. >There are immortal beings--no, not so simple as calling them gods--but >we are not them. >I want to live a better life, and I want as many as you as possible to >live a better life, but the Universe does not work by coddling >failures. There has been an error in the core of mathematics for >over a hundred years, which has steadily been working against the >future of humanity. Friendly reminder of something you seem to forget quite often: When you talk this way you sound like a lunatic. Honest. >Look around you. If you wish, you can be given *more* evidence that >reality works by consistency and mathematics. That evidence will >involve a significant amount of human misery above and beyond what has >been seen up until now. >It's not punishment. It's a consequence of the Laws of Physics. >If you cannot accept the math, then you are failures. >It's that simple. If you are incapable of accepting the math, then >the Universe, as She is perfect, will simply allow situations beyond >your ability to handle. >So a Bush thinks there are weapons of mass destruction in Iraq. All >the evidence necessary to realize there aren't at the level he claims >are there, and I even warn him, but does he listen? >No. It's failures like that which will doom you and yours. You must >be able to reason at a level which fits circumstances. >You must learn to THINK at a level which will test you day and night. >You must learn to push your minds to the limits as if your life >depends on it, as the lives of your children, the future of humanity, >do. >It's not a game. The tests are over and humanity failed. Now the >fight is for the survival of humanity's children. Some of you must, >will learn to be more than human. >James Harris David C. Ullrich ************************** As far as I'm concerend you're trying to wait until I die, so I figure maybe you should die instead. How about that, eh? Wouldn't that be a better twist? You refuse to follow the math, so the great Powers that control reality and *speak* in mathematics decide to kill you instead of me. So what do you think about that, eh? Oh, can't hear Them talking? Well, I guess that's because you don't really understand Mathematics, the true language, which is THE language. They're talking about you now, and They agree with my assessment, and will not penalize me as They allowed the others like Galois and Abel to be penalized. They will kill you instead. James Harris speaking on Weird factorization, genius === Subject: Re: JSH: Ok, so now maybe you know > Friendly reminder of something you seem to forget quite often: > When you talk this way you sound like a lunatic. > Honest. Almost every unmoderated newsgroup on usenet has its deranged posters who natter on at interminable length. It is a complete waste of time to try to get them to behave rationally. Such is the case with JSH. The American Math Society used to have a policy of accepting and publishing all abstracts sent in to the NOTICES and almost every issue had something by one crackpot or other. No one tried to publish refutations. What would happen if no one ever responded to JSH's posts? Think how much valuable time of the responders and the readers of the responses would be saved! === Subject: Re: JSH: Ok, so now maybe you know > Friendly reminder of something you seem to forget quite often: > When you talk this way you sound like a lunatic. > Honest. > Almost every unmoderated newsgroup on usenet has its deranged posters > who natter on at interminable length. It is a complete waste of time > to try to get them to behave rationally. Such is the case with JSH. > The American Math Society used to have a policy of accepting and > publishing all abstracts sent in to the NOTICES and almost every issue > had something by one crackpot or other. No one tried to publish > refutations. What would happen if no one ever responded to JSH's > posts? Think how much valuable time of the responders and the readers > of the responses would be saved! Virtually any human activity, if not done, would save valuable time. === Subject: Re: JSH: Ok, so now maybe you know > Some of you may accept Jim Ferry's comments at face value while others > may look for some escape from my discoveries. > But it's mathematics. > Mathematics is not an invention of human minds but a fundamental > constant which defines and shapes all of reality. > It defines and controls all of you as well, so you do not wish to > properly comprehend just how bad of an idea it is to try and defy the > math. Really? Mathematics doesn't define anyone. Further, the only one here who is insistent on defying the math is you. After all, you have consistently refused to learn any of the mathematics you pretend has errors embedded in it. Every specific example you've proposed has been refuted soundly, and that done quite quickly. However, you have just as consistently refused to accept the validity of proofs presented to you, even when they were simple calculations. > When I talk about the math, I'm talking about the idea construct by > which a human being can get some sense of that which is infinite, as > it shows traces of its activity, which stands outside of time. How come you insist on calling those who raise valid counterexamples, complete with proofs, liars? Why do you maintain that anyone who shows how wrong you are must be some evil, dishonest person working against mathematics? These are not the actions of someone who is truly talking about the math. Instead, they are the pathetic workings of an inferior mind that is intent on establishing its own preeminence over the world. > Because of that, there is no limitation from time or space, no way to > stop it, no way to control it, kill it, or in any way for any of you > to affect it at all. > However, it is consistent. One hopes, in fact assumes, that mathematics is consistent. That this assumption is well-founded is reinforced by the absence of apparent contradiction, after centuries of hard work by thousands upon thousands of individual careers. However, no one can *know* that mathematics is consistent. > Think you understand Goedel if you will, but I *do* understand his > work, and I understand what many of you do not, which is that it > *proves* the finiteness of our world and your inherent (as well as my) > limitations. I doubt that you understand G.9adel's work. You've probably read some popularizations (Hofstadter's book, for instance), but I suspect that you lack both the background and the mental flexibility to comprehend his proofs. Why would I say that? Well, for one, you mention Goedel as though that encompasses something unique. There is a body of work including his completeness theorem, his incompleteness theorem, and some consistency (of CH) and independence (of AC and GCH) results. Another reason you can be seen not to understand G.9adel's work is that you assert the consistency of mathematics. Virtually all of us believe that mathematics is consistent, but it is simply impossible to *know* this, since it is (by G.9adel) impossible to prove. > We are mortal. But we are not alone. That also is unknowable. Unless you're talking about your pet goldfish. > So when I tell you that it is important to accept the math, I'm not > talking mysticism, or giving vague threats or maybe's as it's not > about dying--you're going to die anyway--it's about living a better > life till you inevitably die. > And make no mistake, it's *provable* that immortality, as many of you > unfortunately conceive it, for us is impossible, and you will all > eventually die, as will I. Oh, it's *provable*. It's surely demonstrable, inducing from all of history, but *provable*? > There are immortal beings--no, not so simple as calling them gods--but > we are not them. Your claim of immortal beings is yet another unsupported assertion. For one thing, time still goes by. What hasn't happened yet (death of any purported immortal) may yet occur. > I want to live a better life, and I want as many as you as possible to > live a better life, but the Universe does not work by coddling > failures. There has been an error in the core of mathematics for > over a hundred years, which has steadily been working against the > future of humanity. Did you ever notice that when you run into a major gaffe of yours, in this case the claim that a non-unit algebraic integer can be coprime to a non-unit multiple of itself, you immediately run off to some irrelevant track, whether it's the threat of legal or military action, the Powers that Be rant, or this pseudo cosmic truth crap? > Look around you. If you wish, you can be given *more* evidence that > reality works by consistency and mathematics. That evidence will > involve a significant amount of human misery above and beyond what has > been seen up until now. Right. A significant amount of human misery above and beyond what has been seen up until now. > It's not punishment. It's a consequence of the Laws of Physics. Oh, good. I was afraid you were going to punish all of humankind. However, why settle on such unsupported nonsense, instead of deriving this significant amount of human misery from the Laws of Physics? > If you cannot accept the math, then you are failures. > It's that simple. If you are incapable of accepting the math, then > the Universe, as She is perfect, will simply allow situations beyond > your ability to handle. Sorry, situations are allowed by the Universe all the time. Our understanding or acceptance of the math are irrelevant to that. Why not indicate *what* the acceptance of your bogus math would give us to handle these unspecified situations? > So a Bush thinks there are weapons of mass destruction in Iraq. All > the evidence necessary to realize there aren't at the level he claims > are there, and I even warn him, but does he listen? That guy has a lot of nerve, ignoring the likes of you. I think you should have taken my advice and visited the FBI with your sheaf of notes and printouts. BTW, I thought you were planning on *voting* for this guy, when the Presidential elections were coming around next time. > No. It's failures like that which will doom you and yours. You must > be able to reason at a level which fits circumstances. I suppose that's why you refuse to learn any more algebra than the amount you already think you know. After all, learning math must destroy your reasoning level. Kind of like how weight training makes it harder for a person to handle heavy physical effort. > You must learn to THINK at a level which will test you day and night. > You must learn to push your minds to the limits as if your life > depends on it, as the lives of your children, the future of humanity, > do. Oh, THIS must be why you refuse to learn any more mathematics. > It's not a game. The tests are over and humanity failed. Now the > fight is for the survival of humanity's children. Some of you must, > will learn to be more than human. You see? Pseudo-cosmic claptrap. This is more properly fodder for that bozoidal Mega discussion area than here. > James Harris Dale. === Subject: Re: JSH: Ok, so now maybe you know > Think you understand Goedel if you will, but I *do* understand his > work, and I understand what many of you do not, which is that it > *proves* the finiteness of our world and your inherent (as well as my) > limitations. How does it prove this? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: logarithmic regression >I'm looking for a method to find variables to minimize the summed >squared error in the following: v(x) = a + b exp (-c * x). I'm >familiar with a method when a = 0, but in this data, that's not the >general case. Any help is appreciated. > It becomes a problem in nonlinear regression, and you have to use > iterative methods. What math software do you have available to you? > - Randy Randy, Unfortunately, I need to be able to integrate this regression into a larger program. We're dealing with many subjects and lots of data, so this needs to be automated to a degree. It's part of a larger analysis, so many different regressions need to be run on the same set before the answer finally falls out. I've thought and heard it a few times that educated guessing at a, subtracting that from the data points, and doing regression on b exp (-c * x), which can be converted into a linear problem, might not be too bad an idea. Everything is well-behaved and the values will always fall within certain ranges, so I don't see this as too difficult. Aside from the above, the only iterative method I know is a genetic algorithm that, while interesting, really isn't suited for this problem, it seems. Do you have any algorithm names that I could research? Justin === Subject: Re: logarithmic regression > >I'm looking for a method to find variables to minimize the summed >squared error in the following: v(x) = a + b exp (-c * x). I'm >familiar with a method when a = 0, but in this data, that's not the >general case. Any help is appreciated. > > It becomes a problem in nonlinear regression, and you have to use > iterative methods. What math software do you have available to you? > > - Randy > Randy, > Unfortunately, I need to be able to integrate this regression into a > larger program. We're dealing with many subjects and lots of data, so > this needs to be automated to a degree. It's part of a larger > analysis, so many different regressions need to be run on the same set > before the answer finally falls out. > I've thought and heard it a few times that educated guessing at a, > subtracting that from the data points, and doing regression on b exp > (-c * x), which can be converted into a linear problem, might not be > too bad an idea. Everything is well-behaved and the values will always > fall within certain ranges, so I don't see this as too difficult. I was going to suggest this as a manual method. But this can also be used to get a good starting guess for a full iterative method. > Aside from the above, the only iterative method I know is a genetic > algorithm that, while interesting, really isn't suited for this > problem, it seems. Do you have any algorithm names that I could > research? Descent methods, Nonlinear least squares. The simplest approach would be Steepest descent which also goes by the name Hill-climbing (when the descent is turned upside down) and Gradient descent. Actually, your function is simple enough that you can probably work out the necessary derivatives analytically (in general purpose software it's done numerically). Error function you're trying to minimize: E(a,b,c) = (1/2)*sum(over data) [y - a - b*exp(c x)]^2 = (1/2)*sum [y - v(x)]^2 The 1/2 serves no purpose except to eliminate some irritating 2's in the derivatives. Gradient (in terms of a, b, c): dE/da = -sum(over x) (y - v(x)) dE/db = -sum(over x) (y - v(x))*exp(c*x) dE/dc = -sum(over x) (y - v(x))*b*x*exp(c*x) The basic gradient descent method is: Evaluate the gradient, and do some sort of line search along the direction -grad(a,b,c) to find a good next step. That is, search for some small alpha such that (a,b,c) - alpha*(dE/da, dE/db, dE/dc) is sufficiently downhill. There are many many variants on this. If you don't want to worry about line search you can just choose a small fixed alpha that gives you convergence for your data (try alpha = 0.5 and work your way down), but usually you'll end up with a very small fixed alpha and then the procedure might take hundreds of iterations. Also, you need a stopping criterion. You're at a minimum when your gradients are zero. You'll never get there exactly, so you can just check for when all the terms are less than some modest value (depends on your data). ----------- Or... you can look for shareware software that implements all the sophistication the nonlinear descent community can throw at it. Netlib is a huge source for such stuff. Check this out if you want to go that route: http://plato.la.asu.edu/topics/problems/nlounres.html (I'd go with the gradient of f is available or finite difference approximations are applicable). - Randy === Subject: Re: Folk Psychology and Social Convention >> http://www.behavior.org/journals_BP/2000/Place.pdf >> There is quite a large literature on social conformity which could be >> referred to, but whilst of some heuristic value (in the helpful, >> illustrative sense) I really find that social psychological literature >> relatively trivial compared to the rather detailed and powerful work >> which has been done for decades in what I rather loosely refer to in >> this newsgroup as behaviour analysis and learning theory. >find that very conforming to hear again. >Longley Citation Statistics [8 long years and counting] >------------------------------------------------------- >folk psychology [derogatorily] in over 1000 posts >Quine [glowingly] in over 1800 posts >silly [regarding others' ideas] over 400 times >Skinner [in awe] in over 800 posts >behaviorism [as accepted dogma] in over 500 posts >[pernicious] mentalism over 100 times >[pernicious] cognitivism over 250 times >intensional [as wrongish] over 1600 times >extensional [as correctish] over 1850 times >frag.html [self-promotion] over 300 times Yet despite this, I bet we see yet more of your asinine misrepresentations of popular neuroscience in posts to comp.ai.philosophy. If you actually read some of the material you irreverently *count* and disparage above, you might learn why your posts amount to little more than science fiction and why they have next to nothing to do with the philosophy of AI. The above terms are exactly those which should occur in a forum which is specifically for the discussion of the philosophy of AI, something you'd discover for yourself if you actually read and understood any of the relevant literature. -- David Longley === Subject: Re: Folk Psychology and Social Convention >> http://www.behavior.org/journals_BP/2000/Place.pdf >> >> There is quite a large literature on social conformity which could be >> referred to, but whilst of some heuristic value (in the helpful, >> illustrative sense) I really find that social psychological literature >> relatively trivial compared to the rather detailed and powerful work >> which has been done for decades in what I rather loosely refer to in >> this newsgroup as behaviour analysis and learning theory. >find that very conforming to hear again. >Longley Citation Statistics [8 long years and counting] >------------------------------------------------------- >folk psychology [derogatorily] in over 1000 posts >Quine [glowingly] in over 1800 posts >silly [regarding others' ideas] over 400 times >Skinner [in awe] in over 800 posts >behaviorism [as accepted dogma] in over 500 posts >[pernicious] mentalism over 100 times >[pernicious] cognitivism over 250 times >intensional [as wrongish] over 1600 times >extensional [as correctish] over 1850 times >frag.html [self-promotion] over 300 times > Yet despite this, I bet we see yet more of your asinine > misrepresentations of popular neuroscience in posts to > comp.ai.philosophy. If you actually read some of the material you > irreverently *count* and disparage above, you might learn why your posts > amount to little more than science fiction and why they have next to > nothing to do with the philosophy of AI. > The above terms are exactly those which should occur in a forum which is > specifically for the discussion of the philosophy of AI, something you'd > discover for yourself if you actually read and understood any of the > relevant literature. It's very clear you are stuck in a loop. Why not learn something new and interesting: http://webvision.med.utah.edu/VisualCortex.html#pathways http://white.stanford.edu/~heeger/psych202/lecture-notes/ visual-cortex/visual-cortex.html === Subject: Re: Folk Psychology and Social Convention > http://www.behavior.org/journals_BP/2000/Place.pdf > > There is quite a large literature on social conformity which could be > referred to, but whilst of some heuristic value (in the helpful, > illustrative sense) I really find that social psychological literature > relatively trivial compared to the rather detailed and powerful work > which has been done for decades in what I rather loosely refer to in > this newsgroup as behaviour analysis and learning theory. >> >> >>find that very conforming to hear again. >> >>Longley Citation Statistics [8 long years and counting] >>------------------------------------------------------- >>folk psychology [derogatorily] in over 1000 posts >>Quine [glowingly] in over 1800 posts >>silly [regarding others' ideas] over 400 times >>Skinner [in awe] in over 800 posts >>behaviorism [as accepted dogma] in over 500 posts >>[pernicious] mentalism over 100 times >>[pernicious] cognitivism over 250 times >>intensional [as wrongish] over 1600 times >>extensional [as correctish] over 1850 times >>frag.html [self-promotion] over 300 times >> Yet despite this, I bet we see yet more of your asinine >> misrepresentations of popular neuroscience in posts to >> comp.ai.philosophy. If you actually read some of the material you >> irreverently *count* and disparage above, you might learn why your posts >> amount to little more than science fiction and why they have next to >> nothing to do with the philosophy of AI. >> The above terms are exactly those which should occur in a forum which is >> specifically for the discussion of the philosophy of AI, something you'd >> discover for yourself if you actually read and understood any of the >> relevant literature. >It's very clear you are stuck in a loop. Why not learn something new >and interesting: >http://webvision.med.utah.edu/VisualCortex.html#pathways >http://white.stanford.edu/~heeger/psych202/lecture-notes/ >visual-cortex/visual-cortex.html And the reason you think I need to learn about visual neuroscience is? (I thought I'd learned a fair bit of that back at NIMR years ago but hey, what do I know about neuroscience eh?) And the relevance of visual functional-neuroanatomy/neuroscience to the *philosophy* of AI is? Can you please tell us why these frequent references to empirical work in the neuroscience of vision are appropriate posts to a newsgroup which is dedicated to the philosophy of AI - and why you aren't posting it to comp.ai or comp.ai.vision instead? Would the functional-neuroanatomy/neuroscience of the monoamine systems, or neuropeptides or the cerebellum, the hypothalamus, amygdala or basal ganglia be as relevant - if so how? if not why not? Hell, why not make it bionet.neuroscience.gee.whizz.ai.somehow and be damned? I've tried to explain what philosophy *is* these days and what might well be worth discussing as the philosophy of AI - it even fits in with the books that I have on the philosophy of AI oddly enough - but .. what do I know eh? The last time *I* looked, folk psychology was all about the attribution of properties (intensional) such as intelligence.... but, it seems, that, like intensions more generally -(something one sees for example in the contemporary work of McCarthy) this is not something which *you* and Larry etc are interested in except to count how the words have been used by me - especially if I say folk psychology is not the stuff one would want in AI anyway! - but hey - you have the 1st Amendment over there in the USA don't you....? Bill of Rights and all that, and it would seem that some of you believe that means you can decide for yourselves that philosophy of AI is about making cottage-cheese if you want - As Eray says - who am *I* to tell you otherwise??? Over the past week, some of these threads have been cross posted to talk.bizarre and other inappropriate groups. I tried to claw some of the original purpose back by referring to the logical structure of the posts themselves given the heat - what I was referring to is something which is central to the management of lots of human behaviour, but the point has probably gone unnoticed because several people in this newsgroup seem to think that psychology is irrelevant to AI. I find this absolutely bizarre. Beginning with said that as canonical, I have used other examples of intensional contexts to point out the way that intensions actually get in the way of what we call intelligent behaviour, yet this seems to have been lost on just about everyone who has responded. The reason? Lack of insight? Intelligence? Failure of connection (failure of Leibniz's Law in intensional contexts) - who cares eh? Hey, guess what I saw in http://www.newandexcitingideas.com/today.htm thsimorning! (follow-ups to comp.ai.philosophy) -- David Longley === Subject: Re: Folk Psychology and Social Convention >> http://www.behavior.org/journals_BP/2000/Place.pdf >> >> There is quite a large literature on social conformity which could be >> referred to, but whilst of some heuristic value (in the helpful, >> illustrative sense) I really find that social psychological literature >> relatively trivial compared to the rather detailed and powerful work >> which has been done for decades in what I rather loosely refer to in >> this newsgroup as behaviour analysis and learning theory. > > >find that very conforming to hear again. > >Longley Citation Statistics [8 long years and counting] >------------------------------------------------------- >folk psychology [derogatorily] in over 1000 posts >Quine [glowingly] in over 1800 posts >silly [regarding others' ideas] over 400 times >Skinner [in awe] in over 800 posts >behaviorism [as accepted dogma] in over 500 posts >[pernicious] mentalism over 100 times >[pernicious] cognitivism over 250 times >intensional [as wrongish] over 1600 times >extensional [as correctish] over 1850 times >frag.html [self-promotion] over 300 times > Yet despite this, I bet we see yet more of your asinine > misrepresentations of popular neuroscience in posts to > comp.ai.philosophy. If you actually read some of the material you > irreverently *count* and disparage above, you might learn why your posts > amount to little more than science fiction and why they have next to > nothing to do with the philosophy of AI. > The above terms are exactly those which should occur in a forum which is > specifically for the discussion of the philosophy of AI, something you'd > discover for yourself if you actually read and understood any of the > relevant literature. Michaels in replying to my initial post to comp.ai.philosophy, crossposted this thread comp.ai.genetic, sci.math, talk.bizarre. I presume, in order to foment further confusion and mayhem (as these other readers, having no idea of the relevant history, might indeed think these exchanges were bizarre). But that is precisely the sort of out of context rhetoric which Michaels has been charged with in other guises - even in his own reading, thinking and posting. This behaviour has been highlighted for him in terms of the canonical form which intensional contexts take - said that (much as one might in an undergraduate philosophy or psychology tutorial) - but he doesn't understand the nature of the criticism, any more than Larry, or Eray does. The reason? Lack of training in the relevant fields - it's as simple as that. And who am *I* or anyone else to tell these guys *anything*? (30 years of experience ignored for what? - Sadly, like many people in my field, I've seen this pattern before students usually, who come back years later and apologise for their behaviour and say how they wish they'd seen it at the time so they could have made better use of their time) I think Michael's response to what I post shows no understanding of what is important, or relevant. I post here assuming that people have a reasonably good background in the history of AI and philosophy. Sadly. it's just a fact that this is not the case, which means those who have can make *no* assumptions whatsoever - something which will force dilution of anything said to such an extent that nothing much of interest can be said in a short post. So long as it *is* the case (and it is sadly missing in the case of most of the people who write silly things about what I post) I reckon those *with* such a background, will not post here - it's not worth the effort. I posted a link to a paper by U T Place, perhaps the father of modern mind-brain identify theory. I chose it because it's an interesting illustration of how social conformity operates in folk-psychology. Whilst Place has always been a behaviourist in the Skinnerian tradition. I recall monopolizing tutorials with him in the late 1970s when he took my year for Philosophy of Psychology. Ironically[#1], at that time I was arguing that there were important compatibilities between Husserlian phenomenology (the science of intension if ever there was one - cf. Quine, Word and Object 1960) and radical behaviourism, but that this was being lost by misrepresentation of phenomenology and behaviourism at the time. In my view, this is still widely the case, and if people can get the latter so wrong it is not surprising that the former is almost totally misunderstood! The reason for starting the thread was not only to highlight the dynamics of social conformity (and what Skinner had to deal with when he took his work out of the laboratory context), nor just to highlight what might be going on when I am regarded as challenging folk psychology (which of course I'm not), but primarily to bring out the extent to which the extensional and intensional distinction has figured not only in 20th century psychology, but also in everyday life, and how failure to fully appreciate this is likely to make any assessment of the nature or viability of AI extremely difficult to assess. My last 30 odd years have been spent getting as broad and as detailed a knowledge as I could in psychology, neuroscience and computer science and applied work, and what I've been trying to do in this newsgroup is pull some of that together for others. Intelligence as a folk psychological attribution refers to something that people, at least, do. It's an attribute which is made on the basis of perceived individual differences, and has, along with many of these notions, been made the subject of analysis by empirical psychology since at least the early decades of this century. In one form this has been the study of learning, in another, social judgement, and in a more controversial sense still, it has been studied as I.Q, refined from the construction of tests of individual differences which explicitly conform to the normal distribution. When these have been constructed as culture fair tests which purport to be free of biases which might be attributed to linguistic disadvantages or cultural knowledge, they reduce to tests of symbolic manipulation, the operations of which are specifically based on the logical rules of the Propositional and First Order Predicate Calculus. Most, by definition do average on such tests, a small proportion of the population do either exceptionally or poorly (again, by design and definition), and the test is designed specifically to provide this range of discrimination as the tests were largely used as selection and sorting instruments in the applied professions. (The fact that people misapply this concept to other situations is related to the overall issue I am discussing here, but not central, so let that lie for now). What's important to appreciate is that, leaving aside the mysteries of Factor Analysis, such operations, are, *fundamentally* measures of behaviours (albeit often as data reductions), and the logic of those skills or behaviours is entirely extensional. We spend many years and great resources, trying to learn and shape up these skills in ourselves and our young, and we construct extensive and complex contingencies to maintain such behaviours, often with only a vague knowledge of the fact that there is another side to this behaviour which requires such contingencies to be put into place and managed. Some of those contingencies are precisely the sorts of forces which Place talks about in the link provided above. It is these dynamics which are widely studied by psychologists working in the field of Attribution Theory (called other names over the decades) and in other areas it is called Learning and Conditioning when this work is done in more restricted paradigms. The *names* are not what is important - although *naming* and *reference* are key notions in what I have been talking about in these threads and in Fragments of course.. The point to be taken from the above is that our *language* is both intensional and extensional, in different places. We know that the intensional contexts are problematic, and especially when the reference is intelligent behaviour. There is indeed room for work into the intensional, it is a domain which has much to do with the generation of novel behaviours, but whilst *creativity*, and many notions like this would appear to be something to do with *intelligence*, we have ample evidence that this is NOT the case at all, and whilst it may be stimulating and interesting to read about contexts of discovery [#2] I have been trying to get users of this newsgroup to see that these contexts are NOT ultimately relevant to the contexts of justification and intelligence. If anything, they are anathema, and we need to understand the relationship between the two primarily so we can better focus resources on the extensional stance. Getting this *clear* is what I think the philosophy of AI is all about. #1 Ironic because my undergraduate years were largely spent looking at modern empirical psychology from the perspective of Husserlian phenomenology - a perspective which (though perhaps with a touch of Heidegger) is in many circles seen to be (unfortunately in my view) as an influential force behind modern cognitivism. #2 Much of what one reads in Cognitive Science or its variants where *cognition* is emphasised *instead of* behaviour - e.g. in much of Cognitive Psychology, Computational Neuroscience, Cognitive Neuroscience, cognitive-neuro-endrocrinological-computational-gobbledegook-sc ience etc is not in fact science at all, it's good literature and rhetoric as metaphor has run amok - this material is liked for its literary appeal not for its scientific value) - what's more, many people who have worked in these fields would probably acknowledge this if pressed, the authors of popular books almost invariably admit this to colleagues. They're just meeting a market demand which has arisen over the past 20 years which is quite distinct from those of science per se. Is that to be the fate of this newsgroup? It's certainly going that way if it hasn't got there already. -- David Longley === Subject: Re: Folk Psychology and Social Convention >>Longley Citation Statistics [8 long years and counting] >>------------------------------------------------------- >>folk psychology [derogatorily] in over 1000 posts >>Quine [glowingly] in over 1800 posts >>silly [regarding others' ideas] over 400 times >>Skinner [in awe] in over 800 posts >>behaviorism [as accepted dogma] in over 500 posts >>[pernicious] mentalism over 100 times >>[pernicious] cognitivism over 250 times >>intensional [as wrongish] over 1600 times >>extensional [as correctish] over 1850 times >>frag.html [self-promotion] over 300 times >> >It's very clear you are stuck in a loop. Why not learn something new >and interesting: >http://webvision.med.utah.edu/VisualCortex.html#pathways >http://white.stanford.edu/~heeger/psych202/lecture-notes/ >visual-cortex/visual-cortex.html > And the reason you think I need to learn about visual neuroscience is? > (I thought I'd learned a fair bit of that back at NIMR years ago but > hey, what do I know about neuroscience eh?) > And the relevance of visual functional-neuroanatomy/neuroscience to the > *philosophy* of AI is? The peephole you view your world through is very narrow, Longley. There are many here who think the following matters are the crux of building AI's: How do you recognize what you see? How do you know how to move your arm? How do you choose which words to say? How do you understand what they mean? How does commonsense reasoning work? ............. Every such function is known to engage dozens or hundreds of parts of the brain, each specialized to do various jobs so we cannot expect to explain them all on the basis of only a few simple laws. We still know very little about just what each of those separate brain-parts do and still less about how they cooperate. What is thinking and how does it work? What decides upon your next state of mind? What determines which ideas you'll think about, and which processes you'll apply to them? What chooses which decisions you'll make, and what budgets how much time these will take? ... what sorts of processes do we possess, how might each of those processes work, and how might all of them be combined, to produce the constructions that we call our minds? .............. ... suppose you were given the job of constructing an artificial animal. You could start by making a list of things that your animal-robot might need to do. It might need to find sources of water and food. It might need defenses against attacks, and against extremes of temperature. Then once you have assembled that list, you could start to instruct your engineers to find separate ways to meet each of those needs. Any other questions, Longley? === Subject: Re: Folk Psychology and Social Convention DM: How do you recognize what you see? How do you know how to move your arm? How do you choose which words to say? How do you understand what they mean? How does commonsense reasoning work? GS: And some people think the crux is: What is recognition? What is knowledge? What does it mean to choose? What does it mean to understand? What is commonsense reasoning? DM: Every such function[...] GS: Which functions? DM: is known to engage dozens or hundreds of parts of the brain, each specialized to do various jobs so we cannot expect to explain them all on the basis of only a few simple laws. We still know very little about just what each of those separate brain-parts do and still less about how they cooperate. GS: You don't even know what you are trying to explain. DM: What is thinking and how does it work? GS: What are the options? And what are the assumptions that underlie this list? DM: What decides upon your next state of mind? GS: What is a decision? What is mind? Despite the gratuitous mentalism, a potent answer is: natural selection, cultural selection, ontogenic forces such as operant selection (conditioning) and the current environment. At least, these are the loci of the important selection activities that result in the behavior said to require mentalistic entities. The effects of these histories are mediated by genes and the nervous system but it is not at all clear how. And this is partly due to the way that many behavioral neurobiologists talk about behavior. A good example of such misguided concepts are those provided by everything that Dan has said so far. DM: What determines which ideas you'll think about, and which processes you'll apply to them? What chooses which decisionsyou'll make, and what budgets how much time these will take? GS: Despite the gratuitous mentalism, a potent answer is: natural selection, cultural selection, ontogenic forces such as operant selection (conditioning) and the current environment. At least, these are the loci of the important selection activities that result in the behavior said to require mentalistic entities. The effects of these histories are mediated by genes and the nervous system but it is not at all clear how. And this is partly due to the way that many behavioral neurobiologists talk about behavior. A good example of such misguided concepts are those provided by everything that Dan has said so far. DM: ... what sorts of processes do we possess, GS: Reflexes, habituation, sensitization, classical conditioning and the processes known as generalization, discrimination, extinction, operant conditioning and the processes of operant stimulus control, extinction etc. DM: how might each of those processes work, and how might all of them be combined, to produce the constructions that we call our minds? GS: And also, what cultural forces are at work such that we come to talk about minds? .............. DM: ... suppose you were given the job of constructing an artificial animal. GS: OK. DM: You could start by making a list of things that your animal-robot might need to do. GS: Not exactly. Plus, this ignores the difficult issue of how different sorts of behavior are to be classified. Is opening the door with the left hand and right hand the same 'thing?' Is talking aloud and silently the same thing? Is a prairie dog's warning cry the same as my lookout Bill! and by what criteria does one make these classifications? As long as one cannot answer these questions, it is extraordinarily unlikely that any amount of looking at the brain, or making guesses about it, will come close to a description of how genes and the nervous system mediate the effects of natural selection and ontogenic variables including conditioning history, the action of the current environment, and other variables like deprivation and satiation. DM: It might need to find sources of water and food. It might need defenses against attacks, and against extremes of temperature. Then once you have assembled that list, you couldstart to instruct your engineers to find separate ways to meet each of those needs. GS: If one cannot provide some sort of classificatory scheme relevant to the phenomena that one wants to explain in a mechanistic fashion, then one cannot possibly begin to say how many sorts of mechanisms are required. What Michaels has written does not address these issues and implies that one may merely look at behavior and say what kind it is. It suggests a profound naivety with respect to the subject matter for which he wants to provide a mechanistic explanation. Unfortunately, Michaels is, in my opinion, not much less sophisticated than most psychologists who are also blissfully unaware of the critical nature of these issues and their relation to what comprises the core concepts of their alleged science. DM: Any other questions, Longley? GS: I have some: What is recognition? What is knowledge? What does it mean to choose? What does it mean to understand? What is commonsense reasoning? What is behavior? What kinds of behavior are there? What are the criteria by which we classify kinds of behavior? And there are many, many others that must be answered in more than a provisional way in order to begin to answer the mechanistic questions in which Michaels is interested, and in which he claims to be a knowledgeable amateur. As I implied before, however, few psychologists would disagree with how Michaels talks about behavior, and this is the very reason why, in my opinion, behavioral neuroscience has been largely unable to provide neurobiological, mechanistic descriptions of even the most rudimentary behavioral processes that are responsible for the phenomena said to require a mind. === Subject: Re: Folk Psychology and Social Convention >How do you recognize what you see? >How do you know how to move your arm? >How do you choose which words to say? >How do you understand what they mean? >How does commonsense reasoning work? How do we keep wankfests like this out of sci.math? Lee by repeatedly setting followups, I suppose Rudolph === Subject: A test proposal for Herc (was Re: To prove psychic powers, someone has to sit this test!) > Look, I appreciate your time, but there's no need to go agro over it. I assure you I am not an agressive poster, but, as you have done, I WILL stress some words with uppercase or ** to draw your attention to the point I am trying to make to you. > you' ve got some good points, and a host of errors, I'll go through each . I await your corrections with bated breath. > This is what I want you to consider..... > IF I did have a paranormal ability that was evident in peoples replies to me, > and you must agree that is not impossible, although not possible as we know it, > however unlikely, then HOW would I demonstrate it? its seems if it doesn't work > beyond 100 to 1 chance every go everyone is lost. You have to > try to see the paranormal, don't kill the messenger boy. I generally agree with your stated approach Herc, but as we saw in my last post discussing your example of 50 posts where 6 *might* be able to guessed and the odds table you quoted. 6 to 7 posts that might be *possible* to guess do not demonstrate anything like a significant occurance better than chance. You already agreed that in that scenario you'd need to guess *correctly* a minimum of 26 of the 50 posts for the phenomenon to be significant. Since you have not been able to demonstrate anything like a statistical significance in your postings over the last couple of years on this subject there is little point in me trying to see the paranormal As yet you have not presented any data that > If I'm doing everything wrong then *how* would I show it if I did have powers? > It seems I would get the same response if I was putting it on to if I had something > unique to show? > Just think about it, nearly all the time people's names give away what they are writing > to me, completely unique for me, how would I show it? Not, as Wally has correctly pointed out, on Usenet. Perhaps to a private discussion forum set up on an independent person's website. I have a vague memory of someone suggesting this to you once. I would be a simple matter of you posting to (the equivalent of) a website's guestbook explicitly set up for the purpose and announcing such on the newsgroups. Once your message was entered, the site owner then would block you from returning to that site until 10 people had responded. He could then reply to your usenet announcement with 1. the message and 2. the list of respondents. You would then have an agreed amount of time to peform your magic. The website owner would not in any correspondence confirm hits or misses until the end of the trial. According to your table of chances you would only need to get 7 or better out of 10 individual posts to show that anything significant was happening I would be happy to see that set up as at least a *preliminary* trial to see if there was anything to your claim, and then tighten up the testing protocols for a more rigourous test if indeed there was anything worth testing. I'll respond to the rest and leave this renamed thread for discussion of a test protocol that you might agree to - or not. -- Eric Hocking === Subject: Re: A test proposal for Herc (was Re: To prove psychic powers, someone has to sit this test!) > Look, I appreciate your time, but there's no need to go agro over it. > I assure you I am not an agressive poster, but, as you have done, I > WILL stress some words with uppercase or ** to draw your attention to > the point I am trying to make to you. > you' ve got some good points, and a host of errors, I'll go through each . > I await your corrections with bated breath. > This is what I want you to consider..... > IF I did have a paranormal ability that was evident in peoples replies to me, > and you must agree that is not impossible, although not possible as we know it, > however unlikely, then HOW would I demonstrate it? its seems if it doesn't work > beyond 100 to 1 chance every go everyone is lost. You have to > try to see the paranormal, don't kill the messenger boy. > I generally agree with your stated approach Herc, but as we saw in my > last post discussing your example of 50 posts where 6 *might* be able > to guessed and the odds table you quoted. 6 to 7 posts that might be > *possible* to guess do not demonstrate anything like a significant > occurance better than chance. You already agreed that in that > scenario you'd need to guess *correctly* a minimum of 26 of the 50 > posts for the phenomenon to be significant. > Since you have not been able to demonstrate anything like a > statistical significance in your postings over the last couple of > years on this subject there is little point in me trying to see the > paranormal As yet you have not presented any data that I presented 3 posts, you haven't rated them AT ALL! > If I'm doing everything wrong then *how* would I show it if I did have powers? > It seems I would get the same response if I was putting it on to if I had something > unique to show? > Just think about it, nearly all the time people's names give away what they are writing > to me, completely unique for me, how would I show it? > Not, as Wally has correctly pointed out, on Usenet. Perhaps to a > private discussion forum set up on an independent person's website. I > have a vague memory of someone suggesting this to you once. > I would be a simple matter of you posting to (the equivalent of) a > website's guestbook explicitly set up for the purpose and announcing > such on the newsgroups. > Once your message was entered, the site owner then would block you > from returning to that site until 10 people had responded. > He could then reply to your usenet announcement with > 1. the message and > 2. the list of respondents. > You would then have an agreed amount of time to peform your magic. The > website owner would not in any correspondence confirm hits or misses > until the end of the trial. > According to your table of chances you would only need to get 7 or > better out of 10 individual posts to show that anything significant > was happening > I would be happy to see that set up as at least a *preliminary* trial > to see if there was anything to your claim, and then tighten up the > testing protocols for a more rigourous test if indeed there was > anything worth testing. possible but unnessesary#$%?. When Jesus came up to the city and said come to the ocean to watch me walk on water, they didn't start building a swimming pool. The post works best in basic newsgroups, skeptics had a lot of good themes, knives, gardens, maths the matches are highly technical, in art bell hardly noticable. There's no trick to getting a score, anyone can do it. The paranormal is recorded, you don't have to hide it, the scoring is trivial afterwards, you just have to conceil the judges from the data. Usenet is ideal, forums work, even the avatars follow the theme of the message, but I doubt anyone owns a large enough forum who would moderate out a heap of posts. put it this way, you're confusing subjectiveness with identification. when I roll 3 dice, the 1st 2 add up to the 3rd. I can do it in front of you, and say look 2 + 3 = 5. Then you set up a test, I roll the dice blind folded, you tell me the first 2, 2 and 3, then I say 5! Good test, but can you think of a simpler one? Herc === Subject: Re: A test proposal for Herc (was Re: To prove psychic powers, someone has to sit this test!) > Look, I appreciate your time, but there's no need to go agro over it. > I assure you I am not an agressive poster, but, as you have done, I > WILL stress some words with uppercase or ** to draw your attention to > the point I am trying to make to you. > you' ve got some good points, and a host of errors, I'll go through each . > I await your corrections with bated breath. > This is what I want you to consider..... > IF I did have a paranormal ability that was evident in peoples replies to me, > and you must agree that is not impossible, although not possible as we know it, > however unlikely, then HOW would I demonstrate it? its seems if it doesn't work > beyond 100 to 1 chance every go everyone is lost. You have to > try to see the paranormal, don't kill the messenger boy. > I generally agree with your stated approach Herc, but as we saw in my > last post discussing your example of 50 posts where 6 *might* be able > to guessed and the odds table you quoted. 6 to 7 posts that might be > *possible* to guess do not demonstrate anything like a significant > occurance better than chance. You already agreed that in that > scenario you'd need to guess *correctly* a minimum of 26 of the 50 > posts for the phenomenon to be significant. > Since you have not been able to demonstrate anything like a > statistical significance in your postings over the last couple of > years on this subject there is little point in me trying to see the > paranormal As yet you have not presented any data that > I presented 3 posts, you haven't rated them AT ALL! Which is merely more proof that you've got a power that I do not possess. Let's try the controlled trial I've proposed, to see how good you ARE? Are you willing? > If I'm doing everything wrong then *how* would I show it if I did have powers? > It seems I would get the same response if I was putting it on to if I had something > unique to show? > > Just think about it, nearly all the time people's names give away what they are writing > to me, completely unique for me, how would I show it? > Not, as Wally has correctly pointed out, on Usenet. Perhaps to a > private discussion forum set up on an independent person's website. I > have a vague memory of someone suggesting this to you once. > I would be a simple matter of you posting to (the equivalent of) a > website's guestbook explicitly set up for the purpose and announcing > such on the newsgroups. > Once your message was entered, the site owner then would block you > from returning to that site until 10 people had responded. > He could then reply to your usenet announcement with > 1. the message and > 2. the list of respondents. > You would then have an agreed amount of time to peform your magic. The > website owner would not in any correspondence confirm hits or misses > until the end of the trial. > According to your table of chances you would only need to get 7 or > better out of 10 individual posts to show that anything significant > was happening > I would be happy to see that set up as at least a *preliminary* trial > to see if there was anything to your claim, and then tighten up the > testing protocols for a more rigourous test if indeed there was > anything worth testing. > possible but unnessesary#$%?. You post to sci.skeptic moaning that we wont' take you seriously, and when a test protocol is proposed you say a test is not necessary? What do you think the JREF Challenge is? It's a trial of your claim. > When Jesus came up to the city and said come > to the ocean to watch me walk on water, they didn't start building a swimming pool. It *would* have been unnessesary as he was addressing a bunch of blokes sitting in boats fishing at the time if I remember this particular anectdote correctly. > The post works best in basic newsgroups, skeptics had a lot of good themes, > knives, gardens, maths the matches are highly technical, in art bell hardly > noticable. Your missing a crucial point that has been made to you continually. It is *possible* for responses on usenet to be viewed before you preform your trial. I'm proposing a trial where you, or anyone else who wishes to attempt this, will not be able to have foreknowledge of the poster's identity. > There's no trick to getting a score, anyone can do it. The paranormal is recorded, > you don't have to hide it, the scoring is trivial afterwards, you just have to > conceil the judges from the data. No, we have to conceal the *testers* from the data. The judging will be self-evident. You either achieve better than chance, or, if you prefer, you achieve better than any of the other's that make the attempts. > Usenet is ideal, forums work, even the avatars follow the theme of the message, > but I doubt anyone owns a large enough forum who would moderate out a heap > of posts. Usenet is not ideal. A private, moderated forum may work though. The deal would be for you to post and then be blocked from the content of the forum until a sufficient number of responses were gathered. THEN, these responses would be shown to you with the author's name replaced by a choice of names for you to select from. This removes the suspicion that you *could* have used a search engine to glean the author's name prior to the trial beginning. > put it this way, you're confusing subjectiveness with identification. > when I roll 3 dice, the 1st 2 add up to the 3rd. > I can do it in front of you, and say look 2 + 3 = 5. > Then you set up a test, I roll the dice blind folded, you tell me > the first 2, 2 and 3, then I say 5! Good test, but can you think > of a simpler one? I'd be impressed if this worked. What is your claimed success rate for this feat? -- Eric Hocking === Subject: Re: A test proposal for Herc (was Re: To prove psychic powers, someone has to sit this test!) >Not, as Wally has correctly pointed out, on Usenet. Perhaps to a >private discussion forum set up on an independent person's website. I >have a vague memory of someone suggesting this to you once. >I would be a simple matter of you posting to (the equivalent of) a >website's guestbook explicitly set up for the purpose and announcing >such on the newsgroups. >Once your message was entered, the site owner then would block you >from returning to that site until 10 people had responded. >He could then reply to your usenet announcement with >1. the message and >2. the list of respondents. >You would then have an agreed amount of time to peform your magic. The >website owner would not in any correspondence confirm hits or misses >until the end of the trial. >According to your table of chances you would only need to get 7 or >better out of 10 individual posts to show that anything significant >was happening >I would be happy to see that set up as at least a *preliminary* trial >to see if there was anything to your claim, and then tighten up the >testing protocols for a more rigourous test if indeed there was >anything worth testing. > >I'll respond to the rest and leave this renamed thread for discussion >of a test protocol that you might agree to - or not. I'll watch this. Herc will duck and weave. -- Find out about Australia's most dangerous Doomsday Cult: http://users.bigpond.net.au/wanglese/pebble.htm You can't fool me, it's turtles all the way down. === Subject: Re: A test proposal for Herc (was Re: To prove psychic powers, someone has to sit this test!) > >Not, as Wally has correctly pointed out, on Usenet. Perhaps to a >private discussion forum set up on an independent person's website. I >have a vague memory of someone suggesting this to you once. >I would be a simple matter of you posting to (the equivalent of) a >website's guestbook explicitly set up for the purpose and announcing >such on the newsgroups. >Once your message was entered, the site owner then would block you >from returning to that site until 10 people had responded. >He could then reply to your usenet announcement with >1. the message and >2. the list of respondents. >You would then have an agreed amount of time to peform your magic. The >website owner would not in any correspondence confirm hits or misses >until the end of the trial. >According to your table of chances you would only need to get 7 or >better out of 10 individual posts to show that anything significant >was happening >I would be happy to see that set up as at least a *preliminary* trial >to see if there was anything to your claim, and then tighten up the >testing protocols for a more rigourous test if indeed there was >anything worth testing. > >I'll respond to the rest and leave this renamed thread for discussion >of a test protocol that you might agree to - or not. > I'll watch this. Herc will duck and weave. Hasn't started well. -- Eric Hocking === Subject: Take any mass anywhere Take any object; body or mass of matter anywhere; anywhere on Earth, or even to another planet: Its inertia won't change, but its weight will: Howcome? === Subject: Re: Take any mass anywhere > Take any object; body or mass of matter anywhere; anywhere on Earth, or even > to another planet: Its inertia won't change, but its weight will: Howcome? Because it will be in gravitational fields of varying strengths. But you knew that. -- G.C. === Subject: Re: Take any mass anywhere >>Take any object; body or mass of matter anywhere; anywhere on Earth, or even >>to another planet: Its inertia won't change, but its weight will: Howcome? > Because it will be in gravitational fields of varying strengths. But > you knew that. Not necessarily. In all likelihood, he *didn't* know that. Don sHead is an aging eighth-grade dropout. He has no understanding of math beyond grade-school arithmetic. Even basic concepts of high-school physics are well beyond his comprehension. === Subject: Re: Take any mass anywhere >>Take any object; body or mass of matter anywhere; anywhere on Earth, or even >>to another planet: Its inertia won't change, but its weight will: Howcome? >> > Because it will be in gravitational fields of varying strengths. But > you knew that. > Not necessarily. In all likelihood, he *didn't* know that. > Don sHead is an aging eighth-grade dropout. He has no understanding of math > beyond grade-school arithmetic. Even basic concepts of high-school physics are > well beyond his comprehension. Amen! Isn't there a saying that Arguing with an idiot is often a symmetric operation! ? Bob Pease === Subject: Re: Take any mass anywhere > >>Take any object; body or mass of matter anywhere; anywhere on Earth, or > even >>to another planet: Its inertia won't change, but its weight will: > Howcome? >> > > Because it will be in gravitational fields of varying strengths. But > you knew that. > Not necessarily. In all likelihood, he *didn't* know that. > Don sHead is an aging eighth-grade dropout. He has no understanding of > math > beyond grade-school arithmetic. Even basic concepts of high-school > physics are > well beyond his comprehension. > Amen! > Isn't there a saying that > Arguing with an idiot is often a symmetric operation! ? > Bob Pease Yes, that's true enough, but an unbiased referee might be able to tell who's lying; who's the idiot and who isn't. === Subject: Re: Take any mass anywhere >>Take any object; body or mass of matter anywhere; anywhere on Earth, or even >>to another planet: Its inertia won't change, but its weight will: Howcome? >> > Because it will be in gravitational fields of varying strengths. But > you knew that. > Not necessarily. In all likelihood, he *didn't* know that. > Don sHead is an aging eighth-grade dropout. He has no understanding of math > beyond grade-school arithmetic. Even basic concepts of high-school physics are > well beyond his comprehension. You dumb lunkhead Mark: Where do you get all that garbage? Besides even if it were true and you weren't such a low down liar: It's still like I say: Elementary school arithmetic, with an understanding of ratios and fractions is all anyone needs to see that a body's inertia is the measure of its mass, and no one even needs that to see that the mutual force exerted between the terra firma surface of an Earthlike planet and things resting thereon varies; depending on the massiveness, size, density and rotation of that planet. That's probably too darn simple for the likes of you. === Subject: Re: Take any mass anywhere > Take any object; body or mass of matter anywhere; anywhere on Earth, or even > to another planet: Its inertia won't change, but its weight will: Howcome? > Because it will be in gravitational fields of varying strengths. But > you knew that. Yes: That's why its weight varies; but how come its inertia doesn't? I confess that I know that too; do you? > -- > G.C. === Subject: Re: Take any mass anywhere > > Take any object; body or mass of matter anywhere; anywhere on Earth, or > even > to another planet: Its inertia won't change, but its weight will: > Howcome? > Because it will be in gravitational fields of varying strengths. But > you knew that. > Yes: That's why its weight varies; but how come its inertia doesn't? I > confess that I know that too; do you? Conservation of mass--in so far as it is conserved! -- G.C. === Subject: Re: Take any mass anywhere message > > Take any object; body or mass of matter anywhere; anywhere on Earth, or > even > to another planet: Its inertia won't change, but its weight will: > Howcome? > > Because it will be in gravitational fields of varying strengths. But > you knew that. > > Yes: That's why its weight varies; but how come its inertia doesn't? I > confess that I know that too; do you? > Conservation of mass--in so far as it is conserved! > -- > G.C. Sure George: A body's mass and inertia will be conserved if something doesn't occur that adds or subtracts matter from it. === Subject: Re: Take any mass anywhere I just told you why... >Because locally the variance in the quantum state > ,which is the causation of the effects aspect of > gravitational variance, due to the number and separation > distance of all matter locally. Weight is Relative > Mass is not.... > M1 ----------- M2------------M3 > If Mass 1 (M1) is weighed in relation to M2 and M3 we get > a weight which is relative to the Mass and distance from > other masses.... > But if I Move Mass 2 (M2) > The Quantum State local value (------------) is reduced in one > direction and increased in the other. Due to the inverse Tensor > value of the Quantum State varying between relative masses its > local gravitational interplay values will be dependent on > the inherent mass of the object placed in motion... > -- -> > M1 ------------------------------M2----------------M3 > QS Value QS Value > Reduced Increased > > Take any object; body or mass of matter anywhere; anywhere on Earth, or > even > to another planet: Its inertia won't change, but its weight will: > Howcome? > Because it will be in gravitational fields of varying strengths. But > you knew that. > Yes: That's why its weight varies; but how come its inertia doesn't? I > confess that I know that too; do you? > -- > G.C. === Subject: Re: Take any mass anywhere In sci.math, Donald G. Shead : > Take any object; body or mass of matter anywhere; anywhere on Earth, or even > to another planet: Its inertia won't change, but its weight will: Howcome? I'm not sure there's a good answer to that sort of question; it just does. Of course it gets weirder. Take that object and start it moving. Its mass *changes*. (This is a consequence of SR and a little weird to assimilate; fortunately for most it's not significant until v is a significant fraction of lightspeed.) (I'll admit I don't know how to handle angular energy but suspect a formula similar to E = m0 c^2 / sqrt(1 - v^2/c^2), except v is replaced by omega and m0 is replaced by something relating to moment of inertia.) -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Take any mass anywhere > In sci.math, Donald G. Shead > > : > Take any object; body or mass of matter anywhere; anywhere on Earth, or even > to another planet: Its inertia won't change, but its weight will: Howcome? > I'm not sure there's a good answer to that sort of question; it > just does. > Of course it gets weirder. Take that object and start it moving. > Its mass *changes*. (This is a consequence of SR and a little > weird to assimilate; fortunately for most it's not significant > until v is a significant fraction of lightspeed.) > (I'll admit I don't know how to handle angular energy but > suspect a formula similar to E = m0 c^2 / sqrt(1 - v^2/c^2), > except v is replaced by omega and m0 is replaced by something > relating to moment of inertia.) > -- > #191, ewill3@earthlink.net > It's still legal to go .sigless. Like I just told Paul: It's not nearly that complicated Goulie? It's just that there is _no reason_ for its inertia to change; which will only happen if matter is added to, or taken from it: The reason its weight changes with a change in location; is because the acceleration of free fall varies at various locations; so that at _any_ particular location the ratio of the weight [w] of any mass, to the acceleration [g] at which it will free fall there; is a constant [w/g]: Where _this_ constant [w/g] is called its gravitational mass. === Subject: Re: Take any mass anywhere > In sci.math, Donald G. Shead > > : > Take any object; body or mass of matter anywhere; anywhere on Earth, or > even > to another planet: Its inertia won't change, but its weight will: > Howcome? > > I'm not sure there's a good answer to that sort of question; it > just does. > Of course it gets weirder. Take that object and start it moving. > Its mass *changes*. (This is a consequence of SR and a little > weird to assimilate; fortunately for most it's not significant > until v is a significant fraction of lightspeed.) > (I'll admit I don't know how to handle angular energy but > suspect a formula similar to E = m0 c^2 / sqrt(1 - v^2/c^2), > except v is replaced by omega and m0 is replaced by something > relating to moment of inertia.) > -- > #191, ewill3@earthlink.net > It's still legal to go .sigless. > Like I just told Paul: > It's not nearly that complicated Goulie? It's just that there is _no reason_ > for its inertia to change; which will only happen if matter is added to, or > taken from it: > The reason its weight changes with a change in location; is because the > acceleration of free fall varies at various locations; so that at _any_ > particular location the ratio of the weight [w] of any mass, to the > acceleration [g] at which it will free fall there; is a constant [w/g]: > Where _this_ constant [w/g] is called its gravitational mass. I won't try to explain why I think your incorrect ...others have done that quite well so I'll not debate the issue I will only point out that all you are doing is looking at the effects aspect of matter in relation to other matter ... nothing you have ever written has attempted to address causation,, you do as all others ... throw in a constant here or a infinity there and think some great epiphany has occurred... I say only Why is there a constant and if it varies at a rate that occurs at a scale that the variance cannot be verified on human scale how can you claim a constant .. Constants and Infinities are but illusion that occur to the limited knowledge base of man when the only rules he knows fail to apply due to scale... The difference between you and I is that I know my postulates are goofy.... and you really think your on to sumpthin..... Have you noticed... When you write of your views how so many jump you with all the counter posts and sometimes just plain meanness... While I post mine and hardly ever get any major screaming ...sometime even stuff I thinks a bit loony doesn't get beat like your stuff does.... With mine everyone says that just don't sound right but hell if I can figure out where you screwing up in your thought processes... with you they see it clearly...... ;) === Subject: Re: Take any mass anywhere Cut< > I won't try to explain why I think your incorrect ...others have done that > quite well so I'll not debate the issue I will only point out that all you > are doing is looking at the effects aspect of matter in relation to > other matter ... nothing you have ever written has attempted to address > causation,, you do as all others ... throw in a constant here or a > infinity there and think some great epiphany has occurred... I say only > Why is there a constant and if it varies at a rate that occurs at a scale > that the variance cannot be verified on human scale how can you claim > a constant . 'Cuz a body's inertia _is_ constant! It's a measure of the amount of matter in it. Which will vary only if some matter is added, or some is taken away. If that isn't reality then what is? . Constants and Infinities are but illusion that occur to > the limited knowledge base of man when the only rules he knows fail to > apply due to scale... The difference between you and I is that I know > my postulates are goofy.... and you really think your on to sumpthin..... > Have you noticed... When you write of your views how so many > jump you with all the counter posts and sometimes just plain meanness... Yes, but they have no choice, they've got to dissagree; to suppress and nip these unorthodox ideas in the bud, because my views are threatening to the status quo. > While I post mine and hardly ever get any major screaming ...sometime > even stuff I thinks a bit loony doesn't get beat like your stuff does.... I'm almost crying tears for you; your loony stuff isn't disputed because it fits into and is part of the looniness of the statis quo. > With mine everyone says that just don't sound right but hell if > I can figure out where you screwing up in your thought processes... > with you they see it clearly...... ;) That's what I'm striving for; not trying to win popularity, just trying to make the truth clear. === Subject: Re: Take any mass anywhere > Cut< > I won't try to explain why I think your incorrect ...others have done that > quite well so I'll not debate the issue I will only point out that all you > are doing is looking at the effects aspect of matter in relation to > other matter ... nothing you have ever written has attempted to address > causation,, you do as all others ... throw in a constant here or a > infinity there and think some great epiphany has occurred... I say only > Why is there a constant and if it varies at a rate that occurs at a scale > that the variance cannot be verified on human scale how can you claim > a constant . > 'Cuz a body's inertia _is_ constant! It's a measure of the amount of matter > in it. Which will vary only if some matter is added, or some is taken away. > If that isn't reality then what is? Caused you missed my point... If the inertia causation is that the massive object is tensor impacted by the conditional state between all other matter then the inertia can vary at a rate based on universal distribution of all matter and will vary locally with expansion at a rate that would not be detectable within experimental error within the life span of our solar system and therefore an apparent observable constant on our scale of observation.... > . Constants and Infinities are but illusion that occur to > the limited knowledge base of man when the only rules he knows fail to > apply due to scale... The difference between you and I is that I know > my postulates are goofy.... and you really think your on to sumpthin..... > Have you noticed... When you write of your views how so many > jump you with all the counter posts and sometimes just plain meanness... > Yes, but they have no choice, they've got to dissagree; to suppress and nip > these unorthodox ideas in the bud, because my views are threatening to the > status quo. Dude.. You have a mighty high opinion of your value in the arena of ideas....... No ones threatened and there's no vast right wing conspiracy..... > While I post mine and hardly ever get any major screaming ...sometime > even stuff I thinks a bit loony doesn't get beat like your stuff does.... > I'm almost crying tears for you; your loony stuff isn't disputed because it > fits into and is part of the looniness of the statis quo. > With mine everyone says that just don't sound right but hell if > I can figure out where you screwing up in your thought processes... > with you they see it clearly...... ;) > That's what I'm striving for; not trying to win popularity, just trying to > make the truth clear. And there's the Rub..... === Subject: Re: Take any mass anywhere Cut, cut ca-docket< > Have you noticed... When you write of your views how so many > jump you with all the counter posts and sometimes just plain meanness... > Yes, but they have no choice, they've got to dissagree; to suppress and > nip > these unorthodox ideas in the bud, because my views are threatening to the > status quo. > Dude... You have a mighty high opinion of your value in the > arena of ideas...... No ones threatened and there's no vast right > wing conspiracy..... You've never heard of the have's and have not's I suppose: The have's are united in spirit to keep the upstarts of the have not's in their place: To keep the poor and ignorant too dumb to qualify for positions beside them, at the pork barrel and on the gravy train: There's only so much of that to go around you know; not enough for those who haven't earned it. Some of them - including the do gooders - may not even realize it, and justify the whole thing on the fact that the have not's have equal oportunity, and are just too dumb and lazy to help themselves. > While I post mine and hardly ever get any major screaming ...sometime > even stuff I thinks a bit loony doesn't get beat like your stuff > does.... > > I'm almost crying tears for you; your loony stuff isn't disputed because > it > fits into and is part of the looniness of the statis quo. > With mine everyone says that just don't sound right but hell if > I can figure out where you screwing up in your thought processes... > with you they see it clearly...... ;) > > > That's what I'm striving for; not trying to win popularity, just trying to > make the truth clear. > And there's the Rub..... Yes if you play with fire, expect to get burnt. === Subject: Re: Take any mass anywhere > Take any object; body or mass of matter anywhere; anywhere on Earth, or even > to another planet: Its inertia won't change, but its weight will: Howcome? Because locally the variance in the quantum state ,which is the causation of the effects aspect of gravitational variance, due to the number and separation distance of all matter locally. Weight is Relative Mass is not.... M1 ----------- M2------------M3 If Mass 1 (M1) is weighed in relation to M2 and M3 we get a weight which is relative to the Mass and distance from other masses.... But if I Move Mass 2 (M2) The Quantum State local value (------------) is reduced in one direction and increased in the other. Due to the inverse Tensor value of the Quantum State varying between relative masses its local gravitational interplay values will be dependent on the inherent mass of the object placed in motion... -- -> M1 ------------------------------M2----------------M3 QS Value QS Value Reduced Increased === Subject: Re: Take any mass anywhere > Take any object; body or mass of matter anywhere; anywhere on Earth, or > even > to another planet: Its inertia won't change, but its weight will: Howcome? > Because locally the variance in the quantum state > ,which is the causation of the effects aspect of > gravitational variance, due to the number and separation > distance of all matter locally. Weight is Relative > Mass is not.... > M1 ----------- M2------------M3 > If Mass 1 (M1) is weighed in relation to M2 and M3 we get > a weight which is relative to the Mass and distance from > other masses.... > But if I Move Mass 2 (M2) > The Quantum State local value (------------) is reduced in one > direction and increased in the other. Due to the inverse Tensor > value of the Quantum State varying between relative masses its > local gravitational interplay values will be dependent on > the inherent mass of the object placed in motion... > -- -> > M1 ------------------------------M2----------------M3 > QS Value QS Value > Reduced Increased Really now Paul can't you see that it's not nearly that complicated? It's just that there is _no reason_ for its inertia to change; which will only happen if matter is added to, or taken from it: The reason its weight changes with a change in location; is because the acceleration of free fall varies at various locations; so that at _any_ particular location the ratio of the weight [w] of any mass, to the acceleration [g] at which it will free fall there is a constant [w/g]. === Subject: Re: Take any mass anywhere > Take any object; body or mass of matter anywhere; anywhere on Earth, or even > to another planet: Its inertia won't change, but its weight will: Howcome? Why not take *your* mass to another planet and then let us know via laser beams. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Take any mass anywhere > Take any object; body or mass of matter anywhere; anywhere on Earth, or even > to another planet: Its inertia won't change, but its weight will: Howcome? > Why not take *your* mass to another planet and then let us know via laser beams. > -- > There are two things you must never attempt to prove: the unprovable -- and the > obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com you have added an unneeded extra letter to mass Bob Pease === Subject: Re: Take any mass anywhere > Take any object; body or mass of matter anywhere; anywhere on Earth, or > even > to another planet: Its inertia won't change, but its weight will: > Howcome? > Why not take *your* mass to another planet and then let us know via laser > beams. > -- > There are two things you must never attempt to prove: the unprovable -- > and the > obvious. > -- > Democracy: The triumph of popularity over principle. > -- > http://www.crbond.com > you have added an unneeded extra letter to mass > Bob Pease My worry is that equus asinus may die from lack of oxygen. -- G.C. === Subject: Re: Take any mass anywhere > My worry is that equus asinus may die from lack of oxygen. The specimen in question thrives on flatulence gases (methane and carbon dioxide with traces of organic sulfides), and is fortunately able to produce enough to sustain itself. Tom Davidson Richmond, VA === Subject: Re: Take any mass anywhere Cut< > Why not take *your* mass to another planet and then let us know via laser > beams. > you have added an unneeded extra letter to mass > Bob Pease You guys are just saying that because you don't know the answer; you only think you know; but you don't want to make m^asses of yourselves: Better to make me out as one. === Subject: Re: Take any mass anywhere >Take any object; body or mass of matter anywhere; anywhere on Earth, or even >to another planet: Its inertia won't change, but its weight will: Howcome? This must be a trick question. ************************ David C. Ullrich === Subject: Re: Take any mass anywhere >Take any object; body or mass of matter anywhere; anywhere on Earth, or even >to another planet: Its inertia won't change, but its weight will: Howcome? > This must be a trick question. Nah: It's just too tough. > ************************ > David C. Ullrich === Subject: Re: Q: Product code generator polynomial > I just want to know if I have worked correctly (using GF(2^4)) > For C_1: > n_1 = 5 > k_1 = 3 > r_1 = 2 -> the amount of redundant symbols > g_1 = alpha^6 + alpha^13X +X^2 > and C_2: > n_2 = 3 > k_2 = 2 > r_2 = 1 > g_2 = alpha^10 + X > (I have checked that this time g_1(X) is a factor of X^5 - 1 and > g_2(X) is a factor of X^3 - 1, but please verify this...) > 1 = (-1)*5 + 2*3 > Thus, we have g_1(X^6) and g_2(X^10) (-> 10 = -5 mod 15) > Also because we know the resulting code will be a cyclic code of > length n_1*n_2, I divided the product g_1(X^6)*g_2(X^10) with > X^{n_1*n_2} - 1 and obtained a polynomial g_r(X) > g_p(X) (the generator of the product code) will then be: > g_p(X) = gcd[X^{n_1*n_2} - 1,g_r(X)] > Using MATLAB, I got the following polynomial: > g_p(x) = [12 9 11 13 5 7 4 6 8 0] (i.e. alpha^12 + alpha^9X + ... + > alpha^0X^9) > which can be a generator for a (15,6) cyclic code. > Why does this code have a minimum weight of w_1*w_2 = 2*1 = 2? > (According to Lin and Costello) Why is that? It seems to me that the length 5 cyclic code generated by your g_1(x) has minimum weight 3 (BCH bound tells you it is at least 3, and we manifestly have words of weight 3 in the code). Similarly the code generated by g_2(x) has minimum weight 2. So the product code should have minimum weight 6, not 2. I think that that is also the correct minimum weight. Consider the following multiplication table alpha^6 alpha^13 1 0 0 _____________________________________________ alpha^10 | alpha alpha^8 alpha^10 0 0 1 | alpha^6 alpha^13 1 0 0 0 | 0 0 0 0 0 with g_2 in the first column and g_1 in the first row, and the rest formed as the product of these entries. Now, when you read this 3x5 table in the order indicated by Robin Chapman's table, you get a polynomial of weight 6 (obviously). And this polynomial IS in the code you produced above. Therefore the minimum weight cannot be 9. The product code theorem tells you that the minimum weight cannot be less than six, so the minimum weight is exactly six. As a conclusion you should observe that the product code construction will usually not produce as good codes as the RS codes (for this alphabet). They can be useful in other places, when you want longer codes, but it seems to that the product of two (shorter) RS codes is always worse than a longer RS-code (when its length doesn't exceed q-1). This is hardly surprising given that RS-codes are on the Singleton bound and hence optimal. You may get some reasonable codes as products of longer cyclic codes (e.g. large alphabet BCH-codes). I've also seen the product code used in McEliece cryptosystem. May be you know of/find a way to use them for burst error correction! Inspecting g_p(X), the cyclic code > should have minimum weight 9? As I said above, I don't see, how you arrive at that conclusion. Jyrki Lahtonen, Turku, Finland === Subject: Force and displacement Every body in the universe is in a constant state of being displaced! There are two kinds of displacement; which both consist of changes in position: _Inertial Displacement_; is due to a body's own inertia, where everything is moving and changing position relative to everything else. _Forced Displacement_; is due to impenetrability; the fact that no two pass through the exact same place; so that they mutually thrust against each other, and displace each other from their previous motions. Galileo made many discoveries; not the least of which is the fact that all bodies free fall at the same rate; about 16'/sec: Exactly half of the acceleration of free fall. Which is attributed to the 'force of gravity'. Neither he nor anyone else has yet discovered that the mutual force exerted between Earth and bodies resting thereon; _is_ the force of gravity: That realization may have to wait for the next revolutionary change in people's beliefs about science. === Subject: Did mathematicians know? One of the odder things I keep facing is my ability to explain a rather basic argument in a very short space, which has *enormous* implications, only to see mathematicians mostly walk away. See When is a proof? http://www.maa.org/devlin/devlin_06_03.html Here's an excerpt: What is a proof? The question has two answers. The right wing (right-or-wrong, rule-of-law) definition is that a proof is a logically correct argument that establishes the truth of a given statement. The left wing answer (fuzzy, democratic, and human centered) is that a proof is an argument that convinces a typical mathematician of the truth of a given statement. While valid in an idealistic sense, the right wing definition of a proof has the problem that, except for trivial examples, it is not clear that anyone has ever seen such a thing. Maybe mathematicians now believe there are multiple errors in core. Possibly they've found LOTS of problems with proofs so what I've found is just one of many. Now they're setting up for a new position where proof does not exist, and they don't even bother claiming it does. Possibly my biggest problem is timing, as mathematicians have already decided to move on with the knowledge that the foundations they previously touted, are riddled with errors. James Harris === Subject: Re: Did mathematicians know? James Harris scribbled the following on sci.math: > One of the odder things I keep facing is my ability to explain a > rather basic argument in a very short space, which has *enormous* > implications, only to see mathematicians mostly walk away. James, have you ever considered the possibility that the reason why everyone who has looked at your proofs and said that you are wrong is that *you* *are* *wrong*? -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ And according to Occam's Toothbrush, we only need to optimise the most frequent instructions. - Teemu Kerola === Subject: Re: Did mathematicians know? Devlin appears to have committed himself to Sophistry, which is really the same as the Copenhagen School of QM (the reification of the skware of the probability of an event, and a mystical view of the Uncertainty Principle); so, it looks to me, like they are setting up for a new position where proof does not exist, as you say. does Devlin mean to say that (e.g.) the Pythagorean Theorem has never been proven, in spite of hundreds of variants? > What is a proof? The question has two answers. The right wing > (right-or-wrong, rule-of-law) definition is that a proof is a > logically correct argument that establishes the truth of a given > statement. The left wing answer (fuzzy, democratic, and human > centered) is that a proof is an argument that convinces a typical > mathematician of the truth of a given statement. > While valid in an idealistic sense, the right wing definition of a > proof has the problem that, except for trivial examples, it is not > clear that anyone has ever seen such a thing. > Now they're setting up for a new position where proof does not exist, > and they don't even bother claiming it does. --les ducs de Buffet; vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall & e-Dereg! http://larouchepub.com === Subject: Re: Did mathematicians know? > Devlin appears to have committed himself to Sophistry, > which is really the same as the Copenhagen School of QM > (the reification of the skware of the probability > of an event, and a mystical view of the Uncertainty Principle); > so, it looks to me, like they are setting up > for a new position where proof does not exist, > as you say. > does Devlin mean to say that (e.g.) > the Pythagorean Theorem has never been proven, > in spite of hundreds of variants? The way I read it, the right wing approach corresponds to something like formalised mathematics such as http://mizar.uwb.edu.pl/JFM/index.html and the left wing approach to the more common approach of convincing people that a proof is formalisable. Jon === Subject: Re: Did mathematicians know? >One of the odder things I keep facing is my ability to explain a >rather basic argument in a very short space, which has *enormous* >implications, only to see mathematicians mostly walk away. That's a strange thing to say, considering the number of mathematicians who continue to explain that this is wrong and that is wrong, no matter how many times you repeat them... >See When is a proof? http://www.maa.org/devlin/devlin_06_03.html >Here's an excerpt: > >What is a proof? The question has two answers. The right wing >(right-or-wrong, rule-of-law) definition is that a proof is a >logically correct argument that establishes the truth of a given >statement. The left wing answer (fuzzy, democratic, and human >centered) is that a proof is an argument that convinces a typical >mathematician of the truth of a given statement. >While valid in an idealistic sense, the right wing definition of a >proof has the problem that, except for trivial examples, it is not >clear that anyone has ever seen such a thing. > >Maybe mathematicians now believe there are multiple errors in core. Huh? How in the world do you get from that quote to this conjecture about what mathematicians think? >Possibly they've found LOTS of problems with proofs so what I've >found is just one of many. Possibly indeed - if you could name one example that would be a little more compelling. >Now they're setting up for a new position where proof does not exist, >and they don't even bother claiming it does. Huh? >Possibly my biggest problem is timing, as mathematicians have already >decided to move on with the knowledge that the foundations they >previously touted, are riddled with errors. No, I don't think that timing is your biggest problem. David C. Ullrich ************************** As far as I'm concerend you're trying to wait until I die, so I figure maybe you should die instead. How about that, eh? Wouldn't that be a better twist? You refuse to follow the math, so the great Powers that control reality and *speak* in mathematics decide to kill you instead of me. So what do you think about that, eh? Oh, can't hear Them talking? Well, I guess that's because you don't really understand Mathematics, the true language, which is THE language. They're talking about you now, and They agree with my assessment, and will not penalize me as They allowed the others like Galois and Abel to be penalized. They will kill you instead. James Harris speaking on Weird factorization, genius === Subject: Re: Did mathematicians know? > One of the odder things I keep facing is my ability to explain a > rather basic argument in a very short space, / / ( // ) // / _||||//_/ / _ _ /|(O)(O)| / | | ___________________/ / // // |____| // || / //| | 0 0 / // ) V / ____/ // / ( / /_________| |_/ / / / | || / / / / || | | | | | || | | | | | || |_| |_| |_|| _ _ _ James Harris === Subject: Re: Did mathematicians know? > One of the odder things I keep facing is my ability to explain a > rather basic argument in a very short space, > / / > ( // ) > // / > _||||//_/ > / _ _ > /|(O)(O)| > / | | > ___________________/ / > // // |____| > // || / > //| | 0 0 / > // ) V / ____/ > // / ( / > /_________| |_/ > / / / | || > / / / / || > | | | | | || > | | | | | || > |_| |_| |_|| > _ _ _ James Harris Well done Mr Pig! or may I call you Porky? -- G.C. === Subject: Re: Did mathematicians know? > Well done Mr Pig! or may I call you Porky? Call me PPJ as we all strive for a *very short space*. === Subject: [JSH] Re: Did mathematicians know? Note subject change. In sci.math, Porky Pig Jr >> One of the odder things I keep facing is my ability to explain a >> rather basic argument in a very short space, > / / > ( // ) > // / > _||||//_/ > / _ _ > /|(O)(O)| > / | | > ___________________/ / > // // |____| > // || / > //| | 0 0 / > // ) V / ____/ > // / ( / > /_________| |_/ > / / / | || > / / / / || > | | | | | || > | | | | | || > |_| |_| |_|| > _ _ _ James Harris Looks like a donkey h.99te to me... :-) But where's the windmill? -- #191, ewill3@earthlink.net -- insert random very weird pun here It's still legal to go .sigless. === Subject: Re: [JSH] Re: Did mathematicians know? man, what's hote mean ?!? > _ _ _ James Harris > Looks like a donkey h.99te to me... :-) > But where's the windmill? --les ducs de Buffet; vote NONE OF THE BELOW on Trickier Dick Cheney's California Recall & e-Dereg! http://larouchepub.com http://members.tripod.com/~american_almanac/ === Subject: Hallmark of error in core Now it's worth pointing out a hallmark of a problem or error in core mathematics: the ability to prove two different and contradictory things. The problem with algebraic integers isn't about axioms, but about definitions and you see the anti-mathematician Arturo Magidin wielding those definitions with calm precision. By the definition of unit, certain factors that another method seems to prove to be units are not units *in the ring of algebraic integers* though they are in fact units in a more inclusive ring. By the definition those same numbers are therefore non-unit factors, so Arturo Magidin proclaims them to be non-unit factors showing lack of coprimeness, but that relies on the definition. All I actually have to do is show that you have a contradiction, and in fact I have repeatedly, by using very short algebraic arguments. In retaliation, Arturo Magidin has repeatedly tried to cast doubt on that argument, but his objection comes down to claiming that the constant terms of factors of a polynomial are in fact varying, so are NOT constant terms. Such an obvious error in thinking requires slight-of-hand to present, and notice how Arturo Magidin works. It is strong evidence that he *knows* that he is lying, and I wish to emphasize that to you now. I will continue to answer the posts of Arturo Magidin, and point out where he is lying, but for this to end, some of you need to step forward, as make no mistake, an error which allows contradictory things to supposedly be proven, is not an error that should be left alone. James Harris === Subject: Re: Hallmark of error in core > Now it's worth pointing out a hallmark of a problem or error in core > mathematics: the ability to prove two different and contradictory > things. > The problem with algebraic integers isn't about axioms, but about > definitions and you see the anti-mathematician Arturo Magidin wielding > those definitions with calm precision. > By the definition of unit, certain factors that another method seems > to prove to be units are not units *in the ring of algebraic integers* > though they are in fact units in a more inclusive ring. Well, 1 is not a unit in the ring of even numbers, but it is a unit in the larger ring of integers. Is that what you're talking about? === Subject: Re: Hallmark of error in core Visiting Assistant Professor at the University of Montana. >Now it's worth pointing out a hallmark of a problem or error in core >mathematics: the ability to prove two different and contradictory >things. If that were true, then everything is provable. And therefore, when you claim that someone is making false claims in proving something, you are making an error. >The problem with algebraic integers isn't about axioms, but about >definitions and you see the anti-mathematician Arturo Magidin wielding >those definitions with calm precision. Unsupported ad hominem. Based on your incorrect assertions that I have been wrong, and in your decision to not apologize for those prior attacks even when you admit I was right. >By the definition of unit, certain factors that another method seems >to prove to be units are not units *in the ring of algebraic integers* >though they are in fact units in a more inclusive ring. This is true of EVERY element in every integral domain. It's even true of many elements in rings that are not integral domains. >By the definition those same numbers are therefore non-unit factors, >so Arturo Magidin proclaims them to be non-unit factors in the ring of algebraic integers >showing lack of coprimeness, in the ring of algebraic integers See? You are not to be trusted to accurately report my objections. You misrepresent them, time and again. > but that relies on the definition. But that relies on the definition. So, in your opinion, it is bad to use a word to mean what the definition says it means? How should we use the word, then? [.snip.] >In retaliation, Arturo Magidin has repeatedly tried to cast doubt on >that argument, but his objection comes down to claiming that the >constant terms of factors of a polynomial are in fact varying, so are >NOT constant terms. No, my objection is that your claim that the behavior of complicated functions at m=0 tells you their behavior at arbitrary points is unfounded, and in this case, false. You are propping up a strawman. All your personal attacks are based on this strawman. >Such an obvious error in thinking requires slight-of-hand to present, >and notice how Arturo Magidin works. It is strong evidence that he >*knows* that he is lying, and I wish to emphasize that to you now. I know I am lying, like when I told you that if r_1 is a non-unit factor of 2 in the ring of algebraic integers, then r_1 and 2 cannot be coprime in the ring of algebraic integers? Yes, you told me many times I was lying. You told EVERYONE in the newsgroup I was lying. You said I was doing wacky stuff, anti-algebra, voodoo math, that I was a despicable evil person. What was the end result of all that? Ah, yes: YUCK! I hate it when I screw-up. It turns out that I was wrong and 'a' and 'b' can't both be algebraic integers, which--when cleaned up--Arturo Magidin's argument shows. Followed up by saying that you would not apologize nor retract any of your false accusations, because you care about the truth. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Hallmark of error in core > Now it's worth pointing out a hallmark of a problem or error in core > mathematics: the ability to prove two different and contradictory > things. You have failed to do that. There are fatal flaws in your argument. Furthermore, if you can 'prove two different and contradictory things' then all proofs are vacuous and without meaningful content, including yours. If you believe that is the case, you have no standing to argue with anyone about any conclusion, and ought to look for work as a human shield somewhere. > The problem with algebraic integers isn't about axioms, but about > definitions and you see the anti-mathematician Arturo Magidin wielding > those definitions with calm precision. > By the definition of unit, certain factors that another method seems > to prove to be units are not units *in the ring of algebraic integers* > though they are in fact units in a more inclusive ring. So what? A 'more inclusive ring' already exists -- the algebraic numbers. > By the definition those same numbers are therefore non-unit factors, > so Arturo Magidin proclaims them to be non-unit factors showing lack > of coprimeness, but that relies on the definition. > All I actually have to do is show that you have a contradiction, and > in fact I have repeatedly, by using very short algebraic arguments. > In retaliation, Arturo Magidin has repeatedly tried to cast doubt on > that argument, but his objection comes down to claiming that the > constant terms of factors of a polynomial are in fact varying, so are > NOT constant terms. > Such an obvious error in thinking requires slight-of-hand to present, > and notice how Arturo Magidin works. It is strong evidence that he > *knows* that he is lying, and I wish to emphasize that to you now. > I will continue to answer the posts of Arturo Magidin, and point out > where he is lying, but for this to end, some of you need to step > forward, as make no mistake, an error which allows contradictory > things to supposedly be proven, is not an error that should be left > alone. > James Harris SINGLE NUMBER which 'should' be in the ring of algebraic integers, but which is not. This is extraordinary, in view of your continued insistence that elmentary algebra supports your claim. Why don't you give us a single number (numeric value) which settles the issue. Just one, James Harris. Just one. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Desire for fame is a bitch > > |> > |> [...] > |> > |> |> However, Professor McKenzie began claiming my work is algebraic > |> |> geometry which is out of his field. So I explained the entire thing > |> |> over more than an hour, having driven up four hours from Atlanta > |> |> metro, and he tells me it's out of his field. > |> > |> [...] > |> > |> In other words, you talk stuff that is orthogonal to real mathematics, > |> not even wrong, and you were wasting the guy's time and so he sent you > |> on. > |> > |> Which defies the fact that I explained it point-by-point. The > |> professor challenged me repeatedly on points and I handled all > |> challenges. > |> > |> James, imagine that while you were waiting to see the guy, > |> you saw some psychotic wander in off the street, clutching > |> a battered briefcase full of napkins and toilet paper > |> covered with scribbling. Imagine you saw this guy walk > |> into Prof. Mackenzie's office and scribble madly on the > |> blackboard for an hour. Imagine that every time the professor > |> asked him a question, he answered... something incoherent, > |> but something. > |> > |> Would you say that the fact that the psychotic explained > |> it point-by-point and handled all challenges proved > |> that he was right about whatever his messianic message > |> was? > |> > |> What would you say to the psychotic to convince him > |> otherwise? > I graduated from Vanderbilt University, so I'm not some psychotic > wandering off the street. That's why I asked you to imagine watching this happening in front of you. I didn't say you were the psychotic. > As an alumnus of the school I *expect* to be treated a certain way. What did my hypothetical situation have to do with that? I'm saying that this psychotic can also say (traditional random capitalization added) I EXPLAINED my proof point BY point AND I handled ALL CHALLENGES I'm not saying you're the guy with the briefcase full of toilet paper. I am asking you what I explained my proof point by point is supposed to prove to anybody. My hypothetical psychotic did the same. I'm asking you what I answered all questions is supposed to mean. So did my raving lunatic. I'm asking you what I spent an hour writing on his blackboard is supposed to show. So did my nutcase. What are you saying about your presentation that my hypothetical psychotic can't also say? > I spoke with him in *two* sessions over a period of more than an hour > as he took a coffee break and I wandered off to speak to a Professor > Wikswo in the physics department. And this shows what exactly? Does everybody who walks into a professor's office twice thereby demonstrate he has an exciting new result? > What's interesting to me is your use of the word think as it looks > more like you are not thinking clearly. > I am a Vanderbilt alumnus. My going to the school to talk to a math > professor in that sense is not a big deal. Pushing the notion that I > wandered in off the street, and even got in to speak to a professor at > the school, though supposedly a babbling psychotic from the example, > indicates a baffling refusal to acknowledge the facts. Nobody said that. This inability to grasp somebody's point is the kind of thing that makes many of us suspicious that you conveyed Arturo's (or is it Nora's) objections correctly. I said nothing of the sort. How can I make this more clear? One last time: I ask you to ponder someone who is clearly raving, but offers as proofs of the truth of his ravings all the things you're saying about your presentation. Not only am I not saying you're a raving looney off the street, the entire POINT of my hypothetical was to give you a situation with which you wouldn't identify. So you read the exact opposite. I ask you to imagine this, and be very very clear that this guy is not you, OK? But he claims, as you do, I talked for an hour, I explained point by point, and I answered all challenges. He says to you what you have said to us. You can even add the professor took a coffee break. What do you say to the raving lunatic to argue that he has not proved whatever his raving message is? - Randy === Subject: Re: Desire for fame is a bitch James, Don't you realize what a load of B.S. all of this is. Suppose he did think you had this great proof. Then why isn't he inviting you to give a talk at the math dept, wanting to get your paper published in a refereed journal. I mean, this is what any person that I know would do if they really thought that someone had a great idea. The reality is he was nice enough to talk to you. No doubt, you were persistant, and maybe he had an hour to kill, and thought he'd see you and get you off of his back. Maybe he thought you might really have something (the Ramanujan from the hills of TN). BUT, his response CLEARLY shows that at the end, it came to nothing in his mind. Although I know you despise the mathematical community and the authorities, in the end, here you are trying to use _his_ professional stature to try to justify _your_ work. Talk about desperate, hypocritical, and deceptive. The reality is that he's not an authority in the field (as you've admitted), and so despite all of his stellar achievements, how are they even relevant? (Another hint that all of this is BS is that even if he wasn't an authority in the field, he still should have been able to follow your work, which is roughly at the level of undergrad algebra). Got it? Mark === Subject: Re: What is the time component of a four vector? === >Subject: What is the time component of a four-vector? >The time component of a four-vector is nonsense; just >like a lot of other far-out mathematical concepts. > Really? Care to expound on that. > Do you mean that time itself is nonsense? Or do you > just have some aversion to groupings of four. > By the way, the last time I heard of something referred > to as far-out was the sixties, man. > adam Shead is too ignorant of anything beyond Sixth grade to discuss this intelligently. His idea of Far out Math is Algebga II. He's not really a troll, but this one is REALLY a Scarlet Salmon. The concept of Space-time is even older than Shead. I've got him plonked, as I don't need sleeping aids when I can depend on Guinness If needed. Whoever said that all Four Vectors are supposed to have a time component.? What about the metric space R4? Groupings of Four have Yang Karma, which is somehow inappropriate for Shead. BTW the expression Far out and Hip replaced the earlier Way out and Hep What are the latest PC terms for this concept? I lost touch with the Pop Music lingo in the early 80's, after I saw the movie The Decline and Fall of Practically Everything, which was in English, but with English subtitles. Dr. Sidethink Hp.D === Subject: Re: Example for winter---was Re: Vale Robert palmer > IIRC, there are two common forms of the haiku. I can't recall what > the alternative to 575 is, though. > I can't find a description of another form, though I notice that > English haiku on the Internet often deviates from 5-7-5. One site > gives a fair justification of this: it suggests that 17 Japanese > syllables conveys roughly the same information as 11 English > syllables, and suggests 3-5-3 for English haiku. That may be the one I was thinking of. It was about 1976 when I had to do those for school. > The same page points out that 7- and 5-syllable combinations are > fairly common in Japanese as an artifact of grammar, and gives a > cute example of a naturally occurring haiku: > kono dote-ni > noboru-bekarazu > keishichou > Do Not Climb This Levee - The Police Department > > So I guess > Great freedom rewards > the daring It would seem Herc has many thoughts all insane === Subject: Re: Example for winter---was Re: Vale Robert palmer > I challenge winter > read all this without laughing > at least internally thought I recognised you, I preferred this one LOL. If you're looking for substance, Herc is worth plonking. If you're looking for This is a post. This is a post on drugs., then keep occasionally reading him, as I do. I once had a boss who said the view out his window was more surreal than that out my window. The view out Herc's window must be beyond surreal. I keep forgetting there's newbies all around posting freely that I haven't disciplined yet. Herc === Subject: Re: Example for winter---was Re: Vale Robert palmer > I challenge winter > read all this without laughing > at least internally > thought I recognised you, I preferred this one > LOL. If you're looking for substance, Herc is worth plonking. If you're > looking for This is a post. This is a post on drugs., then keep > occasionally reading him, as I do. I once had a boss who said the view out > his window was more surreal than that out my window. The view out Herc's > window must be beyond surreal. Yes! That was a great description (if I must say so myself) and I thank you for finding and reposting it. > I keep forgetting there's newbies all around posting freely that I haven't disciplined yet. You lack the means. Your own mind shows the lack. As to newbies, I started on this group about 1995, so I'm not sure I qualify. === Subject: Re: yet another real analysis question (about the difference between C^infty-ness and analyticity) >> I'm sorry to keep adding new posts to this thread, but I have one more >> question: suppose (ii) above is replaced by the relaxed condition that >> the Taylor series has a null radius of convergence at most at isolated >> points. If x is an accumulation point of E, then the radius of >> convergence in x is 0, right? >No. If a < b, you can define f(x) = exp(-1/[(x-a)(b-x)]) on (a,b), f = 0 on >R (a,b). Then f is C^oo(R) with all derivatives 0 on R (a,b). For each >n, choose f_n as above with respect to (1/(n+1), 1/n), n = 1,2, ... There >exist constants c_n > 0 such that sum_(n=1,oo) c_n*f_n converges in C^oo(R) >to some f. For this f, E = {1/n : n = 1,2,...} U {0}, 0 is an accumulation >point of E, and the radius of convergence of the Taylor series is oo at >each point of E, including 0. D'Oh! I guess my questions are over for the moment... Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Re: yet another real analysis question (about the difference between C^infty-ness and analyticity) >>I'm sorry to keep adding new posts to this thread, but I have one more >>question: suppose (ii) above is replaced by the relaxed condition that >>the Taylor series has a null radius of convergence at most at isolated >>points. If x is an accumulation point of E, then the radius of >>convergence in x is 0, right? >>PS: for those who didn't read the rest of the thread, E is the set of >>points where f (smooth) fails to be analytic. >No. For example, let g be a function that's C^infinity on R and >analytic nowhere, and let f(x) = exp(-1/x^2) g(x) for x <> 0, 0 for >x = 0. Then f(x) = O(x^n) as x -> 0 for every n, so the Taylor series >of f at x=0 is 0, which has infinite radius of convergence. But f >is nowhere analytic. If I'm not mistaken this is basically the counterexample D. Ullrich made a few posts ago. However I don't think this function satisfies my requirement; OTOH another poster provided a much simpler one that does, and made me exclaim: D'Oh! Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Re: Algebra proof > Let a, b, r and s be integers with r>1, s>1 and gcd(r,s)=1. Prove > that if a=b(mod r) and a=b(mod s) then a=b(mod rs). > > How do I do this???? > a = b + m*r for some m, and a = b + n*s for some n, right? That means that > m*r = n*s. Now use the fact that gcd(r,s) = 1 to conclude something about m > (or n). I guess since r and s are relatively prime, then r must divide n. so r*x=n. and a=b+r*x*s, so a=b(mod rs), right? === Subject: Re: Algebra proof >> Let a, b, r and s be integers with r>1, s>1 and gcd(r,s)=1. Prove >> that if a=b(mod r) and a=b(mod s) then a=b(mod rs). >> >> How do I do this???? >> a = b + m*r for some m, and a = b + n*s for some n, right? That means >> that m*r = n*s. Now use the fact that gcd(r,s) = 1 to conclude something >> about m (or n). > I guess since r and s are relatively prime, then r must divide n. so > r*x=n. and a=b+r*x*s, so a=b(mod rs), right? Exactly. === Subject: Re: Algebra proof > Let a, b, r and s be integers with r>1, s>1 and gcd(r,s)=1. Prove > that if a=b(mod r) and a=b(mod s) then a=b(mod rs). This is (part of) the Chinese remainder theorem. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: How long must physics put up w/f=ma? > In sci.math, Donald G. Shead > > : > Cut< >> >> F and a *are* vectors. > That sounds like something from some hosses mouth(;^) > Force is a thrust Goulie; between two or more masses of > matter; a push or a pull. Acceleration is the rate of > change in the velocity of a mass of matter; caused by a > force. > I have a pool ball on a frictionless surface with the usual > NESW coordinate grid. I hit it with a force in some direction > for 1 second. Without the direction, can you tell me where it > is after a few seconds? > Or, I hit it with a force for 1 second, and then another force > for 2 seconds. Where's the ball now? > Force can be graphically represented as the component > of the acceleration that it causes; but 'vectors' will > only suffice where the force does not change direction. > When a force is constantly changing its direction, as most > forces do in this tumultuous world's constantly changing > motions; of everything in it, relative to everything else; > then the line graphically depicting it will be variously curved. > Enter integrals. Not really all that much of a problem. >> As for dividing the two, the >> operation is somewhat meaningless in a multiple-dimension >> coordinate space. In fact, multiplying the two is somewhat >> meaningless, unless one distorts the notion of multiply >> (most people use dot-product instead; for 3-space >> there's a cross-product as well, which works to some >> extent as a product despite its anticommutativity, but >> in 4-space the cross-product requires *three* vectors, >> not two; in N-space one can define a cross-product >> with N-1 operands). >> > Cut the dot-product, and 4-space cross-product nonsense > Ghoulie: You know less of what your talking about than > I do, and I say it's all nonsense. > Hah. > [1] Tell me the number of regular hyper-solids in 4-space. > [2a] Envision a 1m side cube attached to 4 rubberbands. > The rubberbands are identical in composition and are of > length 1m, unstretched. The rubberbands are attached > to (0.5,0.5,0.5), (0.5,0.5,-0.5), (0.5,-0.5,0.5), and > (-0.5,0.5,0.5) on the cube, and (3, 3, 3), (3, 3, -3), > (3, -3, 3), and (-3, 3, 3), respectively. Where is the > center of the cube? > [2b] Remove the rubberband between (0.5, 0.5, 0.5) and > (3,3,3). Where is the center of the cube now? > [3] A car is going up the side of a hill at 15 mph. What is > the speed required on the opposite side of the hill > (assuming a symmetrical hill) so that the average speed of > the car from two points at the base of the hill is 30 mph? > [4a] A wandering drunkard somehow gets lost in a flat, otherwise > featureless desert, and starts from a rock and staggers > randomly. However, each stagger is done with equal length (1m) > and equal intervals of time (1s). How far will he be from the > rock after N staggers, on average? (No fair getting > drunk to try this.) > [4b] A drunken fly gets stuck in a large chamber with a tiny > pedestal (from which it starts), and flies drunkenly in a > manner similar to the staggering drunkard of the previous > problem, except in 3 dimensions. Where is the fly after > N staggers? > [5] A car going 180 kph on a flyover of height 20 m over > the roadway below goes over the side. Assuming the fence > of the flyover offers negligible resistance, calculate > how far the car will travel before hitting the ground. > [6] A youngster throws a Superball(tm) with a speed > of 20 m/s against the sidewalk. Assuming the release > was just over the sidewalk, the Superball is traveling > perpendicular thereto, and the Superball(tm) loses no > energy as it collides therewith, calculate the resulting > height and how long the ball will stay in the air. > [7] NASA has just accepted a proposal for a brand-new > vacuum-tube acceleration rail system. This rail system > will (hopefully) accelerate payloads into near Earth orbit > more cheaply and more speedily than conventional rockets. > Assuming that a payload is not accelerated any faster than > 100 m/s/s and that 8 km/s is required to achieve near-Earth > orbit, calculate (a) the length of the track required, > and (b) the maximum amount of electrical power at the very > instant of release for a payload of 10 metric tonnes. > (Neglect atmospheric, frictional, and relativistic effects.) > [8a] An electron has a mass of about 9 * 10^-31 kg. Estimate > the amount of energy required to accelerate it to lightspeed > (c = 3 * 10^8 m/s), assuming Newtonian physics. > [8b] Estimate the speed of the electron once accelerated with > the amount of energy given in [8a], assuming Einsteinian physics. > [rest snipped] > -- > #191, ewill3@earthlink.net > It's still legal to go .sigless. Math Question. Is arguing with an idiot a symmetric operation?? ;) Bob Pease === Subject: Re: How long must physics put up w/f=ma? In sci.math, Bob Pease : [my stuff snipped] > Math Question. > Is arguing with an idiot a symmetric operation?? > ;) Oh, wait, forgot to switch on the more intelligent part of my brain. :-) *click* No. :-) > Bob Pease -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: A language for rapid prototyping of integer math. > The program runs uses java bigintegers, which are arbitrary length > integers, so the calculations should be similar. > > Java BigIntegers are unsatisfactory (e.g. compared to Python longs) due > to their cumbersome COBOL-like syntax. > > You mean because they're objects and not native types? > Yep. Well it's the native type in my language...it's only checked for width on assignment. > I suppose that > could go into a future release. > Doubtful since it would involve a language change and not just a new > package. I was looking it over. Apparently Java has no mechanism for overriding + style operators. That's what is really needed. > By the way, did you know about > Jython? Python language and syntax was ported completely to Java. > Know about, haven't tried it. CPython works well enough for me... === Subject: Re: Limit of (1^p..+n^p)/(n^(p+1)) > For every natural n, let a_n= (1^p+...n^p)/(n^(p+1)), p>=0. I'm almost > quite sure that a_n --> 1/(p+1), what is easy to prove if p is > integer. But so far I couldn't give a proof that this is true for > every p>=0. > > Rewrite as [(1/n)^p + (2/n)^p + ... + (n/n)^p]*(1/n). These Riemann sums -> > int_[0,1] x^p dx = 1/(p+1) as n -> oo. (The result is also valid for -1 < p > < 0.) > > Do you think the convergence of f_n to f is uniform? > I think the convergence of [(1/n)^p + (2/n)^p + ... + (n/n)^p]*(1/n) to > 1/(p+1) is uniform on [p_0, oo) for any p_0 > -1. Haven't checked the > details yet. We see that if we let n and p goes to infinity, then the sum above goes to zero. We also see that, for a fixed n, the sum approaches 1/n when n-> infinity. In addition, if we let n and p goes to infinity, then 1/n ->0 and 1/(p+1) ->0. Therefore, the convergence of [(1/n)^p + (2/n)^p + ... + (n/n)^p]*(1/n) to 1/(p+1) is uniform on [p1, oo) for some p_1>-. Since [(1/n)^p + (2/n)^p + ... + (n/n)^p]*(1/n) decreases with p, 1/(p+1) is continuous for p>-1 and [p_0, p_1] is compact for every -1 -1. I think this proves your assertion. Artur === Subject: Re: Limit of (1^p..+n^p)/(n^(p+1)) > We see that if we let n and p goes to infinity, then the sum above > goes to zero. We also see that, for a fixed n, the sum approaches 1/n > when n-> infinity. In addition, if we let n and p goes to infinity, > then 1/n ->0 and 1/(p+1) ->0. Therefore, the convergence of [(1/n)^p > + (2/n)^p + ... + (n/n)^p]*(1/n) to 1/(p+1) is uniform on [p1, oo) for > some p_1>-. Since [(1/n)^p + (2/n)^p + ... + (n/n)^p]*(1/n) decreases > with p, 1/(p+1) is continuous for p>-1 and [p_0, p_1] is compact for > every -1 a theorem somewhat similar to Dini's theorem). Therefore, the > convergence is uniform on [p_0, oo) for any p_0 > -1. Artur, I'm not sure how you are letting n and p go to infinity together, nor do I understand for a fixed n, the sum approaches 1/n when n-> infinity, so I don't really understand your argument. Here's a different argument that gives some idea how fast f_n(p) -> f(p). Suppose p > -1. Then |f_n(p) - f(p)| = |[(1/n)^p + (2/n)^p + ... + (n/n)^p]*(1/n) - 1/(p+1)| = |sum_(k=1,n) int_[(k-1)/n, k/n] ( (k/n)^p - x^p ) dx| <= sum_(k=1,n) |int_[(k-1)/n, k/n] ( (k/n)^p - x^p ) dx| The first term in the last sum is |p/(p+1)|/n^(p+1). For the other terms we can use Lemma: If f is C^1 on [a,b], then |int_[a,b] (f(b) - f(x)) dx| = |int_[a,b] f'(x)(x-a) dx| <= (b-a)*int_[a,b] |f'(x)| dx Proof: The = comes from integration by parts; the <= is easy. So |f_n(p) - f(p)| is bounded above by |p/(p+1)|/n^(p+1) + sum_(k=2,n) (1/n)*int_[(k-1)/n, k/n] |p*x^(p-1)| dx. The sum is (1/n) times the integral of |p*x^(p-1)| over [1/n,1], which we can evaluate. So |f_n(p) - f(p)| <= |p/(p+1)|/n^(p+1) + |(1/n) - (1/n)^(p+1)|. That gives uniform convergence on [p_0,oo) for any p_0 > -1. === Subject: Definite integral -----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 It is well-known/can be proven that for example: Integral(dx/((x+1)(x+4))) = (1/3)ln((x+1)/(x+4)) + C Now, obviously we can only use this in definite integration to find the area under the graph y = 1/((x+1)(x+4)) when -4 < x < -1, otherwise the ln((x+1)/(x+4)) returns a complex value. We were told when taught this area of maths that we could use ln(|(x+1)/(x+4)|) (where |a| is the absolute value of a (i.e. -a when a < 0 and a when a >= 0)), if we wanted to use this integral to find the area under the graph in the region -4 < x < -1. This is also relatively clear when one plots graphs of the aforementioned functions. However, this step was never formally justified to us. Perhaps someone could show me a formal justification that using the absolute value function in this way does lead to the correct area in the region -4 < x < - -1? - -- Webmaster: http://www.dx-dev.com DXOS (My Operating System): http://www.dx-dev.com/dxos I contend that we are both atheists. I just believe in one fewer god than you do. When you understand why you dismiss all the other possible gods, you will understand why I dismiss yours. - -Stephen Roberts -----BEGIN PGP SIGNATURE----- Version: 6.5.8ckt http://www.ipgpp.com/ Comment: KeyID: 0x779D0625 Comment: Fingerprint: 5642 852D 88BD BD0B 1A88 94B0 BFD2 C47C 779D 0625 iQA/AwUBP32enb/SxHx3nQYlEQJuuACgtaSwnKZG3+ ogcM19Re84jglkyewAoPBm nwMdrd1myApWS84sb6XRm1vA =EzAr -----END PGP SIGNATURE----- === Subject: Re: Definite integral > It is well-known/can be proven that for example: > Integral(dx/((x+1)(x+4))) = (1/3)ln((x+1)/(x+4)) + C > Now, obviously we can only use this in definite integration to find the > area under the graph y = 1/((x+1)(x+4)) when -4 < x < -1, I supose you mean, you can use the formula for x<-4 or x>-1 > otherwise the > ln((x+1)/(x+4)) returns a complex value. We were told when taught this > area of maths that we could use ln(|(x+1)/(x+4)|) (where |a| is the > absolute value of a (i.e. -a when a < 0 and a when a >= 0)), if we wanted > to use this integral to find the area under the graph in the region -4 < x > < -1. This is also relatively clear when one plots graphs of the > aforementioned functions. However, this step was never formally justified > to us. Perhaps someone could show me a formal justification that using the > absolute value function in this way does lead to the correct area in the > region -4 < x < - -1? In this case integration means you use the Riemann-integral, which in fact is finding a function F to a given f, where f is the derivative of F. This is because finding F means, that Integrate_{a to b} f dx = F(b)-F(a). Now you just has to prove, that the absolute value function fulfills this. Suppose f is a continuous derivatable function. What is d ln|f(x)|/dx at point x0? 1) at f(x0)=0: In this case ln|f(x)| is not defined, hence this is not interesting. 2) at f(x0)>0: In this case there is a whole interval around x0 where f is positive hence on this interval ln |f(x)|=ln f(x). Then the first derivative according to chain rule is 1/f(x)*f'(x)=f'(x)/f(x). 3) at f(x0)<0: Once again there is a whole interval around x0 where f is negative hence on this interval ln |f(x)|=ln (-f(x)). According chain rule at differentating leads to 1/(-f(x))*(-f'(x))=f'(x)/f(x). This means that the derivative of ln |f(x)| is f'(x)/f(x) everywhere f(x)!=0. HTH D.8enes === Subject: Re: Definite integral Two facts are needed. The function 1/x, in the region 0 < x, has indefinite integral log(x). The function 1/x, in the region 0 > x, has indefinite integral log(-x). It is convenient for calculus textbooks to abbreviate these two as: integral 1/x = log|x| . But it really means the two previous facts. === Subject: Closed form determinant of Toeplitz-Hessenberg matrix I would like to get the determinant of a matrix typified by (n=4) 1/2! 1/4! 1/6! 1/8! -c 1/2! 1/4! 1/6! -c 1/2! 1/4! -c 1/2! where missing entries are zero and c is a symbolic expression, for arbitrary n. This is both Toeplitz and Hessenberg (TH). There is a closed form for dets of Vandermonde matrices. Is there a similar result for === Subject: Re: A Number Theory Problem >> Fix k. The expression is now an irreducible >> polynomial in n. It is an easy exercise >> to show that any such polynomial is >> prime infinitely often. >Hmmm. Isn't n^2 + 1 an irreducible polynomial >in n? Is it an easy excercise to show that it is >prime infinitely often? For that matter, n^2 + n + 2 is irreducible, and that's only prime for n=-1 or 0. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: To prove psychic powers, someone has to sit this test! > [...] > I have no idea. > I think you're saying that you can divine the identity of a writer from > the message; right? > If I was in test conditions I make a post to some groups, then I'm isolated away from > anything, then I'm given the list of replies and a seperate lists of names, then I > *guess* the peoples names. The paranormal happens when people reply to me, > I'm only using analytic ability to work out the link from the post to the name. > Name Comment Rarety > Rich Shewmaker Go see James Randi 1000 to 1 > Rust attack the knife hand 100 to 1 > See You In Hell It all depends 50 to 1 > Prodigy You mispelled kid 100 to 1 > Odysseus the trip is 64 light years 30 to 1 Please explain how you determined the odds/probability/rarety [sic] of these posts. The ONLY way to determine the odds of selecting the correct responded by chance is as you've done it below. Are you telling us that you correctly selected the names of the authors (Rich, Rush, SYIH, Prodigy and Odysseus) from, respectively, 1000, 100, 50, 100, 30 different responses to each of your posts? If you are not - you need to rethink how you are determining the probabilities of your observed effect. > Why cannot a rudimentary measure be given of the suggested property? > for example like I tried here > ----3--------------------------------------------------------- --------------- ---- > If ever I actually found myself in that situation, I'd hold it upright, > with the intent of attacking my assailant's knife hand. > ----------------- > A cliff86 > B Rust > C Shanx > D NormDePloom > E Rich Shewmaker > F CNote > G Jeff > H See You In Hell My Friend. > I Someone > J Greg Neill > A sample of people try the question, the highest answer is B Rust, (WHICH IS CORRECT) > There were 10 options, so the measure of coincidence is <= 1/10. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! > > [...] > I have no idea. > I think you're saying that you can divine the identity of a writer from > the message; right? > If I was in test conditions I make a post to some groups, then I'm isolated away from > anything, then I'm given the list of replies and a seperate lists of names, then I > *guess* the peoples names. The paranormal happens when people reply to me, > I'm only using analytic ability to work out the link from the post to the name. > Name Comment Rarety > Rich Shewmaker Go see James Randi 1000 to 1 > Rust attack the knife hand 100 to 1 > See You In Hell It all depends 50 to 1 > Prodigy You mispelled kid 100 to 1 > Odysseus the trip is 64 light years 30 to 1 > Please explain how you determined the odds/probability/rarety [sic] of > these posts. The ONLY way to determine the odds of selecting the > correct responded by chance is as you've done it below. > Are you telling us that you correctly selected the names of the > authors (Rich, Rush, SYIH, Prodigy and Odysseus) from, respectively, > 1000, 100, 50, 100, 30 different responses to each of your posts? If > you are not - you need to rethink how you are determining the > probabilities of your observed effect. I haven't, but if my memory was blanked of them I am quite sure I could do so. And I'm quite sure anyone here could. I've done some tests on people with small samples, even when the match is weak, these matches are strong so that implies it would work on a larger sample. > Why cannot a rudimentary measure be given of the suggested property? > for example like I tried here > ----3--------------------------------------------------------- --------------- ---- > If ever I actually found myself in that situation, I'd hold it upright, > with the intent of attacking my assailant's knife hand. > ----------------- > A cliff86 > B Rust attacking my assailant's knife There must be a way to measure this coincidence! > C Shanx > D NormDePloom > E Rich Shewmaker > F CNote > G Jeff > H See You In Hell My Friend. > I Someone > J Greg Neill > A sample of people try the question, the highest answer is B Rust, (WHICH IS CORRECT) > There were 10 options, so the measure of coincidence is <= 1/10. Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > > Are you telling us that you correctly selected the names of the > authors (Rich, Rush, SYIH, Prodigy and Odysseus) from, respectively, > 1000, 100, 50, 100, 30 different responses to each of your posts? If > you are not - you need to rethink how you are determining the > probabilities of your observed effect. > I haven't, but if my memory was blanked of them I am quite sure I could do so. Since we cannot test to see if you can have your memory blanked, this assertion is worthless. This is *also*, I believe, the difficult that Wally et al are having with this test being attempted in usenet. There are just too many ways someone can devine the authors names from a post by other methods easier than paranormal. > And I'm quite sure anyone here could. I've done some tests on people > with small samples, even when the match is weak, these matches are strong > so that implies it would work on a larger sample. OK. I'm prepared to set up a test. Why does it HAVE to be on usenet? Why could I not post 10 of your comments on a webpage and then have you attempt to match the authors of the first replies to those posts? I am proposing that the respondents email to me and *I* strip out the names and repost your initial post and the first response to it for you to guess the author for a list of four names. THis is in essence what you are porposing on usenet - Would this work for you? > Why cannot a rudimentary measure be given of the suggested property? > for example like I tried here > > ----3--------------------------------------------------------- ------------ ------- > If ever I actually found myself in that situation, I'd hold it upright, > with the intent of attacking my assailant's knife hand. > > ----------------- > A cliff86 > B Rust attacking my assailant's knife There must be a way to measure this coincidence! > C Shanx > D NormDePloom > E Rich Shewmaker > F CNote > G Jeff > H See You In Hell My Friend. > I Someone > J Greg Neill > A sample of people try the question, the highest answer is B Rust, (WHICH IS CORRECT) > There were 10 options, so the measure of coincidence is <= 1/10. You misrepresent this test. Only one person attempted the guess and that was from a choice of TWO respondents. A 50/50 chance that he admitted to throwing a coin to choose. As an Aussie you would be aware that you can make money at two-up - and that is less than 50/50 odds. If he had chosen correctly from 10 choices I'd be more impressed, but even with 50/50, with enough trials you can still measure statistically significant variation from chance, admittedly you'd have to get 10 from 10, but it's still measurable. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! > Look, I appreciate your time, As I've replied to this as part of a test proposal to you as you requested. > You're off on a tangent then going back on your word. > Considering the only promise I made was Tell you what - answer the > question asked and I will have a go, I have not *reneged*, as you had > not answered the question. > what is the relationship -- 3 times, none > how does this prove paranormal -- 3 times - this case it doesn't OK - so these are the types of posts you would ignore in your test, or would they be included in your data set? This is one of the problems - YOU get to select which posts are suitable. At the outset, to test if there IS anything to your claim, I'd suggest that you don't drop any posts that don't suit your mood. YOu could try to devise some agreement in your test proposal that allows you to choose which posts you'd try to answer (much like dowsers etc get to choose their success target etc). You'd have to do this before the list of posters was put to you though, and the posts couldn't be stored somewhere that a search engine could interrogate. > > And *you* bang on about going off on a tangent! Where the hell did > that quote come from. *I* was talking about the first three responses > to you in this thread and how their content can be connected to their > author and here's *you* rabbitting on about some post you made back on > the 4th May? > Wally's reply this post is not significant because he has already replied to me elsewhere. > In his 1st post you CAN spot his name. Think of a name for 'see my kid'. Disregarding the fact that you early said that there is no connection whatsoever, I'm afraid my powers aren't as honed as yours and cannot make *any* connection between 'see my kid' and Wally Anglesea. > The claim is not ALWAYS, why do all skeptics take the extreme interpretation of > any claim then set a test based on that? There are more scientific methods like > trying to isolate the observed effect. Trying to observe the effect in the first place seems to be the problem here. Your example of 6 *possible* chances to guess a poster's name from 50 posts does not demonstrate any observable effect. Your claim of success in 50 posts is misguided. If the other respondents here were to give you any leeway, the only way for you to show an effect would be to correctly guess the poster's names ONLY from the 6 you identified as being good possibilities. The problem with this, and you must see the other's points of view on this Herc, is that in order for you to make the selection of the 6 *possibles*, you would need to know in advance, the poster's name. The only other option is to have someone else compile the posts and the selection of poster's names for each, let's say 50 and give you the choice of which one's you thought had good connections. THEN and only then could the target for success be determined for that particular trial. In your scenario of 6 good posts and a choice of 24 posters (4 different posters for each post) you would need to get at least 5 out of 6 correct (from your table). For a first run, I'd even settle for 4 right. > You accuse me of going back on my word? Let's talk about honesty and > put back the paragraphs that you snipped before your 1st reply I put > up that you are renigging on > one > of the reasons is it is impossible for the claim to hold for repeated posts to me from > the same person, reason should be obvious, 2 of these posters this is not their 1st reply > and the claim is for the 1st reply. > I've Googled the thread, both in a.f.a-b and here in s.s., these are > the first replies from each of the posters on this thread. PLease > post a message ID contradicting this. > As I said - please post the message ID that contradicts the statement > that the three responses I listed as the first responses by the > authors to your initiating post of this thread is incorrect. > Viewing message === > Subject: Re: LOOK my claim is > Original Format This is a totally different thread, you know full well that I am talking about the first replies to this thread. Might as well drop this part of the discussion as it is going nowhere. > Its intermitent, I selected HALF of all replies to me from 1st Jan to 28th Feb this year > and > Sounds like data mining to me. > No, 50 posts in 60 days is significant, because there were only 100 total. If I roll a 6 on > a dice 50 out of 100 times its significant to billions to one against chance, that is exactly > the odds I am paralleling. > The two examples above are in no way connected as mathematical > examples of what you are doing. > You haven't seen my 50 question test, it has 4 options but could easily use 6. Then > it uses the same binomial parameters as the example above. > If a sample of people can guess 26 out of 50 right with 4 options, its outside normal range, > i.e. it shows a *correlation* between the names and posts. If there were 26 out of 50 correct then yes, there would be something significant to look at, but, as Wally and others point out, these posts can too readily be Googled now. To get anywhere with this you need to drop these posts as examples of anything and get someone to start a private, secure site where you can post to and people can respond. See my other post for this proposal. > The equivalent example of your dice rolling experiment as an analogy > of your usenet post data-mining is more like: > If I roll a die 100 times (the number of posts you quote), then ignore > 50% of the throws that don't equal 6, based on simple averages, it's > probable that 16 or 17 of the remaining 59 results will be 6. The > Astonishing odds of this refined set are no longer 1 in 6, but 17 > in 59 that you have rolled a 6, about 29%. Note though, that a 29% > success rate of 6s is not significant compared to the normal 18% > chance, as you've eliminated results that don't fit your required > true, but selecting every second post is very light data mining, its not > necessary I could use the whole 100 but the test would be longer than > necessary. I assumed a score of 12 out of 50 for the remaining 50, Your assumption would appear to be incorrect, you said yourself that, for the observed effect to be significant you would need to score at least 26/50. > you shouldn't do worse than random, then took the expected range for > 100, and I think the expected range is only slightly higher at 29/50. If by slightly you mean 43/100. In your 4 option scenario, a score of 29 is significant for 60 trials. Regardless, you cannot use the probability score based on 100 trials when you only do 50. > So correct, I would be satisfied if people scored 26 out of 50 just to > demonstrate a correlation, but strictly you need 30 out of 50 to prove > the entire 100 sample is outside normal range. You cannot use the probabilities for success on 50 trials (it's 26, not 30) and extrapolate that score for the 100 trials. Remember you are IGNORING 50 of these trials as unsuitable for your purposes. In order to demonstrate a significant effect greater than chance you have to apply the test to ALL 100 trials AND achieve a minimum score of 43 correct from 100 sample posts. > Engineering techniques > allow you to remove up to several % of data points to gather statistics, Since you're speaking to an engineer that spends a great deal of his time working on performance bencharking and the like, I'd like to see that process Herc. I could have improved this year bonus no end! > I'm using half the data, if not most, so the difference is only a few questions. No, the difference between testing a sample of 50 and a sample of 100 is a minimum success rate of 26 and 43 respectively. 17 questions is not only a few. > Not satisfied then I can make it a 100 question test covering every reply to me > from 1st Jan to 28th Feb, same result, just half the questions are duds. I'm quite prepared to accept you dropping 50% of the sample - you just need to remember to apply the 50 trial odds to it and NOT extrapolate to the entire 100 posts, half of which were not tested. > result. You then go further and say that Only 6 or 7 of those could > you guess the name from 100 sample though. So *now* you've moved the > goal posts to the equivalent of saying that 6 or 7 in 100 is > significant (ie 7%), but by datamining you can, on average, have a > success rate of 29% if the result is throwing a 6. > yes but the equivalent would be a 100 sided dice for those posts. No, it wouldn't. By your rules, the odds are determined by the list of posters you can choose from. The only way you can use your 100-sided die example is if you gave 100 names to choose from. > Like Rich Shewmaker, I've seen people guess it from a small list, > and they've agreed the experiment should be for a longer list. > I'm quite confident I could post Rich's comments in 100 groups and > a dozen would pick his name from 100 random names. Is there anything significant about Rich's name that would make his posts more readily associated with his posts? Just curious on what your post select process is? > Wally's name was a weak connection, one of the 1 in 6 coincidences, the 2 > other names had nothing to do with 'try it on my kid', but trying to > make meaning out of anglesea only comes down to a few things, one > of which is the phonetic play angel see, and the idea of calling a kid an angel. OK, I understand the word play you're at - the problem really is that totally ignore your claims. You need to get these test posts stored at a site where you cannot possibly search for the posters name and someone else builds the list of possible choices. > You don't want to sit these tests, if they were part of an iq score for some club > everyone would just try to guess them. you have a go, you see a link, no other > links, maybe thats right. subjective but still empirically testable. Just not in an arena that search engines can be employed upon to divine the true poster. You've got to get someone to volunteer to host a secure guestbook where sample replies to your posts can be stored where YOU cannot devine the poster's names before you apply your powers to. > 50 out of 100 total posts were significant, (could be guessed with 10 options) > THAT 50% > 6% could be guessed with over 100 options. > 44 good, 6 excellent out of 100 OK, read it properly > So, 6% could be guessed readily, then. How is this significant? > expected range for 1% chance (guess from 100 random names) of 100 tries is under 6. No it is not. In 100 trials, with 10 options for each trial, the significant number of successes (from http://www.automeasure.com/chance.html ) is 24%. Have a look at that table again. 10 choices - a 1 in 10 chance of success. In 100 trials, the expected chance range is 0-23. > also its 50% could be guessed readily, 6 percent could be guessed easily. In this case you have to decide whether you are going to do the trial against 100, 50 or 6 samples. If you decide that of the 100 samples only 6 give you any chance of success, then the trial can only be done on those 6. With 10 options open to you you'd have to succeed 5 out of 6 attempts for your result to be significant. If you plumped for 50 trials with 10 choices, your result would need to be 15 or better and for 100, 24 or better. Review that table you quoted, you are making serious errors with your expectations. > > I suggest you brush up on your statistics - in the meantime stay away > from racetracks and casinos. > Get a qualification in statistics before you make half dozen errors in a single post again. > Let's see how you go with your example below, shall we? > For now, Jerome's [Herc's] claim of power goes unproven. > This is by Rich Shewmaker > any idea why? > of course you have, hE ricH our king > Uh? Where does hE ricH our king come from in the above post, and how > does it relate to the posters name? I *really* cannot see the > connection your making here. > Eric H. ricH rich > I post the rich showmaker example asking for the connection, > you reply regarding the connections with the name THAT'S THE ANSWER - HE RICH! > You already KNEW I was going to respond to the post because you > *addressed* it to me in the newsgroup as an example of the > word-numerology thing you're determined to prove! > You then go out and Google a post with a tenuous anagram of my name, > post it and then say that this is proof of your paranormal talent > because I responded? ! ! ! You CANNOT be SERIOUS > > no that quote (rich shewmaker) was the post you replied to. You already knew who was going to respond to the damned post - you directed it at me! Can you not see this point? Because you COULD have searched for a post that you had already worked out had a connection to my name and posted it knowing that I would respond, others less generous than myself COULD point to this as a fairly disingenuous method to force connections that are NOT random. My replying to the post you put forward as an example of connection was NOT a random choice of mine, you knew BEFORE the fact that I would reply to it, so could selected a post that you had already engineered a connection to. If you cannot see the point I'm trying to make here you will not see the point other's are making with regard to your data to find the author's name. > 02 02 2002 I predicted paranormal proof, then 1 year later... Your connection process get's more tenuous by the minute Herc > The reason I'm using this example is people have already spotted it from a list in sci.math, > its 1000 to 1 post in itself, > Why? > because Rich Shewmaker is an EXTREMELY odd name for someone to write > go and see James Randi. That does not determine the 1:1000 odds you quote. How did you determine these odds. > the measure_of_coincidence is proportional (somewhat) to the number of names it > can be picked out from. So there were 1000 replies from 1000 individual posters on sci.math to your single post was there? This is the only way you can apply this logic to get your 1/1000 odds Herc. === > http://tinyurl.com/pd9g Subject: proof of god 0202 2002 (& premonition) > Where in that muddled rant is the shuttle alluded to? > the date, one year to the day before a disaster, and the word premonition, Herc, have you MET $ollog and N0str0damus? > not worth going in to, the fact is I put the date 02 02 2002 and proof of god > in the subject of half dozen posts in major newsgroups one year before the 3 paranormal What was so paranormal about the replies - and you only got 3 paranormal ones from 6 posts? Using your own chart of probablilities you'd need a MINIMUM of 50 replies to each of your 6 posts for 3 of them to statistically significant. Or were you not doing your author/post matching magic trick with these posts? Even if you were, what has this to do with your prediction? > I ALSO posted in alt.magic the day before that I would do magic with the newsgroups. Since your 3 replies to 6 posts is only significant if 50 individuals replied to each of your posts, I presume then that this is the case. If not, where do you derive your proof that you did magic with the newsgroups. > So not only can you measure the coincidence in the 3 replies on 02 02 2002 I clearly > predicted there would be proof. No, not clear at all... > With a little thought, you could have guessed the name! > But not using that logic Herc, and anyway, I thought you said that as > this was Wally's first reply, that why it COULD NOT WORK this thread, > simple hey?. > It could not work on this reply of Wallys What has *that* got to do with my question? And why are you introducing a new post? > in this thread that you posted up : Uh, no, sparky. You have introduced this tangent (remember that statement of yours earlier, Herc?) to this thread. > Wally Anglesea™ > because THIS was Wally Angel See's first reply to me some time ago : You're all over the shop today. The question was not related to a reple some time ago but to the first 3 replies in this interminable thread. > So if I get one of my kids to post under my name- because my name is > obvious (and only one kid, always the same one), you will be able to > guess *his* or *her* name, right? not the surname, but the first and > middle names? > Answer your question? No. But it certainly confirms my thoughts on your style of discussion and why you're having such trouble proving your paranormal capabilities. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! === > http://tinyurl.com/pd9g Subject: proof of god 0202 2002 (& premonition) > Where in that muddled rant is the shuttle alluded to? > the date, one year to the day before a disaster, and the word premonition, > Herc, have you MET $ollog and N0str0damus? > not worth going in to, the fact is I put the date 02 02 2002 and proof of god > in the subject of half dozen posts in major newsgroups one year before the 3 paranormal > What was so paranormal about the replies - and you only got 3 > paranormal ones from 6 posts? Using your own chart of probablilities > you'd need a MINIMUM of 50 > replies to each of your 6 posts for 3 of them to statistically > significant. Or were you not doing your author/post matching magic > trick with these posts? Even if you were, what has this to do with > your prediction? OK I see the problem. These 3 happen to be in the 50 test I keep mentioning. But it is its own proof, I didn't mean to mention the 50 at all, just these 3. You must have looked up there were 6 replies to those threads or something? The fact is this is a much simpler test, the only data mining is I selected ONE PARTICULAR DAY, the 3 all happened on 02 02. Plus I predicted this day, one coincidence, anyone could do that. I pinpointed the date before hand, then got 3 better than 100 to 1 posts. Its a million to 1. > I ALSO posted in alt.magic the day before that I would do magic with the newsgroups. > Since your 3 replies to 6 posts is only significant if 50 individuals > replied to each of your posts, I presume then that this is the case. > If not, where do you derive your proof that you did magic with the > newsgroups. > So not only can you measure the coincidence in the 3 replies on 02 02 2002 I clearly > predicted there would be proof. > No, not clear at all... Its in alt.magic in the same thread as Rich Shewmaker, someone says use rec.knives as the group, the subject is I'd like to do a magic trick here, and I say, don't follow me to rec.knives saying I'm doing a trick. Its also implied from the PROOF OF GOD 02 02 2002 posts I made one year to the day before. Anyway, there was a reply next to Rust and See You In Hell named Chas, who talked about knife chases. Here are the 2 times I tried to do it live with an audience monitoring, both times it worked, but only to small level of coincidence. These are parts of my attempts to prove a power on the internet. The claim is so remarkable only some people say they can see it. When people pass a message to me I can see it in their name. Here are some tests I began to show it live. Anyway I got 3 replies to my rec.gardens post, no comments just give the answer like 1a, 2c, 3b Put vermiculture in your browser and you'll get all kinds of info and worms, worm bins, compost, etc. and where you can order them, if you so choose. > You can't get worms at garden stores, is there much to know > about using them? > Herc >You can't get worms at garden stores, is there much to know >about using them? Gee, I don't think so. You just turn them on, and they play! Gardens Alive might have red wigglers > You can't get worms at garden stores, is there much to know > about using them? > Herc a/ Jonathan Sachs b/ madgard c/ Valkyrie _____________________________________________________ All right, I'll play. I'll try to do it properly, although frankly I think the whole thing is nonsense. No googling, of course. I *will* include some comments, but I'll add some spoiler space in case anyone is really wanting to give it a go. My guess is 1c, 2a, 3b 'Reasons' follow... . > Put vermiculture in your browser and you'll get all kinds of info and > worms, worm bins, compost, etc. and where you can order them, if you so > choose. 'Vermiculture' is the first significant word, so Valkyrie seems to fit. >You can't get worms at garden stores, is there much to know >about using them? > Gee, I don't think so. You just turn them on, and they play! (okay, I chose this one mostly because the other two were taken) > Gardens Alive might have red wigglers First word is 'Gardens', so 'madgard' seemed appropriate Right. Now, off to Google to see if I was right. Pretty good, Herc! I was spot on with my guess. Danny __________________________________________________ all I did was make 50 hyperlinks to most the replies to me at to about 50 just from replies this year. There is an empirical statistical anomaly. See You In Hell My Friend: Rust: Chas: This is a subset of 3 replies I did as part of a live test, to measure how I defeat random chance you take a small subset from the 50 replies and present the message stripped of the signature to a person or people, and they should be data of quantum entanglement operating on a macro scale, i.e. coordinating how people interact with me. >can have an unbiased audience first. Just the name of one >rec newsgroup, and the magic will unfold. > What the hell... > rec.knives good for you young man, now while I make a post to rec.knives I must ask the audience for absolute silence, i.e. don't go there spouting I'm doing a trick! Herc > rec.knives Ok, we'll just leave the suggested group up there for everyone to see. Now I received 2 replies in rec.knives and I want to present it as a puzzle for all of you to see. No need to go to rec.knives just yet, but you can go there later to check, I wan't you all to have The first comment was : 1/ It really all depends on the situation. and the second comment was : 2/ If ever I actually found myself in that situation, I'd hold it upright, with the intent of attacking my assailant's knife hand. Brutal, shear brutal. Now the contenders for these replies are : A/ Rust B/ See You In Hell My Friend A rather long handle on contender B, but remember you must make a it 1A 2B, or was it 1B 2A? Take your time to think about it, this one took me a few minutes to work out. Take handle B for instance, what kind of scenerio does 'see you in hell' imply? What does rust do? Herc > Okay... Now What? Look here ladies and gentlemen, a kind volunteer from the audience to take part in the live magic before you today.... that must be hard to maintain live!! Well, without looking at rec.knives what's your guess? 1A 2B or 1B 2A that is, did Rust write 'it all depends' and See You In Hell write 'attacking my assailant's knife' OR did Rust write 'attacking my assailant's knife' and See You In Hell write 'it all depends' No looking ladies and gentlemen, watch as a volunteer from the audience, we've never met before have we young man? makes a Now for your guess Trent! Herc Is that because rust attacks metal? and going to hell really does depend? You saw it here live folks ! magic in front of your eyes ! here is one of the real posts to confirm... - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - If ever I actually found myself in that situation, I'd hold it upright, with the intent of attacking my assailant's knife hand. -Rust Trent was right! BELIEVE IT OR NOT Herc > Yay! Bravo! > The only possible explanation is, of course, that I have latent psychic > powers that allow me to be 100% accurate 50% of the time when I have to > choose from two choices! > Seriously though, good work. I want more Usenet magic! ------------------------------------------------------------ === Subject: Re: To prove psychic powers, someone has to sit this test! === > http://tinyurl.com/pd9g Subject: proof of god 0202 2002 (& premonition) > Where in that muddled rant is the shuttle alluded to? > the date, one year to the day before a disaster, and the word premonition, > Herc, have you MET $ollog and N0str0damus? > not worth going in to, the fact is I put the date 02 02 2002 and proof of god > in the subject of half dozen posts in major newsgroups one year before the 3 paranormal > What was so paranormal about the replies - and you only got 3 > paranormal ones from 6 posts? Using your own chart of probablilities > you'd need a MINIMUM of 50 > replies to each of your 6 posts for 3 of them to statistically > significant. Or were you not doing your author/post matching magic > trick with these posts? Even if you were, what has this to do with > your prediction? > OK I see the problem. These 3 happen to be in the 50 test I keep mentioning. Yes you seem confused - I asked what was paranormal about the replies. So, what *was* paranormal about the replies? > But it is its own proof, I didn't mean to mention the 50 at all, just these 3. > You must have looked up there were 6 replies to those threads or something? Haven't bothered to look up the thread, I''m only attempting to understand the assertion you are making in this thread. In the above I, you said from 100 posts you selected 50, 6-7 of which could be guessed readily. I'm only analysing your comments so far. My comment relates to your chance calculating table which indicates that for your 3 posts to be significant, you would have had to choose the correct author from 50 choices to each post. You quoted that table as the one you wish to determine a statistical significance to your effect, I am only playing by your rules. I suggest that you reconsult that table to see what *I'm* talking about. > The fact is this is a much simpler test, the only data mining is I selected ONE > PARTICULAR DAY, the 3 all happened on 02 02. Plus I predicted this day, to > one coincidence, anyone could do that. For the types of guessing you are doing on these posts, 1000 posts does not make the odds 1/1000. It is the number of replies that you are divining the author's names from that make up the odds of guessing. ie 4 responses = 1/4 chance of success. > I pinpointed the date before hand, then > got 3 better than 100 to 1 posts. Its a million to 1. No it is not. You have not established that the chance of guessing the author's names from the 3 posts is 100 to 1. For the odds of matching the author to his/her post to be 100 to 1, you would have to be choosing from a list of 100 authors. As I've said, until you understand this, the numbers you quote as odds will be refuted or ignored by anyone you want to be interested in testng you. > Its in alt.magic in the same thread as Rich Shewmaker, someone says use rec.knives > as the group, the subject is I'd like to do a magic trick here, and I say, don't follow > me to rec.knives saying I'm doing a trick. > Anyway, there was a reply next to Rust and See You In Hell named Chas, who talked > about knife chases. A simple 50/50 chance and the one person who attempted it said he used a coin to decide. Again, for significance in a 50/50 chance (using your table) you'd need to get it 10 out of 10. In 20 posts you'd have to get 17 or better for it to be significant. > Here are the 2 times I tried to do it live with an audience monitoring, > both times it worked, but only to small level of coincidence. OK, two trials. First one with 50/50 odds the second with 1/3 odds. Even though they succeeded this is *still* not significant as it is still in the bounds of chance that someone could manage it. If you could demonstrate it for 10 posts, we'd be getting somewhere. > Pretty good, Herc! I was spot on with my guess. > Danny > The only possible explanation is, of course, that I have latent psychic > powers that allow me to be 100% accurate 50% of the time when I have to > choose from two choices! Herc, you SERIOUSLY have to come to grips with probability. At the very least consult that Laws of Chance Table URL you quote. Give two alternatives, your chance of choosing correctly is 50% This is the same as saying you will be 100% accurate 50% of the time. There is nothing latent, psychic nor powerful about this. Flip a coin 50 times and chances are you'll be 100% accurate 50% of the time. For this to be significant you would have to get 39/50 correct, not 25. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! > With a little thought, you could have guessed the name! > But not using that logic Herc, and anyway, I thought you said that as > this was Wally's first reply, that why it COULD NOT WORK this thread, > simple hey?. > It could not work on this reply of Wallys > What has *that* got to do with my question? And why are you > introducing a new post? Wally only has one theme to his name, and he's used it up. Now his destiny to speak of seeing children is over and he has free will again. > in this thread that you posted up : > Uh, no, sparky. You have introduced this tangent (remember that > statement of yours earlier, Herc?) to this thread. > Wally Anglesea™ > > because THIS was Wally Angel See's first reply to me some time ago : > You're all over the shop today. The question was not related to a > reple some time ago but to the first 3 replies in this interminable > thread. for 10 times now, 1st replies EVER only. > So if I get one of my kids to post under my name- because my name is > obvious (and only one kid, always the same one), you will be able to > guess *his* or *her* name, right? not the surname, but the first and > middle names? > Answer your question? > No. But it certainly confirms my thoughts on your style of discussion > and why you're having such trouble proving your paranormal > capabilities. Hence the breadth technique. Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > With a little thought, you could have guessed the name! > But not using that logic Herc, and anyway, I thought you said that as > this was Wally's first reply, that why it COULD NOT WORK this thread, > simple hey?. > > It could not work on this reply of Wallys > What has *that* got to do with my question? And why are you > introducing a new post? > Wally only has one theme to his name, and he's used it up. Now his > destiny to speak of seeing children is over and he has free will again. [...backs sloowly away from computer screen...] > in this thread that you posted up : > Uh, no, sparky. You have introduced this tangent (remember that > statement of yours earlier, Herc?) to this thread. > Wally Anglesea™ > > because THIS was Wally Angel See's first reply to me some time ago : > You're all over the shop today. The question was not related to a > reple some time ago but to the first 3 replies in this interminable > thread. > for 10 times now, 1st replies EVER only. Surely you're going to run out of respondents? So you're saying if we WERE to set up a test on usenet and any of the currrent participants current threads with you was one of the respondents you would not be able to guess their names from their posts? Am I getting this right? > So if I get one of my kids to post under my name- because my name is > obvious (and only one kid, always the same one), you will be able to > guess *his* or *her* name, right? not the surname, but the first and > middle names? > Answer your question? > No. But it certainly confirms my thoughts on your style of discussion > and why you're having such trouble proving your paranormal > capabilities. > Hence the breadth technique. I hesitate to ask. Please describe what you have dubbed the breadth technique. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! > With a little thought, you could have guessed the name! > But not using that logic Herc, and anyway, I thought you said that as > this was Wally's first reply, that why it COULD NOT WORK this thread, > simple hey?. > > It could not work on this reply of Wallys > What has *that* got to do with my question? And why are you > introducing a new post? > Wally only has one theme to his name, and he's used it up. Now his > destiny to speak of seeing children is over and he has free will again. Kooky! Might be time to break out the woo-woo guns. -=-=-=-=- === Subject: Re: To prove psychic powers, someone has to sit this test! The engine on my bulldozer just fried while plowing through your most recent wall of fertilizer. -- Felony case 02-CR-0617 9/1/03: Oregon Department of Justice V. Raymond Ronald Karczewski, Defendant. The defendant's name is NOT copyrighted. === Subject: Re: To prove psychic powers, someone has to sit this test! >> With a little thought, you could have guessed the name! >> But not using that logic Herc, and anyway, I thought you said that as >> this was Wally's first reply, that why it COULD NOT WORK this thread, >> simple hey?. >> >> It could not work on this reply of Wallys >> What has *that* got to do with my question? And why are you >> introducing a new post? >Wally only has one theme to his name, and he's used it up. Now his >destiny to speak of seeing children is over and he has free will again. Kook. >> in this thread that you posted up : >> Uh, no, sparky. You have introduced this tangent (remember that >> statement of yours earlier, Herc?) to this thread. >> Wally Anglesea™ >> >> because THIS was Wally Angel See's first reply to me some time ago : Actually it wasn't. IIRC I called you a kook long before I questioned you about having my kids post messages. You have selective memory it seems. -- Find out about Australia's most dangerous Doomsday Cult: http://users.bigpond.net.au/wanglese/pebble.htm You can't fool me, it's turtles all the way down. === Subject: Re: To prove psychic powers, someone has to sit this test! > The claim is not ALWAYS, why do all skeptics take the extreme interpretation of > any claim then set a test based on that? There are more scientific methods like > trying to isolate the observed effect. > Trying to observe the effect in the first place seems to be the > problem here. Your example of 6 *possible* chances to guess a > poster's name from 50 posts does not demonstrate any observable > effect. Your claim of success in 50 posts is misguided. If the other > respondents here were to give you any leeway, the only way for you to > show an effect would be to correctly guess the poster's names ONLY > from the 6 you identified as being good possibilities. The problem > with this, and you must see the other's points of view on this Herc, > is that in order for you to make the selection of the 6 *possibles*, > you would need to know in advance, the poster's name. The only other > option is to have someone else compile the posts and the selection of > poster's names for each, let's say 50 and give you the choice of which > one's you thought had good connections. THEN and only then could the > target for success be determined for that particular trial. In your > scenario of 6 good posts and a choice of 24 posters (4 different > posters for each post) you would need to get at least 5 out of 6 > correct (from your table). For a first run, I'd even settle for 4 > right. I don't see that, if I was mining or someone else was mining for the top 5% of replies, then I would have to have a much larger sample of choices, you could probably mine 5% of *your* replies and due to writing style you could pick the name from one in 4. The top 5% are like Rich Shewmaker, I estimate people could guess the name Rich from a random list of 20 names, and people could also guess the name Mr Shewmaker from a list of 20 names. Together Rich Shewmaker could easily be guessed from almost unlimited number of names. To my eyes the post is phenominal just by itself. You said you'd seen more extraordinary posts??? Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > The claim is not ALWAYS, why do all skeptics take the extreme interpretation of > any claim then set a test based on that? There are more scientific methods like > trying to isolate the observed effect. > Trying to observe the effect in the first place seems to be the > problem here. Your example of 6 *possible* chances to guess a > poster's name from 50 posts does not demonstrate any observable > effect. Your claim of success in 50 posts is misguided. If the other > respondents here were to give you any leeway, the only way for you to > show an effect would be to correctly guess the poster's names ONLY > from the 6 you identified as being good possibilities. The problem > with this, and you must see the other's points of view on this Herc, > is that in order for you to make the selection of the 6 *possibles*, > you would need to know in advance, the poster's name. The only other > option is to have someone else compile the posts and the selection of > poster's names for each, let's say 50 and give you the choice of which > one's you thought had good connections. THEN and only then could the > target for success be determined for that particular trial. In your > scenario of 6 good posts and a choice of 24 posters (4 different > posters for each post) you would need to get at least 5 out of 6 > correct (from your table). For a first run, I'd even settle for 4 > right. > I don't see that, This is obvious. > if I was mining or someone else was mining for the top 5% of replies, > then I would have to have a much larger sample of choices, you could probably > mine 5% of *your* replies and due to writing style you could > pick the name from one in 4. Herc, I have no problem with you selecting which posts you think are going to be the easiest for you to get a connection with. > The top 5% are like Rich Shewmaker, I estimate > people could guess the name Rich from a random list of 20 names, and people > could also guess the name Mr Shewmaker from a list of 20 names. And for it to be significantly better than chance, 4 out of 5 tries would need to be correct. This applies to EACH of your assertions, that is 4 out of 5 for guessing Rich and 4 out of and different 5 to guess on Shewmaker. > Together > Rich Shewmaker could easily be guessed from almost unlimited number of names. It's not neccesary for you to have unlimited options in the trials, merely repeated trials. You could do it with only three options, but for your effect to be greater than random chance, 10 out of 10 respondents would have to choose correctly. If you had 1000 options you could do it in 5 trials - you'd only need one respondent to get it right, although that's the same odds for a 10 trials. Even if you got TWO respondents choosing correctly, you'd have to run 80 trials before you ruled out random chance. > To my eyes the post is phenominal just by itself. Unfortunately it is not phenominal to people who understand calculating random chance and probability, and if you're going to lay your hands on JREF's $, you are going to have to demonstrate statistical significance and not just Hey I think it's keen. > You said you'd seen more > extraordinary posts??? I merely said that I've encountere mored extraordinary names than Shewmaker, but yes, now that you say it, I HAVE seen more extraordinary posts... -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! > Wally's reply this post is not significant because he has already replied to me elsewhere. > In his 1st post you CAN spot his name. Think of a name for 'see my kid'. > Disregarding the fact that you early said that there is no connection > whatsoever, I'm afraid my powers aren't as honed as yours and cannot > make *any* connection between 'see my kid' and Wally Anglesea. Sure you can. Imagine you're on the space ship enterprise, the Klingons have left a space mine trap and you have 3 directions you can take. 2 will lead to instant death of you and all your crew. The correct solution will get you out of the trap. Each direction to select is marked by a klingon vessel, one of which contains the klingon pilot who set the trap, the other 2 vessels are automated and selecting their path will ensure instant death. You try to open communications with each klingon vessel but only get the call sign of the operator, Wally Anglesea, Irony Alert, and Carl R. Osterwald. One is the real klingon operator and the other 2 are duds. What do you do? Roll a dice? Wait, your communications crew deciphered a klingon message from one of the vessels with the wording : Ha, the enterprise is doomed, the destruction of the enterprise will ensure my promotion in the Klingon Empire, my name will echo the top 40 klingon assasins if only the captain of the enterprise would know the power this victory will endow to my kid, that itself fulfills my ancestors name and our line will carry forth ever more... I think you'd make it. You're just in a skeptic frame of mind you can't do it for a proof, which follows on to you can't do invidual examples. Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > If a sample of people can guess 26 out of 50 right with 4 options, its outside normal range, > i.e. it shows a *correlation* between the names and posts. > If there were 26 out of 50 correct then yes, there would be something > significant to look at, but, as Wally and others point out, these > posts can too readily be Googled now. To get anywhere with this you > need to drop these posts as examples of anything and get someone to > start a private, secure site where you can post to and people can > respond. See my other post for this proposal. > The equivalent example of your dice rolling experiment as an analogy > of your usenet post data-mining is more like: > If I roll a die 100 times (the number of posts you quote), then ignore > 50% of the throws that don't equal 6, based on simple averages, it's > probable that 16 or 17 of the remaining 59 results will be 6. The > Astonishing odds of this refined set are no longer 1 in 6, but 17 > in 59 that you have rolled a 6, about 29%. Note though, that a 29% > success rate of 6s is not significant compared to the normal 18% > chance, as you've eliminated results that don't fit your required > true, but selecting every second post is very light data mining, its not > necessary I could use the whole 100 but the test would be longer than > necessary. I assumed a score of 12 out of 50 for the remaining 50, > Your assumption would appear to be incorrect, you said yourself that, > for the observed effect to be significant you would need to score at > least 26/50. > you shouldn't do worse than random, then took the expected range for > 100, and I think the expected range is only slightly higher at 29/50. > If by slightly you mean 43/100. In your 4 option scenario, a score of > 29 is significant for 60 trials. Regardless, you cannot use the > probability score based on 100 trials when you only do 50. > So correct, I would be satisfied if people scored 26 out of 50 just to > demonstrate a correlation, but strictly you need 30 out of 50 to prove > the entire 100 sample is outside normal range. > You cannot use the probabilities for success on 50 trials (it's 26, > not 30) and extrapolate that score for the 100 trials. Remember you > are IGNORING 50 of these trials as unsuitable for your purposes. In > order to demonstrate a significant effect greater than chance you have > to apply the test to ALL 100 trials AND achieve a minimum score of 43 > correct from 100 sample posts. Yes but I mine the posts before the tests, so the remaining 50 are still worth 12 right by averages. If a sample of people sat the test closed book and scored 43 out of 50, then a sample of those sit the ordinary 50 afterwards and would score another 12, making 55 / 100 getting multi choice 4 options right. That's more than a correlation thats a credit. > Engineering techniques > allow you to remove up to several % of data points to gather statistics, > Since you're speaking to an engineer that spends a great deal of his > time working on performance bencharking and the like, I'd like to see > that process Herc. I could have improved this year bonus no end! Austalia Standards for compressive concrete strength is one, usually where a small number of results can distort the mean which isn't related to this example. > I'm using half the data, if not most, so the difference is only a few questions. > No, the difference between testing a sample of 50 and a sample of 100 > is a minimum success rate of 26 and 43 respectively. 17 questions is > not only a few. I'm not mining out the wrong ones with 0 score, I'm mining out the average ones, which anyone could use a randomizer on to guess 12 more right. > Not satisfied then I can make it a 100 question test covering every reply to me > from 1st Jan to 28th Feb, same result, just half the questions are duds. > I'm quite prepared to accept you dropping 50% of the sample - you just > need to remember to apply the 50 trial odds to it and NOT extrapolate > to the entire 100 posts, half of which were not tested. I doubt that is correct. Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > You cannot use the probabilities for success on 50 trials (it's 26, > not 30) and extrapolate that score for the 100 trials. Remember you > are IGNORING 50 of these trials as unsuitable for your purposes. In > order to demonstrate a significant effect greater than chance you have > to apply the test to ALL 100 trials AND achieve a minimum score of 43 > correct from 100 sample posts. > Yes but I mine the posts before the tests, so the remaining 50 are still worth 12 right > by averages. No they are not. The remaining 50 trials were never performed and therefore do not have any play in this test. Oh, and for my information, how do you calculate 12 right by averages? Perhaps then we might be able to clear up your confusion with the laws of probability. > If a sample of people sat the test closed book and scored 43 out of 50, then a sample > of those sit the ordinary 50 afterwards and would score another 12, making The score on the first 50 trials has no bearing on the outcome of the second trial of 50. There is nothing that says 12 correct answers *would* be achieved, chance merely says that you could expect a score between 1 and 25 from random chance alone. > 55 / 100 getting multi choice 4 options right. That's more than a correlation thats a credit. It would be if you performed 100 trials. But to only conduct 50 and extrapolate is incorrect. I believe that this is what you are doing with the other examples you've given. You've had three successes and therefore presume (without testing) that any further results will be 100%. By this logic if you get three heads in a row in a game of two-up you claim that you should be able to ask the pit boss for the bank as you've clearly shown that you are going to clean them out. By the way - unless you're in to public humiliation I don't suggest you try this approach. ... oops, too late. -- Eric Hocking > Engineering techniques > allow you to remove up to several % of data points to gather statistics, > Since you're speaking to an engineer that spends a great deal of his > time working on performance bencharking and the like, I'd like to see > that process Herc. I could have improved this year bonus no end! > Austalia Standards for compressive concrete strength is one, usually where a small number of > results can distort the mean which isn't related to this example. > I'm using half the data, if not most, so the difference is only a few questions. > No, the difference between testing a sample of 50 and a sample of 100 > is a minimum success rate of 26 and 43 respectively. 17 questions is > not only a few. > I'm not mining out the wrong ones with 0 score, I'm mining out the average ones, which > anyone could use a randomizer on to guess 12 more right. > Not satisfied then I can make it a 100 question test covering every reply to me > from 1st Jan to 28th Feb, same result, just half the questions are duds. > I'm quite prepared to accept you dropping 50% of the sample - you just > need to remember to apply the 50 trial odds to it and NOT extrapolate > to the entire 100 posts, half of which were not tested. > I doubt that is correct. > Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > You accuse me of going back on my word? Let's talk about honesty and > put back the paragraphs that you snipped before your 1st reply I put > up that you are renigging on > one > of the reasons is it is impossible for the claim to hold for repeated posts to me from > the same person, reason should be obvious, 2 of these posters this is not their 1st reply > and the claim is for the 1st reply. > I've Googled the thread, both in a.f.a-b and here in s.s., these are > the first replies from each of the posters on this thread. PLease > post a message ID contradicting this. > As I said - please post the message ID that contradicts the statement > that the three responses I listed as the first responses by the > authors to your initiating post of this thread is incorrect. > Viewing message === > Subject: Re: LOOK my claim is > Original Format > This is a totally different thread, you know full well that I am > talking about the first replies to this thread. Might as well drop > this part of the discussion as it is going nowhere. You only get a limited amount of themes out of a name, but you can get many posts, so its usually the first one *ever*. When writing : 2 of these posters this is not their 1st reply I meant 1st reply ever. If the effect worked every thread then I would need 10 meanings for Wally Anglesea. Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > what is the relationship -- 3 times, none > how does this prove paranormal -- 3 times - this case it doesn't > OK - so these are the types of posts you would ignore in your test, or > would they be included in your data set? This is one of the problems > - YOU get to select which posts are suitable. At the outset, to test > if there IS anything to your claim, I'd suggest that you don't drop > any posts that don't suit your mood. YOu could try to devise some > agreement in your test proposal that allows you to choose which posts > you'd try to answer (much like dowsers etc get to choose their success > target etc). You'd have to do this before the list of posters was put > to you though, and the posts couldn't be stored somewhere that a > search engine could interrogate. These would be ignored, in the set of 100 they would make up 10 of the 50 I didn't use. You only get a limited amount of themes out of a name, but you can get many posts, so its usually the first one. The statistics is trivial, I'm just trying to show the effect to people. No use waving my arms - hey everyone look at this ordinary reply to me! Practically any kind of marking system will show that I can compete against other people and always score higher over a small amount of trials. Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > what is the relationship -- 3 times, none > how does this prove paranormal -- 3 times - this case it doesn't > OK - so these are the types of posts you would ignore in your test, or > would they be included in your data set? This is one of the problems > - YOU get to select which posts are suitable. At the outset, to test > if there IS anything to your claim, I'd suggest that you don't drop > any posts that don't suit your mood. YOu could try to devise some > agreement in your test proposal that allows you to choose which posts > you'd try to answer (much like dowsers etc get to choose their success > target etc). You'd have to do this before the list of posters was put > to you though, and the posts couldn't be stored somewhere that a > search engine could interrogate. > These would be ignored, in the set of 100 they would make up 10 of the 50 > I didn't use. You only get a limited amount of themes out of a name, but you > can get many posts, so its usually the first one. I'm willing to give you the chance to pick and choose as you like. A you willing to try this if I set up a website for replies? > The statistics is trivial, I'm just trying to show the effect to people. No use waving > my arms - hey everyone look at this ordinary reply to me! No there is not any use doing this. What you must do is show that there is any significance greater than chance that you can connect an author to his/her post on a forum where you are not able to see the respondents nor be able to interrogate the replies. At the moment *you* are choosing the posts therefore already have prior knowledge to the respndents' names. Are you willing to discuss a trial I could put together for you? > Practically any kind of marking system will show that I can compete against other > people and always score higher over a small amount of trials. Ah, but can you score better than the result expected by random chance alone? Now THAT would be writing to JREF about. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! > This is by Rich Shewmaker > any idea why? > of course you have, hE ricH our king > Uh? Where does hE ricH our king come from in the above post, and how > does it relate to the posters name? I *really* cannot see the > connection your making here. > Eric H. ricH rich > I post the rich showmaker example asking for the connection, > you reply regarding the connections with the name THAT'S THE ANSWER - HE RICH! > > You already KNEW I was going to respond to the post because you > *addressed* it to me in the newsgroup as an example of the > word-numerology thing you're determined to prove! > You then go out and Google a post with a tenuous anagram of my name, > post it and then say that this is proof of your paranormal talent > because I responded? ! ! ! You CANNOT be SERIOUS > > no that quote (rich shewmaker) was the post you replied to. > You already knew who was going to respond to the damned post - you > directed it at me! Can you not see this point? Because you COULD > have searched for a post that you had already worked out had a > connection to my name and posted it knowing that I would respond, > others less generous than myself COULD point to this as a fairly > disingenuous method to force connections that are NOT random. My > replying to the post you put forward as an example of connection was > NOT a random choice of mine, you knew BEFORE the fact that I would > reply to it, so could selected a post that you had already engineered > a connection to. If you cannot see the point I'm trying to make here > you will not see the point other's are making with regard to your data > to find the author's name. I'm talking out your 1st reply to me, to my thread with this at the start. _________________________________________________ use only your common sense to answer..... IT'S EASY Herc ----1--------------------------------------------------------- -------------- ---- Randi will test you when you properly apply to be tested __________________________________________________ You gave a direct reply to the post with the quiz on it, the 1st question was that above, you're name is He's Rich. How did I force you to reply or direct this quiz at you? Herc sorry, I have to phrase that you're name is LIKE He's Rich or a big debate ensues. Funny a lecturer in sci.math David C Ullrich first post was : SEE If I had paranormal powers I'd predict a coin sequence : HTHTHHHTHTHTHTTHTHTHT I'LL b RICH Then take bets. === Subject: Re: To prove psychic powers, someone has to sit this test! > result. You then go further and say that Only 6 or 7 of those could > you guess the name from 100 sample though. So *now* you've moved the > goal posts to the equivalent of saying that 6 or 7 in 100 is > significant (ie 7%), but by datamining you can, on average, have a > success rate of 29% if the result is throwing a 6. > yes but the equivalent would be a 100 sided dice for those posts. > No, it wouldn't. By your rules, the odds are determined by the list > of posters you can choose from. The only way you can use your > 100-sided die example is if you gave 100 names to choose from. try reading this word for word first : ?? Only 6 or 7 of those could you guess the name from 100 sample though > Like Rich Shewmaker, I've seen people guess it from a small list, > and they've agreed the experiment should be for a longer list. > I'm quite confident I could post Rich's comments in 100 groups and > a dozen would pick his name from 100 random names. > Is there anything significant about Rich's name that would make his > posts more readily associated with his posts? Just curious on what > your post select process is? Only his 1st post. He is Rich Shewmaker, he said effectively see the rich show maker. Rich is a good fit, Shewmaker is a good fit. With some resources people will see just what an odd coincidence it is. I'll put the questionnaire in the newspaper then the next week the urls and date. I've seen people select his name from a list, I've posted it 100 times now, I'm yet for one person to say what a coincidence, good match to the name. Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > result. You then go further and say that Only 6 or 7 of those could > you guess the name from 100 sample though. So *now* you've moved the > goal posts to the equivalent of saying that 6 or 7 in 100 is > significant (ie 7%), but by datamining you can, on average, have a > success rate of 29% if the result is throwing a 6. > yes but the equivalent would be a 100 sided dice for those posts. > No, it wouldn't. By your rules, the odds are determined by the list > of posters you can choose from. The only way you can use your > 100-sided die example is if you gave 100 names to choose from. > try reading this word for word first : ?? > Only 6 or 7 of those could you guess the name from 100 sample though Just because you originally had 100 potential trials to perform does not make the odds 100:1. It is the number of options of EACH TRIAL that will determine the probability of success greater than random chance. > Like Rich Shewmaker, I've seen people guess it from a small list, > and they've agreed the experiment should be for a longer list. > I'm quite confident I could post Rich's comments in 100 groups and > a dozen would pick his name from 100 random names. > Is there anything significant about Rich's name that would make his > posts more readily associated with his posts? Just curious on what > your post select process is? > Only his 1st post. He is Rich Shewmaker, he said effectively see the rich show maker. > Rich is a good fit, Shewmaker is a good fit. With some resources people will > see just what an odd coincidence it is. I'll put the questionnaire in the newspaper > then the next week the urls and date. > I've seen people select his name from a list, I've posted it 100 times now, I'm yet > for one person to say what a coincidence, good match to the name. So you're saying that you have posted this trial 100 times and had replies? How many of the respondents correctly selected his name from these 100 trials? If it was more than 43, you're on to something. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! > also its 50% could be guessed readily, 6 percent could be guessed easily. > In this case you have to decide whether you are going to do the trial > against 100, 50 or 6 samples. If you decide that of the 100 samples > only 6 give you any chance of success, then the trial can only be done > on those 6. With 10 options open to you you'd have to succeed 5 out > of 6 attempts for your result to be significant. If you plumped for 50 > trials with 10 choices, your result would need to be 15 or better and > for 100, 24 or better. > Review that table you quoted, you are making serious errors with your > expectations. The stats are trivial, I could guess from 2 names as long as I do better than 50%. You just tally up the odds afterwards until you're outside the expected range. I could guess from 3 names as long as i do better than 33.3% I've got a 2 months sample where I have 50 notable examples, analyse it correctly its so much above one in 10,000. I've got 3 all on the one day I stated, each can be picked from 100 names, that's one in a million. ATLEAST Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > also its 50% could be guessed readily, 6 percent could be guessed easily. > In this case you have to decide whether you are going to do the trial > against 100, 50 or 6 samples. If you decide that of the 100 samples > only 6 give you any chance of success, then the trial can only be done > on those 6. With 10 options open to you you'd have to succeed 5 out > of 6 attempts for your result to be significant. If you plumped for 50 > trials with 10 choices, your result would need to be 15 or better and > for 100, 24 or better. > Review that table you quoted, you are making serious errors with your > expectations. > The stats are trivial, Probability, not statistics, and yes, the maths IS trivial. > I could guess from 2 names as long as I do better than 50%. True, but you will need to get 15 right from 15 tries, or 39 in 50 trials for it to be significant. As you say, The stats are trivial. > You just tally up the odds afterwards until you're outside the expected range. Whether you work out the probabilities before or after is of no consequence, as the maths to work out the number of correct answeres required to show significance better than chance in 10, 20 or 100 trials with 2, 3 or 1000 options is readily calculated. BUT, if you wish to convince people that there's something to what you claim, it is a lot easier to clearly set out the rules of engagement before you attempt a test. > I could guess from 3 names as long as i do better than 33.3% And again, you'd need to get 13 rigth out of 15 trials or 31 in 50 for your guesses to be significantly better than chance. No, you haven't, because you have not grasped how to calculate probability yet. > I've got a 2 months sample where I have 50 notable examples, analyse it > correctly its so much above one in 10,000. What is the number of responses to these 50 posts? For you to claim better than 10,000 to 1 odds, you would have to have guessed the author correctly 39 times for 2 options, 31 times for 3 options, 26 times for 4 options or 23 times for 5 options. Which of the above scenarios supports your claim on these 50 posts? > I've got 3 all on the one day I stated, each can be picked from 100 names, > that's one in a million. > ATLEAST There were NOT 100 options to choose from in those instances though. Even if there was, you'd need (at a minimum) to achieve 3 right from FIVE trials just to show better than 10,000 to 1 chance. Your one million *guess* is orders of magnitude out in this example. Consult the table that you quoted to me to confirm that what I've just said is not inaccurate. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! > 02 02 2002 I predicted paranormal proof, then 1 year later... > Your connection process get's more tenuous by the minute Herc > The reason I'm using this example is people have already spotted it from a list in sci.math, > its 1000 to 1 post in itself, > Why? > because Rich Shewmaker is an EXTREMELY odd name for someone to write > go and see James Randi. > That does not determine the 1:1000 odds you quote. How did you > determine these odds. Guessing. Rich would be spotted from 20 to 100 names, Mr Shewmaker would be spotted from 20 to 100 names, so its a LOT better than those odds. People on sci.math had said I should do the question with a lot of names for that example. > the measure_of_coincidence is proportional (somewhat) to the number of names it > can be picked out from. > So there were 1000 replies from 1000 individual posters on sci.math to > your single post was there? This is the only way you can apply this > logic to get your 1/1000 odds Herc. No, a few correct replies with 1,000 name options would do it. Herc === Subject: Re: To prove psychic powers, someone has to sit this test! > 02 02 2002 I predicted paranormal proof, then 1 year later... > these 3 posts on the small test (rich, rust, hell) happened on 02 02 > Your connection process get's more tenuous by the minute Herc > The reason I'm using this example is people have already spotted it from a list in > sci.math, > its 1000 to 1 post in itself, > Why? > because Rich Shewmaker is an EXTREMELY odd name for someone to write > go and see James Randi. > That does not determine the 1:1000 odds you quote. How did you > determine these odds. > Guessing. Herc? Uh, odds, chance, probability, whatever you wish to call it is not determined by your guesses. GUESSING 1:1000 odds is not correct. > Rich would be spotted from 20 to 100 names, Mr Shewmaker would > be spotted from 20 to 100 names, so its a LOT better than those odds. Unfortunately Herc, RIch's name was only chosen from 3 alternatives - 1/3 odds, no biggie. The only way to demonstrate that one of his posts would be spotted, would be to give you the choice of 20-100 to choose from and gauge whether you got it right or not. But, since this effect it can only be demonstrated with the first EVER response to you by a poster, using Rich Shewmaker's posts as a future example of your powers is now completely meaningless. > People on sci.math had said I should do the question with a lot of names for that example. And they were right. Getting his name from a choice of 3 is not impressive. > the measure_of_coincidence is proportional (somewhat) to the number of names it > can be picked out from. > So there were 1000 replies from 1000 individual posters on sci.math to > your single post was there? This is the only way you can apply this > logic to get your 1/1000 odds Herc. > No, a few correct replies with 1,000 name options would do it. Again, YOU do not get to reinvent the caclulations of probabilities and *guess* what is significant, actually, 2 right in 5 tries with a choice of 1000 options would be suffiecient to show there *may* be something to what you think you would do. -- Eric Hocking === Subject: Re: To prove psychic powers, someone has to sit this test! some post you made back on >> the 4th May? >> Wally's reply this post is not significant because he has already replied to me elsewhere. >> In his 1st post you CAN spot his name. Think of a name for 'see my kid'. >Disregarding the fact that you early said that there is no connection >whatsoever, I'm afraid my powers aren't as honed as yours and cannot >make *any* connection between 'see my kid' and Wally Anglesea. he thinks my kids are angels. Now although my name is pronounced Angle then sea we sometimes get called Angel sea So Nitwit Herc is using his mental acrobatics (because it's far more than gymnastics) to think there's a connection. What he doesn't realise is, my kida ain't exactly angels :-)) He still refuses to take *my* test, because he knows he'll flunk. -- Find out about Australia's most dangerous Doomsday Cult: http://users.bigpond.net.au/wanglese/pebble.htm You can't fool me, it's turtles all the way down. === Subject: Re: To prove psychic powers, someone has to sit this test! >Not psychic, but just a garden-variety kook. > You need some Industrial Grade Roundup, and/or a roto-tiller, and/or > some IncindiGel close air support and a new garden. Nah, he just needs to yank that pesky weed out and throw it into the wood chipper. -- Felony case 02-CR-0617 9/1/03: Oregon Department of Justice V. Raymond Ronald Karczewski, Defendant. The defendant's name is NOT copyrighted. === Subject: Re: To prove psychic powers, someone has to sit this test! >> >>Not psychic, but just a garden-variety kook. >> >> You need some Industrial Grade Roundup, and/or a roto-tiller, and/or >> some IncindiGel close air support and a new garden. >Nah, he just needs to yank that pesky weed out and throw it into the wood >chipper. Bah! That's too easy. -- V.G. People are more violently opposed to fur than leather, because it is easier to harrass rich women than it is motorcycle gangs. - Bumper Sticker Sarcasm is my sword, Apathy is my shield. === Subject: Re: Alg Int prob explanation > Now then, say you're a mathematician who doesn't want the world to > know the truth, you can endlessly run circles around what *looks* like > it should be the case. It looks like 'b' should be a unit. It > *looks* like 2a should have 2 as a factor, but because there's this > weird problem, what *looks* correct is wrong. > That answer is that there's a more inclusive ring, which includes 'a', > and in that ring ab=1 and 'a' is a unit as is 'b'. Question for you, James. Consider the more general case of factoring c = ab. Let's suppose I'm working in the ring of integers. If c and b are integers, doesn't it look like I should be able to say that b is a factor of c? If I tell you there are cases where c = ab, and both c and b are integers but b is not a factor of c, does that indicate a core problem with the integers? - Randy === Subject: Re: Alg Int prob explanation > Ok, I've had some trouble explaining the problem with the ring of > algebraic integers, but I think I might have found how to show it to > you. > Consider algebraic integers, x, y and b, where x=2a, y=b, and xy=2. OK. Since a is not required to be an algebraic integer, I could have x = sqrt(2) and y = b = sqrt(2), with a = 2^(-1/2). > Now then, you have (2a)b = 2, and while 'b' does not share non-unit > factors with 2 Among other things, b shares lots of factors of the form 2^(1/k) with 2 (for instance 2^(1/12)) and these are not units in the algebraic integers. > in one sense it *does* share itself as a factor with 2 > in a trivial sense. Why trivial? What's a factor in a trivial sense? You realize that because you didn't require a or ab to be an algebraic integer, that b is not a unit, right? ab = 1, but this is not a factorization of 1 in the algebraic integers. > You may think that ab = 1 makes 'b' a unit, but remember, 'a' is NOT > an algebraic integer, so 'b' is not a unit. Right. Very good. Nobody who understands the definition of unit will think that b is a unit in the algebraic integers. So who are you addressing this to? > What I've proven is that situation *can* occur with the definition of > algebraic integers as roots of monic polynomials with integer > coefficients. > It's a wacky enough problem that it can be hard explaining it. What was the situation? What was the problem? Is it that b should be a unit? Is that what you're trying to say? But units have a definition, b doesn't fit it, so it's not a unit. Who says it should be? Is the problem that the definition of factor requires all terms to be in a certain ring? So the problem in core mathematics is that you want to say ab = c makes b a factor of c in the algebraic integers even if a (or c) is not an algebraic integer? Is that the problem? You know, I think that IS what you're trying to say. - Randy === Subject: Re: Alg Int prob explanation > Ok, I've had some trouble explaining the problem with the ring of > algebraic integers, but I think I might have found how to show it to > you. Here's an even simpler way: ab = 1, where a = 1/2 and b = 2. Clearly 'ab' is an algebraic integer and 'b' is an algebraic integer, but 'a' is *not* an algebraic integer. Also there *should* be some units in there somewhere. Therefore the ring of algebraic is flawed and numbers which *should* be in the ring of algebraic integers (a = 1/2) are missing. Is that simple enough? (Unfortunately, simple as it is, it is 100% WRONG!) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Alg Int prob explanation > Ok, I've had some trouble explaining the problem with the ring of > algebraic integers, but I think I might have found how to show it to > you. > Consider algebraic integers, x, y and b, where x=2a, y=b, and xy=2. > Now then, you have (2a)b = 2, and while 'b' does not share non-unit > factors with 2 in one sense it *does* share itself as a factor with 2 > in a trivial sense. > Now 2a is an algebraic integer, but because 'a' is not, it does NOT > have 2 as a factor of 'a' in the ring of algebraic integers. > You may think that ab = 1 makes 'b' a unit, but remember, 'a' is NOT > an algebraic integer, so 'b' is not a unit. > What I've proven is that situation *can* occur with the definition of > algebraic integers as roots of monic polynomials with integer > coefficients. > It's a wacky enough problem that it can be hard explaining it. > So now to how easy it is to confuse you. > Now then, say you're a mathematician who doesn't want the world to > know the truth, you can endlessly run circles around what *looks* like > it should be the case. It looks like 'b' should be a unit. It > *looks* like 2a should have 2 as a factor, but because there's this > weird problem, what *looks* correct is wrong. > That answer is that there's a more inclusive ring, which includes 'a', > and in that ring ab=1 and 'a' is a unit as is 'b'. Is that more inclusive ring the Object Ring? Clive === Subject: Re: Alg Int prob explanation >Ok, I've had some trouble explaining the problem with the ring of >algebraic integers, but I think I might have found how to show it to >you. >Consider algebraic integers, x, y and b, where x=2a, y=b, and xy=2. >Now then, you have (2a)b = 2, and while 'b' does not share non-unit >factors with 2 in one sense it *does* share itself as a factor with 2 >in a trivial sense. >Now 2a is an algebraic integer, but because 'a' is not, it does NOT >have 2 as a factor of 'a' in the ring of algebraic integers. >You may think that ab = 1 makes 'b' a unit, but remember, 'a' is NOT >an algebraic integer, so 'b' is not a unit. >What I've proven is that situation *can* occur with the definition of >algebraic integers as roots of monic polynomials with integer >coefficients. >It's a wacky enough problem that it can be hard explaining it. No, I think you've explained it very clearly. You make it obvious that there's a problem if a ring contains 2 and 2a but not a itself. But why do you restrict yourself to the algebraic integers? Surely you can say the same about the rational integers: after all, poor old a=5/2 is arbitrarily excluded from Z, even though it contains both 2 and 2a=5. >So now to how easy it is to confuse you. >Now then, say you're a mathematician who doesn't want the world to >know the truth, you can endlessly run circles around what *looks* like >it should be the case. It looks like 'b' should be a unit. It >*looks* like 2a should have 2 as a factor, but because there's this >weird problem, what *looks* correct is wrong. >That answer is that there's a more inclusive ring, which includes 'a', >and in that ring ab=1 and 'a' is a unit as is 'b'. >James Harris John Roberts-Jones === Subject: Re: Alg Int prob explanation >Ok, I've had some trouble explaining the problem with the ring of >algebraic integers, There's a reason for that. The problem does not exist. >but I think I might have found how to show it to >you. >Consider algebraic integers, x, y and b, where x=2a, y=b, and xy=2. >Now then, you have (2a)b = 2, and while 'b' does not share non-unit >factors with 2 in one sense it *does* share itself as a factor with 2 >in a trivial sense. Huh? In exactly what sense is it true that b does not share a factor with 2? >Now 2a is an algebraic integer, but because 'a' is not, it does NOT >have 2 as a factor of 'a' in the ring of algebraic integers. >You may think that ab = 1 makes 'b' a unit, but remember, 'a' is NOT >an algebraic integer, We can't remember that because you never said that, and it doesn't follow from anything you've said. But let's assume that: a is not an algebraic integer, fine. >so 'b' is not a unit. Not a unit _in_ the algebraic integers, no. Fine. >What I've proven is that situation *can* occur with the definition of >algebraic integers as roots of monic polynomials with integer >coefficients. _What_ situation, exactly? The things you say above can indeed happen - so what? How does this indicate that there's a problem? (All I can _guess_ is that you mean you've proved the above can happen in a case where a _is_ an algebraic integer, so a both is and is not an algebraic integer. This is just a guess as to what you're driving at... if that's what you mean then yes, that would be a problem, but no, you have not proved that.) >It's a wacky enough problem that it can be hard explaining it. >So now to how easy it is to confuse you. >Now then, say you're a mathematician who doesn't want the world to >know the truth, you can endlessly run circles around what *looks* like >it should be the case. It looks like 'b' should be a unit. It >*looks* like 2a should have 2 as a factor, but because there's this >weird problem, what *looks* correct is wrong. >That answer is that there's a more inclusive ring, which includes 'a', >and in that ring ab=1 and 'a' is a unit as is 'b'. >James Harris ************************ David C. Ullrich === Subject: Re: Alg Int prob explanation > Ok, I've had some trouble explaining the problem with the ring of > algebraic integers, but I think I might have found how to show it to > you. > Consider algebraic integers, x, y and b, where x=2a, y=b, and xy=2. > Now then, you have (2a)b = 2, and while 'b' does not share non-unit > factors with 2 in one sense it *does* share itself as a factor with 2 > in a trivial sense. What the in one sense means here I have no idea. > Now 2a is an algebraic integer, but because 'a' is not, it does NOT > have 2 as a factor of 'a' in the ring of algebraic integers. > You may think that ab = 1 makes 'b' a unit, but remember, 'a' is NOT > an algebraic integer, so 'b' is not a unit. Do the same with x, y and b normal integers. Why is it a problem with algebraic integers and not with normal integers? > What I've proven is that situation *can* occur with the definition of > algebraic integers as roots of monic polynomials with integer > coefficients. It does not depend on the definition. The same can occur in every ring that is not a field. > Now then, say you're a mathematician who doesn't want the world to > know the truth, you can endlessly run circles around what *looks* like > it should be the case. It looks like 'b' should be a unit. It > *looks* like 2a should have 2 as a factor, but because there's this > weird problem, what *looks* correct is wrong. Take a=1/2, b=2, x=1, y=2. xy = (2a)b = 2. It looks like 'b' should be a unit. It looks like 2a should have 2 as a factor, but because there's this weird problem, what *looks* correct is wrong. > That answer is that there's a more inclusive ring, which includes 'a', > and in that ring ab=1 and 'a' is a unit as is 'b'. Yup. The ring of algebraic numbers for instance. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Alg Int prob explanation > Ok, I've had some trouble explaining the problem with the ring of > algebraic integers, but I think I might have found how to show it to > you. > Consider algebraic integers, x, y and b, where x=2a, y=b, and xy=2. > Now then, you have (2a)b = 2, and while 'b' does not share non-unit > factors with 2 in one sense it *does* share itself as a factor with 2 > in a trivial sense. Please be precise how this case xb = 2, with x,b algebraic integers, allows b *not* to share non-unit factors with 2. IN WHAT SENSE does it not share non-unit factors with 2 in the ring of algebraic integers? Surely you realize that if b is an algebraic integer, sqrt(b) is also an algebraic integer (for whichever square root you choose: both square roots are algebraic integers). Then although you may view b as a trivial common factor between b and 2, you cannot view sqrt(b) as a trivial common factor between b and 2. In fact, for EVERY algebraic integer z, if z divides b, then z also divides 2. > Now 2a is an algebraic integer, but because 'a' is not, it does NOT > have 2 as a factor of 'a' in the ring of algebraic integers. So? This is irrelevant to your mistaken point about b and 2. > You may think that ab = 1 makes 'b' a unit, but remember, 'a' is NOT > an algebraic integer, so 'b' is not a unit. I didn't think that, why did you think anyone would have thought that? > What I've proven is that situation *can* occur with the definition of > algebraic integers as roots of monic polynomials with integer > coefficients. What situation? Your claim regarding 2 and b is clearly false. Your claim regarding a and b is irrelevant. > It's a wacky enough problem that it can be hard explaining it. Wacky. > So now to how easy it is to confuse you. I'm not confused. I'm barely amused. Your claim is refused. > Now then, say you're a mathematician who doesn't want the world to > know the truth, you can endlessly run circles around what *looks* like > it should be the case. It looks like 'b' should be a unit. It > *looks* like 2a should have 2 as a factor, but because there's this > weird problem, what *looks* correct is wrong. This weird problem. So, what you're saying seems to be this: Let xy = 2, x and y algebraic integers, where x is not a multiple of 2, then y should be a unit. After all, there is surely the algebraic number a = x/2, making x = 2a, so your situation is covered. All you have to do is expand the ring of algebraic integers to include the algebraic number 'a', which turns out to be 1/y: xy = (2a)y = 2(ay) = 2. This yields 2ay - 2 = 0 or 2(ay - 1) = 0, so since the algebraic numbers form a field, and since the above is an equation of algebraic numbers, we have (since 2 != 0) ay=1. > That answer is that there's a more inclusive ring, which includes 'a', > and in that ring ab=1 and 'a' is a unit as is 'b'. Hey! I found it! Problem is, you seem to want to do this for *all* integers n, not just n=2, and *all* products xy = n in the ring of algebraic integers. I'm just wondering how you avoid reaching the full field of algebraic numbers. By now, you probably realize that the hand-waving Object Ring answer is no answer at all, unless you can exhibit a construction of that ring. Perhaps you could start there, right after you explain how it is that, given the products: qr = 5 rs = a with a,q,r,s all in the ring of algebraic integers, and with r not a unit, you can claim that a is coprime to 5. After all, we have an explicit common factor r in the ring of algebraic integers, and that common factor is not a unit. > James Harris Dale === Subject: Re: Alg Int prob explanation > Ok, I've had some trouble explaining the problem with the ring of > algebraic integers, but I think I might have found how to show it to > you. > Consider algebraic integers, x, y and b, where x=2a, y=b, and xy=2. > Now then, you have (2a)b = 2, and while 'b' does not share non-unit > factors with 2 in one sense it *does* share itself as a factor with 2 > in a trivial sense. > Now 2a is an algebraic integer, but because 'a' is not, it does NOT > have 2 as a factor of 'a' in the ring of algebraic integers. > You may think that ab = 1 makes 'b' a unit, but remember, 'a' is NOT > an algebraic integer, so 'b' is not a unit. > What I've proven is that situation *can* occur with the definition of > algebraic integers as roots of monic polynomials with integer > coefficients. But so what? It's been said that one doesn't understand a problem until one can explain it. You certainly haven't explained your objection to the accepted definition of the algebraic integers to my satisfaction, and I'm one of those people who reads a lot of your posts. Ok, we have the set of algebraic integers. They are a ring but not a field, which means that some of them have multiplicative inverses that are not algebraic integers. Now what exactly is it that you think the problem is? Just tell me in words. A short declarative sentence. === Subject: Re: Alg Int prob explanation > Now what exactly is it that you think the problem is? Just tell me in > words. A short declarative sentence. My understanding of what he is saying is the following. The ring of algebraic integers does not give him the result that he wants that exactly two factors of an explicit expression have factors in common with f. He needs another ring (his object ring) to get the above result. Thus, the ring of algebraic integers is flawed: it is missing factors or units that are required for his result to be true. Take his example of the the ring of even integers. Here, 2 and 6 are coprime, but in the ring of integers 2 and 6 are not coprime. Thus, the ring of even integers is flawed. The ring of integers corrects this flaw. I don't think he is saying that there is a contradiction in mathematics. Otherwise, as someone already mentioned, it makes no sense to continue writing proofs (like he does in his object mathematics) until the inconsistant axioms are found and are corrected. -- Bill Hale === Subject: Re: Alg Int prob explanation > Now what exactly is it that you think the problem is? Just tell me in > words. A short declarative sentence. > My understanding of what he is saying is the following. > The ring of algebraic integers does not give him the result > that he wants that exactly two factors of an explicit expression > have factors in common with f. Very basic algebra proves that certain numbers *should* have a particular factor in the ring of algebraic integers, but they do not, which leaves a puzzling and contradictory situation, which is why it's an error. Mathematics is consistent. > He needs another ring (his object ring) to get the above result. > Thus, the ring of algebraic integers is flawed: it is > missing factors or units that are required for his result > to be true. > Take his example of the the ring of even integers. > Here, 2 and 6 are coprime, but in the ring of integers 2 and 6 > are not coprime. Thus, the ring of even integers is flawed. > The ring of integers corrects this flaw. > I don't think he is saying that there is a contradiction > in mathematics. Otherwise, as someone already mentioned, > it makes no sense to continue writing proofs (like he > does in his object mathematics) until the inconsistant > axioms are found and are corrected. > -- Bill Hale It's not an axiom problem but a definition problem, as the definition of algebraic integers as roots of monic polynomials with integer coefficients leaves out certain numbers, leading to contradictions, if you assume the ring is complete. For instance, with my example at the start of this thread, with 'a' one of these numbers, you have x=2a, y=b, xy=2, where x and y are algebraic integers, but because 'a' has been arbitrarily excluded by the definition, x does NOT have 2 as a factor *in the ring of algebraic integers*. While some may be surprised that 2a is an algebraic integer, while 'a' is not, as pointed out by Bill Hale in his post, and previously by me, 2(3)=6 gives something of an analogy when considered in the ring of evens. Here 3 is not an even, so 2 is NOT a factor of 6, in the ring of evens. Supposedly the ring of algebraic integers didn't suffer from a similar problem, while I've proven it those, and in fact, it does so in such a way that unlike the ring of evens, it is NOT a complete ring. Explaining can be difficult, but it begins with following the math which proves the problem exists. James Harris === Subject: Re: Alg Int prob explanation > Supposedly the ring of algebraic integers didn't suffer from a similar > problem, Who supposed that and why is it a problem? >while I've proven it those, and in fact, it does so in such a > way that unlike the ring of evens, it is NOT a complete ring. > Explaining can be difficult, No it's not: Here's a succinct statement of the problem you have proven: The ring of algebraic integers is not closed under division. Here's an equivalent statement of the problem: The ring of algebraic integers does not include all the algebraic numbers. Hmmm... so is there a ring you can use that includes everything in the ring of algebraic numbers? Let's see, what could that be? How about, now this is really wacky but... the ring of algebraic numbers? So try to work your proof with the larger ring of algebraic numbers, not just algebraic integers. Does it work? (We know the answer; others have pointed out that the algebraic numbers are too large for what you want to do. But those are lying evil mathematicians making those arguments.) Why not try it yourself? Just replace the words algebraic integers with algebraic numbers in your proof and see what happens. - Randy === Subject: Re: Alg Int prob explanation > What I've proven is that situation *can* occur with the definition of > algebraic integers as roots of monic polynomials with integer > coefficients. situation that can occur with definition - you call it an explanation? > It's a wacky enough problem that it can be hard explaining it. By using the term wacky you're admitting that you don't understand what's going on. If you'd understand, then you'd use the term clear instead. > So now to how easy it is to confuse you. And you. === Subject: Re: Alg Int prob explanation > What I've proven is that situation *can* occur with the definition of > algebraic integers as roots of monic polynomials with integer > coefficients. > situation that can occur with definition - you call it an explanation? That definition with its focus on roots of monic polynomials with integer coefficients is key to the problem. The point is that *certain* roots of non-monics have to be included for a complete ring. > It's a wacky enough problem that it can be hard explaining it. > By using the term wacky you're admitting that you don't understand what's going on. > If you'd understand, then you'd use the term clear instead. It's difficult to explain from the ring of algebraic integers, while I've made a lot of effort to do so because it's something with which many of you are familiar. > So now to how easy it is to confuse you. > And you. I've faced a concerted effort by some posters with a lot of skill at clouding the issue. Their behavior has been reprehensible, but it is something I'm working on handling. Not surprisingly an over hundred year old error *had* to have subtleties. James Harris === Subject: Statistics problem in Excel - need help! I've got a problem I'm just on the edge of solving (or maybe it just feels that way!) and I need some help. Simplified, I have an n column table, where all but the last column are filled with non-numeric values. Then nth column contains a value. I want to determine which of the non-numeric columns is most responsible for change in the numeric columns using Excel. There is a CORREL funciton, but that only works if you are looking for a correlation between two different numeric columns. Any help would be greatly appreciated! John Hoge === Subject: Re: Statistics problem in Excel - need help! > I've got a problem I'm just on the edge of solving (or maybe it > just feels that way!) and I need some help. > Simplified, I have an n column table, where all but the last > column are filled with non-numeric values. Then nth column > contains a value. > I want to determine which of the non-numeric columns is most > responsible for change in the numeric columns using Excel. There > is a CORREL funciton, but that only works if you are looking for > a correlation between two different numeric columns. Any help > would be greatly appreciated! I agree with Dirk that you want some sort of numeric-valued function of strings (or booleans, or what-have-you). Without knowing more about the nature of your non-numerics, though, I think it's hard to tell what would be a reasonable function to use. In particular, I think it's unlikely that the ASCII value of the characters in your (presumed) string non-numeric would mean much. That may not matter if there are only two possible values in each column. Then just arbitrarily assign zero to tall and one to short (or whatever). The easy way to do that in Excel is to use IF in a new column, possibly on a new page. If you have more than two non-numeric values in a column, is there some clear, relevant way to rank them? One idea off the top of my head, if that doesn't work, is to apply CORREL to each of the values for a particular column, and then combine them for an overall evaluation of that column. I mean, if the values are red, blue, and green, use three new columns holding B C D =IF(A2=red,1,0) =IF(A2=blue,1,0) =IF(A2=green,1,0) then get =SQRT(CORRELL(B2:B100,E2:E100)^2+ CORRELL(C2:C100,E2:E100)^2+ CORRELL(D2:D100,E2:E100)^2) Notice that I added the correlation coefficients in quadrature. I don't have a good justification for doing it that way; it just seems like the right thing to do. I'm pretty sure just adding them would just give you zero in all cases. Not very useful. If you've got different numbers of values in different columns, I think you may need to compensate, with a multiplicative constant, most likely. I don't know what the correct constant to use might be. If I was in a hurry, I would do it experimentally. Assign non-numeric values at random over and over and see what kind of distribution the =SQRT(CORRELL...) cells turn out. Actually, I think that last idea has some potential. Perform the =SQRT(CORRELL...) calculation once with the data as originally recorded, then repeatedly scramble the order of the numeric column, keeping track only how many times the randomized correllation value is higher or lower than the real value. The more unlikely a higher value, the more significant a column is. That puts all the columns on an equal footing and gives an understandable meaning to whatever results you come up with. If I were scrambling things in Excel, I probably would generate a column of random numbers (0..1), rank them using LARGE or SMALL, and use LOOKUP to pull the corresponding value out of the original column. Every time you recalculate, you generate new random numbers, if I recall correctly. There ought to be a way to record multiple calculation- cycles/correlation-values in Excel, but, offhand, I don't know of one. A kluge would be to keep your result for a column (#higher) as a value in one cell, add the result of the current cycle to it in a second cell, then _paste special as value_ from the summing cell to the saving cell. *Then* make a macro that does that, put it in a loop and run it a bunch of times. Even though there are some complicated things going on analytically up to this point, at this stage all you have is a Bernoulli distribution. If M is the number of successful trials (random coeff higher than real coeff) and N is the number of all trials, then p = M/N with a standard deviation sqrt(p(1-p)/N). Smaller p <--> more significant. Jim Burns === Subject: Re: Statistics problem in Excel - need help! Jim, The problem is that I can't see how to logically assign values to the nonnumeric values. The data set is an evaluation of a marketing plan. There are 5 inputs - publication, market segment, month of publication, CPM (cost per thousand circulation), and creative used. ROI(Return on investment) is the output column. Month is numeric, but If I use CORREL, excel will tell me how much an increase in the month is proportional to an increase in ROI. I would love it if ROI went up in a linear manner every month, but I know that doesn't always happen. I would expect a negative correlation between CPM and ROI, so I could use CORREL for that. Ideally I would like to use the same function on all input columns, so I can determine statistically what matters most of these 5 inputs. There are 120 data points, and the data is pretty sparse - I don't have the budget for and ad in every pub in every market segment in every month. I appreciate yor time - John Hoge SeaEagle.com === Subject: Re: Statistics problem in Excel - need help! >Jim, >The problem is that I can't see how to logically assign values to the >nonnumeric values. The data set is an evaluation of a marketing plan. >There are 5 inputs - publication, market segment, month of >publication, CPM (cost per thousand circulation), and creative used. >ROI(Return on investment) is the output column. >Month is numeric, but If I use CORREL, excel will tell me how much an >increase in the month is proportional to an increase in ROI. I would >love it if ROI went up in a linear manner every month, but I know that >doesn't always happen. I would expect a negative correlation between >CPM and ROI, so I could use CORREL for that. >Ideally I would like to use the same function on all input columns, so >I can determine statistically what matters most of these 5 inputs. >There are 120 data points, and the data is pretty sparse - I don't >have the budget for and ad in every pub in every market segment in >every month. >I appreciate yor time - >John Hoge >SeaEagle.com Better posted to the stat group, as I have done with this post. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Statistics problem in Excel - need help! > I've got a problem I'm just on the edge of solving (or maybe it just > feels that way!) and I need some help. > Simplified, I have an n column table, where all but the last column > are filled with non-numeric values. Then nth column contains a value. > I want to determine which of the non-numeric columns is most > responsible for change in the numeric columns using Excel. There is a > CORREL funciton, but that only works if you are looking for a > correlation between two different numeric columns. Any help would be > greatly appreciated! You could write a function that takes a non-numeric cell as a parameter and that calculates a value from the characters in the string. A simple (and maybe useless) example: function strcalc( c as string ) as long dim i as long dim j as long j = 0 for i = 1 to len(c) j = j + i * asc(mid(c,i,i)) next strcalc = j end function Of course it depends on the number of characters in the non-numeric data. If you are sure that the max width is 4, you can take j = j + 256^(i-1) * asc(mid(c,i,i)) to make absolutely sure you have a 1-to-1 function. Dirk Vdm === Subject: Re: A certain Dirichlet-sum: question >I should probably give this more thought myself. But this question >does not seem to be as easily answered as I 1st assumed. >Does sum{k=1 to oo} a(k,m)/k^r = >(1 - 1/zeta(r))^m, >where a(k,m) = the number of distinct prime divisors, p, of k, >where p^m divides k, but p^(m+1) does not? I don't think so. For example, a(15,2) = 0 but (1- 1/zeta(r))^2 should contain a term 2/15^r. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: A certain Dirichlet-sum: question [cross-posting reply to rec.puzzles] >I should probably give this more thought myself. But this question >does not seem to be as easily answered as I 1st assumed. >Does sum{k=1 to oo} a(k,m)/k^r = >(1 - 1/zeta(r))^m, >where a(k,m) = the number of distinct prime divisors, p, of k, >where p^m divides k, but p^(m+1) does not? > I don't think so. For example, a(15,2) = 0 but > (1- 1/zeta(r))^2 should contain a term 2/15^r. The reason I ask is that, I THINK: sum{m=1 to oo} (1 - 1/zeta(r))^m *b(m) = sum{m=1 to oo} (sum{k=1 to oo} a(k,m)/k^r) b(m) for a certain (at least one) sequence {b(m)}. Puzzle(??): Find {b(k)}. (I do not know if the sequence {b(k)} is unique.) Leroy Quet === Subject: Test - please ignore testkees === Subject: Multiplying zero by infinity My math teacher can't find what's wrong with this. Can you? (Note: inf = infinity) lim(x->inf) = 1 and 1 = x*(1/x) Okay: lim(x->inf) x = inf lim(x->inf) 1/x = 0 Because lim(x->C) (f(x)*g(x)) = lim(x->C) f(x) * lim(x->C) g(x) then lim(x->inf) x*(1/x) = lim(x->inf) 1 = 1 = lim(x->inf) x * lim(x->inf) 1/x = 1 = (inf) * (0) = 1 So inf * 0 = 1. ( Of course, you could substitute all 1's with, say, N, and get ) ( inf * 0 = N ) Ryan S === Subject: Re: Multiplying zero by infinity > My math teacher can't find what's wrong with this. Can you? > (Note: inf = infinity) > lim(x->inf) 1 = 1 ^ you want this I think > and 1 = x*(1/x) > Okay: > lim(x->inf) x = inf > lim(x->inf) 1/x = 0 > Because > lim(x->C) (f(x)*g(x)) = lim(x->C) f(x) * lim(x->C) g(x) > then > lim(x->inf) x*(1/x) = lim(x->inf) 1 = 1 > = lim(x->inf) x * lim(x->inf) 1/x = 1 > = (inf) * (0) = 1 > So inf * 0 = 1. > ( Of course, you could substitute all 1's with, say, N, and get ) > ( inf * 0 = N ) > Ryan S lim_{x->C} f(x) = infinity might be read as the limit of f(x) as x tends to C is infinity, or as the limit of f(x) as x tends to C does not exist. With the second reading lim (fg) = (lim f)(lim g) looks less tempting. -- G.C. === Subject: Re: Multiplying zero by infinity > My math teacher can't find what's wrong with this. Can you? > (Note: inf = infinity) > lim(x->inf) = 1 > and 1 = x*(1/x) > Okay: > lim(x->inf) x = inf > lim(x->inf) 1/x = 0 > Because > lim(x->C) (f(x)*g(x)) = lim(x->C) f(x) * lim(x->C) g(x) > then > lim(x->inf) x*(1/x) = lim(x->inf) 1 = 1 > = lim(x->inf) x * lim(x->inf) 1/x = 1 > = (inf) * (0) = 1 > So inf * 0 = 1. > ( Of course, you could substitute all 1's with, say, N, and get ) > ( inf * 0 = N ) > Ryan S Look up indeterminate form in the index of any calculus textbook. === Subject: Re: Multiplying zero by infinity >My math teacher can't find what's wrong with this. Can you? You're assuming that the limit of a product is the product of the limits, and that's not true. (It's true under certain conditions: If lim f(x) = a, lim g(x) = b, and a and b are both _finite_ then lim f(x) g(x) = ab. You can't apply that fact here because one of the limits is _not_ finite.) >(Note: inf = infinity) >lim(x->inf) = 1 >and 1 = x*(1/x) >Okay: >lim(x->inf) x = inf >lim(x->inf) 1/x = 0 >Because > lim(x->C) (f(x)*g(x)) = lim(x->C) f(x) * lim(x->C) g(x) >then > lim(x->inf) x*(1/x) = lim(x->inf) 1 = 1 > = lim(x->inf) x * lim(x->inf) 1/x = 1 > = (inf) * (0) = 1 >So inf * 0 = 1. >( Of course, you could substitute all 1's with, say, N, and get ) >( inf * 0 = N ) >Ryan S ************************ David C. Ullrich === Subject: Re: Multiplying zero by infinity > My math teacher can't find what's wrong with this. Can you? > (Note: inf = infinity) > lim(x->inf) = 1 > and 1 = x*(1/x) This statement makes no sense. Perhaps you meant this: lim(x->inf) ( x*(1/x) ) = 1. That is certainly true, since the limit of a constant is equal to that constant. > Okay: > lim(x->inf) x = inf > lim(x->inf) 1/x = 0 > Because > lim(x->C) (f(x)*g(x)) = lim(x->C) f(x) * lim(x->C) g(x) > then > lim(x->inf) x*(1/x) = lim(x->inf) 1 = 1 > = lim(x->inf) x * lim(x->inf) 1/x = 1 > = (inf) * (0) = 1 The first equality lim(x->C) (f(x) g(x)) = lim(x->C) f(x) * lim(x->C) g(x) holds only if both limits on the right side exist. The substitution of lim(x->inf) x for lim(x->C) f(x) violates that condition: the statement lim(x->inf) x = inf is merely one version of the statement lim(x->inf) x does not exist (there are far weirder ways for limits to fail to exist, and a person should encounter a number of those ways in any decent Calculus class) Another way of seeing that this is a fallacy is to note that the product on the right side is defined only if the individual factors are numbers; inf is not a number, and does not follow any of the rules that numbers follow. For instance, for every number x, x+1 is not equal to x. for every nonzero number x, 2x is not equal to x. > So inf * 0 = 1. > ( Of course, you could substitute all 1's with, say, N, and get ) > ( inf * 0 = N ) Infinity is not a number. It does not obey any of the rules for numbers. There *are* ways to make sense of a product of factors, one of which approaches infinity, with the other approaching zero; they all come down to determining the rates at which those two process occur, and establishing just what the balance is between those two rates. For instance, in your example, you can cancel factor of x in the numerator against the factor of x in the denominator, and see what's left. You mentioned that your math teacher couldn't figure this out. I hope this math teacher is not teaching at the college or university level, or if so, is a graduate student who has just come out of college. The question is not really a difficult one (my guess is that it falls close to the edge between too trivial for me to bother with; someone else is sure to catch it and Boy, I gotta set this person straight level of difficulty). A good Calculus student will see most of this right away, and a Mathematics major should worry about his suitability for the field if he finds this to be nontrivial. Maybe your teacher found it difficult to convey the idea. I could buy that, without consigning your teacher to Math Hell. > Ryan S Dale === Subject: Re: Multiplying zero by infinity > My math teacher can't find what's wrong with this. Can you? > (Note: inf = infinity) > lim(x->inf) = 1 > and 1 = x*(1/x) > This statement makes no sense. > Perhaps you meant this: > lim(x->inf) ( x*(1/x) ) = 1. The teacher should have seen that this is the only reasonable statement of the problem. Maybe he was too busy to even get that far. > That is certainly true, since the limit of a constant is equal > to that constant. Schnippers.. > You mentioned that your math teacher couldn't figure this out. I hope > this math teacher is not teaching at the college or university level, > or if so, is a graduate student who has just come out of college. What kind of a Department would hire a person like this, ? Often the TA's are better teachers than the Profs because the Profs are too busy with research ot be interested in the students. >The question is not really a difficult one (my guess is that it falls close > to the edge between too trivial for me to bother with; someone else is > sure to catch it and Boy, I gotta set this person straight level of > difficulty). A good Calculus student will see most of this right away, > and a Mathematics major should worry about his suitability for the > field if he finds this to be nontrivial. > Maybe your teacher found it difficult to convey the idea. I could buy > that, without consigning your teacher to Math Hell. Maybe the teacher is ESL? It Should ( ha ha) be immediately obvious that the point of the problem is to use the Distribution of multiplication in derivatives rule, subject to some conditions. In the last week, three different students have posted here claiming that the teacher didn't understand anything beyond High School Algebra, with supporting anecdotal evidence or misunderstood statements. If this is true, Its scary, but not surprising. Bob Pease === Subject: Re: Multiplying zero by infinity >Okay: >lim(x->inf) x = inf >lim(x->inf) 1/x = 0 >Because > lim(x->C) (f(x)*g(x)) = lim(x->C) f(x) * lim(x->C) g(x) Applies only if both limits exist and are finite. === Subject: Re: Multiplying zero by infinity > Because > lim(x->C) (f(x)*g(x)) = lim(x->C) f(x) * lim(x->C) g(x) This is true only if both lim(x->C)f(x) and lim(x->C)g(x) exist i.e. they are real since lim(x->inf) x =inf the rule does not apply. In terms of 0/0 or inf/inf you have to apply lHospital. In your case this means lim(x->inf) x/x=lim(x->inf) 1/1=1 === Subject: Re: Multiplying zero by infinity > My math teacher can't find what's wrong with this. Can you? > (Note: inf = infinity) > lim(x->inf) = 1 > and 1 = x*(1/x) > Okay: > lim(x->inf) x = inf > lim(x->inf) 1/x = 0 > Because > lim(x->C) (f(x)*g(x)) = lim(x->C) f(x) * lim(x->C) g(x) > then > lim(x->inf) x*(1/x) = lim(x->inf) 1 = 1 > = lim(x->inf) x * lim(x->inf) 1/x = 1 > = (inf) * (0) = 1 > So inf * 0 = 1. > ( Of course, you could substitute all 1's with, say, N, and get ) > ( inf * 0 = N ) The limit of the product is not the product of the limits. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: division by zero...?? > A (potentially) stupid question :) > What would happens if someone discovers a self-consistent new algebra > of numbers in which we could divide by zero?? revolutionnary? > interesting but not useful? > Thnx There's got to be a way to do it, somewhere, even it looks *really* ugly. I think. Maybe. Or maybe not. Ryan (On that topic, see my multiplying zero by infinity) === Subject: Re: division by zero...?? > A (potentially) stupid question :) > > What would happens if someone discovers a self-consistent new algebra > of numbers in which we could divide by zero?? revolutionnary? > interesting but not useful? > > Thnx > There's got to be a way to do it, somewhere, even it looks *really* ugly. > I think. > Maybe. > Or maybe not. > Ryan > (On that topic, see my multiplying zero by infinity) It looks primally beautiful! See message 3 in this thread. My Hoop algebras unify the useful algebras in a way that explains why there are so few real algebras without divisors of zero, and why real and complex algebras (and quaternions, and octonions) do have divisors of zero if they employ complex coefficients (i.e. when they are used over the complex field). Incidentally, reality is the illusion. Primal numbers (the half line) come first, and real numbers are only one of the multitude of other useful number systems. As for complex numbers, they are the flickering images on the back wall of Plato' cave. Roger Beresford. Beauty is the first test: there isno place in the world for ugly mathematics G. H. Hardy. === Subject: Re: division by zero...?? >>A (potentially) stupid question :) >>What would happens if someone discovers a self-consistent new algebra >>of numbers in which we could divide by zero?? revolutionnary? >>interesting but not useful? >The universe would explode. Not implode? /BAH Subtract a hundred and four for e-mail. === Subject: Re: division by zero...?? >A (potentially) stupid question :) >What would happens if someone discovers a self-consistent new algebra >of numbers in which we could divide by zero?? revolutionnary? >interesting but not useful? >>The universe would explode. > Not implode? No, it would implode if someone were to multiply by zero. (Which of course _has_ been done. This proves that what appears to be the universe is just an illusion - it imploded long ago.) You really have to explain _everything_ to some people. >/BAH >Subtract a hundred and four for e-mail. ************************ David C. Ullrich === Subject: Re: division by zero...?? >A (potentially) stupid question :) >What would happens if someone discovers a self-consistent new algebra >of numbers in which we could divide by zero?? revolutionnary? >interesting but not useful? > Depends on what you mean by divide by zero. But the answer is almost > certainly terribly boring. I have done it (sort of), and got no rational response (i.e. Uncle Al replied) in various places such as Alt.sci and http://library.wolfram.com/infocenter/MathSource/4894/ I actually develop partial-fraction division by zero. If A={a1,a2,...am}, it has a Cramer's rule inverse in any conservative m-element algebra (I call these hoops). The factors of the determinant of the multiplication table (I call them sizes; most algebras have several) are denominators in the partial fraction expansion of the inverse. If one of these sizes is zero, the universe does not explode (Sorry, David) because the factor can be constrained to be zero and the partial fraction is indeterminate, not infinity. > For example, the idea of division is usually that: > a/b = c means that c*b= a. That is, division is defined in terms of > multiplication. > So if you could divide by zero, that would mean that when you have > a/0 = c, that means that c*0 = a. But it is easy to check that, under > the usual rules of arithmetic, c*0 = 0: > etc snip The usual rules are a convention (axiom system to the erudite) to facilitate simple algebra with numbers a rather than sets of numbers A. My standard example is conic sections. Constrain a 3D bicone to 2D and the distance from a 2D plane is zero. You work happily on this plane, having factored out the zeroed distance. > You could, perhaps, define some other properties for the addition and > multiplication that do not imply that if a/0 = c then a=0, and in that > case, their interest will depend on whether or not they are useful for > something; just to be able to divide by zero is simply not a good > enough reason. snip If AB=0, all the sizes of AB are zero. This occurs if each size is zero in either (or both) of A & B. The result is annihilation, though neither A nor B need be zero. Clifford(0,3) is a useful case. Roger Beresford. I am too much of a sceptic to deny the possibility of anything. T.H.Huxley === Subject: Re: division by zero...?? === >Subject: Re: division by zero...?? >Message-id: >>A (potentially) stupid question :) >>What would happens if someone discovers a self-consistent new algebra >>of numbers in which we could divide by zero?? revolutionnary? >>interesting but not useful? >The universe would explode. So what happens when you divide by infinity? Does the universe implode? >>Thnx >************************ >David C. Ullrich -- Mensanator Ace of Clubs === Subject: Re: division by zero...?? === >>Subject: Re: division by zero...?? >>Message-id: >A (potentially) stupid question :) >What would happens if someone discovers a self-consistent new algebra >of numbers in which we could divide by zero?? revolutionnary? >interesting but not useful? >>The universe would explode. >So what happens when you divide by infinity? Does the universe implode? Wouldn't it be the other way around? /BAH Subtract a hundred and four for e-mail. === Subject: Re: division by zero...?? === >Subject: Re: division by zero...?? >Message-id: >>A (potentially) stupid question :) >> >>What would happens if someone discovers a self-consistent new algebra >>of numbers in which we could divide by zero?? revolutionnary? >>interesting but not useful? >The universe would explode. > So what happens when you divide by infinity? Does the universe implode? >>Thnx I think the OP is serious. But I don't know whether the term Algebra of numbers is referring to the Mathematics definition regarding classes of subsets , or an invented term consisting of some structure where the operations *, + and selected inverses and identity elements are defined to allow division by the additive identity element(s) ? Bob Pease === Subject: Re: division by zero...?? === >Subject: Re: division by zero...?? >Message-id: > > >>A (potentially) stupid question :) >> >>What would happens if someone discovers a self-consistent new algebra >>of numbers in which we could divide by zero?? revolutionnary? >>interesting but not useful? > >The universe would explode. > So what happens when you divide by infinity? Does the universe implode? > >>Thnx > I think the OP is serious. > But I don't know whether the term Algebra of numbers is referring to the > Mathematics definition regarding classes of subsets , or an invented term > consisting of some structure where the operations *, + and selected > inverses and identity elements are defined to allow division by the > additive identity element(s) ? > Bob Pease Once, during one of my more sleep-deprived undergrad days, I managed to convince myself through similar reasoning that dividing by zero could yield any real number. I really needed a nap. === Subject: Re: division by zero...?? === >Subject: Re: division by zero...?? >Message-id: > > >>A (potentially) stupid question :) >> >>What would happens if someone discovers a self-consistent new algebra >>of numbers in which we could divide by zero?? revolutionnary? >>interesting but not useful? > >The universe would explode. > > So what happens when you divide by infinity? Does the universe implode? > > >>Thnx > I think the OP is serious. > But I don't know whether the term Algebra of numbers is referring to > the > Mathematics definition regarding classes of subsets , or an invented term > consisting of some structure where the operations *, + and selected > inverses and identity elements are defined to allow division by the > additive identity element(s) ? > Bob Pease > Once, during one of my more sleep-deprived undergrad days, I managed to > convince myself through similar reasoning that dividing by zero could yield > any real number. I really needed a nap. Nope, sometimes sleep deprivation produces moments of insight coupled with delusional thinking. With the ordinary operations on real numbers, allowing division by zero is a classical way to prove that any real number equals any other! Bob Pease === Subject: Re: Do Physicists Understand Their Own Peer-Reviewed Literature? Cut< > In reality we know that physical billiard balls bounce off of each > other Because they deform, and reform to regain most of their original shape: Beyond this is conjecture. >due to underlying forces. Claiming that there exist > underlying forces, hardly seems like a physical explanation to me. > Stephen === Subject: Re: Do Physicists Understand Their Own Peer-Reviewed Literature? There was a controversial case not long ago... Twin television personalities in France getting Ph.D.s in theoretical physics, publishing their theses in peer-reviewed journals, then being accused of writing nonsense. Even the doctoral committee members admitted not understanding it. Something like that. === Subject: Re: Do Physicists Understand Their Own Peer-Reviewed Literature? > There was a controversial case not long ago... Twin television > personalities in France getting Ph.D.s in theoretical physics, > publishing their theses in peer-reviewed journals, then being accused > of writing nonsense. Even the doctoral committee members admitted not > understanding it. Something like that. I suppose that you're talking about the Bogdanov brothers: http://math.ucr.edu/home/baez/bogdanov.html Jose Carlos Santos === Subject: Re: Do Physicists Understand Their Own Peer-Reviewed Literature? : Cut< :> In reality we know that physical billiard balls bounce off of each :> other : Because they deform, and reform to regain most of their original shape: : Beyond this is conjecture. Do you think it is conjecture that billiard balls are made of smaller Stephen === Subject: Re: Do Physicists Understand Their Own Peer-Reviewed Literature? > : Cut< > :> > :> In reality we know that physical billiard balls bounce off of each > :> other > : Because they deform, and reform to regain most of their original shape: > : Beyond this is conjecture. > Do you think it is conjecture that billiard balls are made of smaller > Stephen === Subject: Software for Writing Equations Hi I am new to here. I want to know if there are any softwares to write equations. One of my professor said something like letac or something like that. I didnt ask for the spelling. I want to know if a software like that exists. thnks Suresh === Subject: Re: Software for Writing Equations > Hi > I am new to here. I want to know if there are any softwares to write > equations. Others have replied with LaTex and variants. You can also look to Scientific Word (not free) if you don't want to learn LaTex. Also, if it is simple, use equation editor in word. HTH, Flip === Subject: Re: Software for Writing Equations > Hi > I am new to here. I want to know if there are any softwares to write > equations. > One of my professor said something like letac or something like > that. I didnt ask for the spelling. I want to know if a software like > that exists. > thnks > Suresh If you are using Windows there is MiKTeX at http://www.miktex.org/ and TeXnicCenter at http://www.toolscenter.org/products/texniccenter/. -- G.C. === Subject: Re: Software for Writing Equations > Hi > I am new to here. I want to know if there are any softwares to write > equations. > One of my professor said something like letac or something like > that. I didnt ask for the spelling. I want to know if a software like > that exists. LaTeX It's available for download on the net for free. === Subject: Re: Software for Writing Equations LaTeX > Hi > I am new to here. I want to know if there are any softwares to write > equations. > One of my professor said something like letac or something like > that. I didnt ask for the spelling. I want to know if a software like > that exists. > thnks > Suresh === Subject: Re: Short Alg Int Argument Visiting Assistant Professor at the University of Montana. [.snip.] I've split the response in 3 since you tend to complain about length whenever you want some excuse to avoid looking at the math. I'm tired of your red herring. Here is why this example is nothing but misdirection on your part. >I answered this issue in another post where I used P(x) = 11x + 123, >and I'll copy from it to here. >Consider the factorization >(11 + (x+sqrt(x^2 - 8))/2)(11 + (x-sqrt(x^2 - 8))/2) = 11x + 123 >where the constant term is *still* given by setting x=0, which gives >P(0) = (11 + sqrt(-8)/2)(11 - sqrt(-8)/2) = 123 >as it must. This is a red herring. The reason it's a red herring is that your polynomial here does not have the same characteristics as the one in your argument. The way you did this is by taking a polynomial that was quadratic in m and linear in x, Q(x,m) = m^2 + mx + 2 and then factoring it with respect to m; the two roots are, indeed, Q(x,m) = (m-a)(m-b) = (m + (x + sqrt(x^2-8))/2)(m + (x - sqrt(x^2-8))/2) and then plugged in m=11: Q(x,11) = (11 + (x+sqrt(x^2-8))/2)(11 + (x- sqrt(x^2-8))/2) The reason why this is dishonest is that it bears no similarity to what you do in your argument. In your argument, you have a polynomial which is of degree 3 in BOTH m and x: P(m,x) = f^2[(m^3f^4 - 3m^2f^2 + 3m)x^3 - 3(-1+mf^2)u^2x + u^3f] = f^2[(x^3f^4)m^3 - 3(f^3x^2)m^2 + 3(x^3 - f^2u^2)m + (u^3f-3)]. which you then factor with respect to x: P(m,x) = (a_1(m)x+uf) (a_2(m)x+uf) (a_3(m)x+uf) and you try to derive conclusions about a_1, a_2, and a_3 for arbitrary m from the value at m=0. Note, moreover, that each nonzero coefficient of positive degree with respect to one of the variables is a non-constant function of the other; so the leading coefficient (with respect to x^3) is a non-constant function of m, the linear coefficient of x is a non-constant function of m; the coefficient of m^3 is a nonconstant function of x, the coefficient of m^3 is a nonconstant function of x, and the coefficient of m is a nonconstant function of x. In addition, the constant terms with respect to x is not the same as the constant term with respect to m. And none of the terms in one expression is equal to a term in the other. In your dishonest, misdirecting, red-herring pseudo-example, the degrees with respect to x and m are different. In your real application, they are equal. In the pseudo-example, the leading coefficient with respect to m is constant. In the real application, all nonzero coefficients of positive powers with respect to each variable is a non-constant function of the OTHER variable. And one of the terms in one expression is equal to one of the terms in the other expression. Also, the fact that the roots in a degree 2 can be expressed as a constant plus a radical bears striking dissimilarity with the situation in degree 3, where the roots are usually expressed as sums of radicals, with little or nothing outside of them. This may have important consequences with respect to divisibility. In addition, the test case of m=0 in your original example leads to a polynomial of degree 1, strictly smaller than the degree for arbitrary m. Or if you prefer, it leads to the fact that the polynomial that defines the coefficients a_1(m), a_2(m), a_3(m) is completely reducible over Q: it is a product of three linear terms. In this example, when x=0, you get m^2+2, of the same degree as the original polynomial; and irreducible over Q. And irreducibility vs. reducibility is the ->key<- to the problem your argument has. The way the prime factors break up depends on whether the polynomial defining the a's is reducible into three linear terms, into a linear term times a quadratic, or irreducible over Q. When it is reducible into three linear terms (as in the case with m=0) you get the prime factors distributed discretely among the a's. When it is reducible into a linear times an irreducible quadratic, you get some factors going completely to one of the coefficients, and the others are each divided into two non-unit parts, and each of the remaining a's gets part of it. When it is irreducible, each prime factor is broken up into three non-unit factors, and each of the coefficients gets one of the parts. That's why, in your example, say when f=2, u=1, when m=0 you get y^3 + 3(-1)y^2 = y^2 (y-3) you get a_1=a_2=0, a_3=3; all integers, the prime factor 3 having gone to just one. When m=1, however, you get a constant term of 28; the polynomial defining the a's splits into an irreducible quadratic times a linear, y^3 + 3(-1+4)y^2 - 4(16 - 14 + 3) = y^3 + 9y^2 - 28 = (y+2)(y^2 +7y - 14) so the prime factor -2 goes to a_3, and the other two prime factors, 2 and 7, get split between the remaining two coefficients; 7 breaks up as sqrt(7)*sqrt(7), and 2 breaks up as sqrt[ (11-sqrt(105))/2] * sqrt[ (11+sqrt(105))/2] (both algebraic integers) and the two remaining coefficients become a1 = sqrt(7) * sqrt[ (11-sqrt(105))/2] a2 = sqrt(7) * sqrt[ (11+sqrt(105))/2] When the polynomial defining the a's is irreducible of degree 3, the same phenomenon you saw with x^2 + 7x - 14, where the factors 2 and 7 of the constant term get divided among all the roots in equal portions will occur with all the factors. Your example is thus seen to be worthless, and inapplicable as a test case. When you bring it up, you are not clarifying, you are confusing. You are purposely presenting a dis-similar example as if it were an instance of the same situation. It is not. You do not even attempt to present the same argument with it: you never conclude anything about the divisibility of the coefficients based on the values at a single point; not that it would be worth doing, since in your pseudo-example the relevant coefficients are constant. So your example that explains fails to bear even the most obvious similarities to the situation you actually deal with: (1) The degrees in x and in m are different; (2) Not all nonzero coefficients of positive powers with respect to m are functions of the x; (3) The leading coefficient with respect to m, the variable of factorization is constant, not a function of x. (Note also that you reverse the roles of x and y in this pseudo-example, leading to even more possible confusion on the part of the reader). Arguing that this example is like is false; it is misdirection and confusion, and nothing short of dishonesty. [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee fixe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a flash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not find it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Short Alg Int Argument Visiting Assistant Professor at the University of Montana. [.snip.] >The constant terms of the g's are given by setting m=0. That's just >basic algebra as that way m is removed from the picture. So, for example, if we let g(m) = sqrt(1+m) - sqrt(1-m), then the constant term is g(0) = sqrt(1) - sqrt(1) =0. Does that mean that for every value of m, g(m) is NOT coprime to m, since the constant term is 0? Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Short Alg Int Argument > [.snip.] >The constant terms of the g's are given by setting m=0. That's just >basic algebra as that way m is removed from the picture. > So, for example, if we let g(m) = sqrt(1+m) - sqrt(1-m), > then the constant term is g(0) = sqrt(1) - sqrt(1) =0. Yes, as the constant term with respect to m is obtained by setting m=0. > Does that mean that for every value of m, g(m) is NOT coprime to m, > since the constant term is 0? Multiplying g(m) by sqrt(1+m) + sqrt(1-m) gives 1+m -1 + m = 2m, which shows that g(m) is a factor of 2m. Provably for any integer m, both g(m) and sqrt(1+m)+sqrt(1-m) have a factor that is sqrt(2). Trivially, g(m) might be coprime to m if g(m) is a unit. Proving the case for any algebraic integer m, might be more interesting, but I don't doubt the outcome as in fact it *must* be true that the constant term is given by setting m=0, as if not, then it wouldn't be the constant term, now would it? James Harris === Subject: Re: Short Alg Int Argument >> [.snip.] >>The constant terms of the g's are given by setting m=0. That's just >>basic algebra as that way m is removed from the picture. >> So, for example, if we let g(m) = sqrt(1+m) - sqrt(1-m), >> then the constant term is g(0) = sqrt(1) - sqrt(1) =0. >Yes, as the constant term with respect to m is obtained by setting >m=0. >> Does that mean that for every value of m, g(m) is NOT coprime to m, >> since the constant term is 0? >Multiplying g(m) by sqrt(1+m) + sqrt(1-m) gives 1+m -1 + m = 2m, which >shows that g(m) is a factor of 2m. Which is not what I asked. Is g(m) NOT coprime to m for every value of m? >Provably for any integer m, both g(m) and sqrt(1+m)+sqrt(1-m) have a >factor that is sqrt(2). >Trivially, g(m) might be coprime to m if g(m) is a unit. So what? Your argument claims that since g(m) is NOT coprime to m when m=0, then we should conclude that g(m) is NOT coprime to m for any value of m. That is ->exactly<- what you are arguing in your APF. Don't change the claim just because you realize it will be false. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Short Alg Int Argument >> >> [.snip.] >> >>The constant terms of the g's are given by setting m=0. That's just >>basic algebra as that way m is removed from the picture. >> >> So, for example, if we let g(m) = sqrt(1+m) - sqrt(1-m), >> then the constant term is g(0) = sqrt(1) - sqrt(1) =0. >Yes, as the constant term with respect to m is obtained by setting >m=0. > >> Does that mean that for every value of m, g(m) is NOT coprime to m, >> since the constant term is 0? >Multiplying g(m) by sqrt(1+m) + sqrt(1-m) gives 1+m -1 + m = 2m, which >shows that g(m) is a factor of 2m. > Which is not what I asked. Is g(m) NOT coprime to m for every value of m? >Provably for any integer m, both g(m) and sqrt(1+m)+sqrt(1-m) have a >factor that is sqrt(2). >Trivially, g(m) might be coprime to m if g(m) is a unit. > So what? Your argument claims that since g(m) is NOT coprime to m when > m=0, then we should conclude that g(m) is NOT coprime to m for any > value of m. That is ->exactly<- what you are arguing in your APF. Well, actually no, I simply note that setting m=0 gives the constant term. Basically setting m=0 clears m out of the picture, giving a result which does not vary with m, necessarily. Accepting that is just accepting rather basic algebra. > Don't change the claim just because you realize it will be false. What I do is consider factors of P(m) after setting m=0, which necessarily makes them factors of P(0). I have no change in claim. Readers can see the argument at a site where allowal of use of LaTeX makes it easier to read at http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 and verify that for themselves. James Harris === Subject: Re: Short Alg Int Argument Visiting Assistant Professor at the University of Montana. > > [.snip.] > >The constant terms of the g's are given by setting m=0. That's just >basic algebra as that way m is removed from the picture. > > So, for example, if we let g(m) = sqrt(1+m) - sqrt(1-m), > then the constant term is g(0) = sqrt(1) - sqrt(1) =0. >> >>Yes, as the constant term with respect to m is obtained by setting >>m=0. >> > Does that mean that for every value of m, g(m) is NOT coprime to m, > since the constant term is 0? >> >>Multiplying g(m) by sqrt(1+m) + sqrt(1-m) gives 1+m -1 + m = 2m, which >>shows that g(m) is a factor of 2m. >> Which is not what I asked. Is g(m) NOT coprime to m for every value of m? I don't see an answer to the question. Please respond. >>Provably for any integer m, both g(m) and sqrt(1+m)+sqrt(1-m) have a >>factor that is sqrt(2). >> >>Trivially, g(m) might be coprime to m if g(m) is a unit. >> So what? Your argument claims that since g(m) is NOT coprime to m when >> m=0, then we should conclude that g(m) is NOT coprime to m for any >> value of m. That is ->exactly<- what you are arguing in your APF. >Well, actually no, I simply note that setting m=0 gives the constant >term. And then arguing that if that constant term is coprime to m, then g(m) will be coprime to m, thinking that you have g(m) = (something not coprime to m) + (constant term). But you don't. That's what happens with polynomials, and you imagine it's what happens with your functions. But it just is not. You have admitted that it does not work, you just refuse to take the final step from A is true, B is true, and C is true to Each of A, B, and C are true. >> Don't change the claim just because you realize it will be false. >What I do is consider factors of P(m) after setting m=0, which >necessarily makes them factors of P(0). >I have no change in claim. You have avoided making the analogous claim for this instance. So, in this case, we have g(0) = 0. What is your conclusion from that What is the conclusion in general, from g(0) = a? I know we can write g(m) = r(m) + g(0), by setting r(m) = g(m)-g(0). But the question is: what is the general property that r(m) has that allows you to conclude that if g(0) is coprime to f, then g(m) will also be coprime to f for every m? Because that IS Your conclusion in your argument. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Short Alg Int Argument > Why do you take so much trouble to expose such a reasoner as > Mr. Smith? I answer as a deceased friend of mine used to answer > on like occasions - A man's capacity is no measure of his power > to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more; > and a long purse, which does most of all. He has made at least > ten publications, full of figures few readers can critize. A great > many people are staggered to this extend, that they imagine there > must be the indefinite something in the mysterious all this. > They are brought to the point of suspicion that the mathematicians > ought not to treat all this with such undisguised contempt, > at least. > -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan If you liked A Budget of Paradoxes, you'll probably enjoy Pseudoscience and the Paranormal by Terence Hines. The second edition is just out. << Popular culture fills the mind with a steady diet of fantasy, from tales of UFO landings and alien abductions, haunted houses, and communication with the dead to claims of miraculous cures by spiritual healers and breakthrough treatments in alternative medicine. The paranormal - and the pseudoscience that attempts to validate it - is so ubiquitous that many people lose sight of the distinction between the real and the imaginary, and some never learn to make the distinction in the first place. >> -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Short Alg Int Argument Visiting Assistant Professor at the University of Montana. >> [.snip.] >Then it follows from the constant terms that g_1 and g_2 each have a >factor that is f. > > There is no constant terms of g_1 and g_2. The constant term of > g_1(0) and the constant term of g_2(0) are multiples of f, yes: > g_1(0)=uf, and g_2(0)=uf. > > But you have NOT shown that g_1, g_2, or g_3, as functions of m that > they are, have a well defined constant term. For all we know, g_1(1) > is a polynomial with a DIFFERENT constant term from g_1(0); same for > g_2 and g_3. You have your big hunking gap here. >> >>So then the position is that the constant term might change with m, >>which is nonsensical on its face. >> Well, let's see. When u=1, f=2, we have >I answered this issue in another post where I used P(x) = 11x + 123, >and I'll copy from it to here. Red herring. >> P(0) = 4(3x+2) = (0x+2)(0x+2)(3x+2), >> with a1=0, a2=0, a3=3. >> So g1=2, g2=2, g3=3x+2; the first two are multiples of 2, the last one >> is 'coprime to 2' when x is an algebraic integer coprime to 2. >> When m=1, we get >> P(1) = 4*(7x^3 - 9x + 2) >> = [ ((-7-sqrt(105))/2)x +2][ ((-7+sqrt(105))/2)x+2][-2x+2] >> with >> a1 = (-7-sqrt(105))/2 >> a2 = (-7+sqrt(105))/2 >> a3 = -2, >> obtained from your defining polynomial >> a^3 + 3(-1+mf^2)a^2 - f^2(m^3f^4-3m^2f^2 + 3m) = >> = a^3 +3(-1+4)a^2 - 4(16 - 12 + 3) >> = a^3 + 9a^2 - 28 >> = (a+2)(a^2 +7a - 14). >> a1 and a2 are the roots of a^2+7a-14. >> Then we can also write >> a1 = (-7-sqrt(105))/2 = sqrt(7) * sqrt[ (11+sqrt(105))/2] >> a2 = (-7+sqrt(105))/2 = sqrt(7) * sqrt[ (11-sqrt(105))/2]. >> Note that sqrt[ (11-sqrt(105))/2] and sqrt[ (11+sqrt(105))/2] >> are roots of the monic, irreducible, primitive polynomial with integer >> coefficients x^4 - 11x^2 + 4, and since the constant term is not 1 or >> -1 they are not units. Also, their product is: >> sqrt[ (11-sqrt(105))/2]*sqrt[ (11+sqrt(105))/2] = >> = sqrt[ (121-105)/4] >> = sqrt[ 16/4] >> = sqrt[4] >> = 2, >> so their are both factors of 2. >> So we have >> g1 = a1*x+2 >> = [(-7-sqrt(105))/2]*x + 2 >> = [sqrt(7) * sqrt[ (11+sqrt(105))/2]]*x + 2 >> = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] >> g2 = a2*x + 2 >> = [(-7+sqrt(105))/2]*x + 2 >> = [sqrt(7) * sqrt[ (11-sqrt(105))/2]]*x + 2 >> = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] >> g3 = a3*x + 2 >> = -2x + 2 >> = -2(x-1) >> So the factors of 2 now distribute as follows: >> sqrt[(11+sqrt(105))/2] into g1 >> sqrt[(11-sqrt(105))/2] into g2 >> -2 into g3. >And in fact one of the g's does not share non-unit factors with 2, but >how do you tell by physically looking? No, they don't. ALL of them shre non-unit factors with 2: g3 shares the non-unit factor -2. g2 shares the non-unit factor sqrt[(11-sqrt(105))/2]. And g1 shares the non-unit factor sqrt[(11+sqrt(105))/2]. There they are. Explicitly stated, explicitly factored out. >> I don't know what it is you think is the constant term of g1, g2, or >> g3. But as you can plainly see, the factors of 2 get distributed to >> ALL of a1, a2, a3 when m=1, and to ALL of g1, g2, g3; when m=0, only >> g1 and g2 are multiples of 2. >The constant terms of the g's are given by setting m=0. That's just >basic algebra as that way m is removed from the picture. No, that's what happens with a polynomial. But g is NOT a polynomial. Constant term has no meaning for arbitrary functions, so it is not just basic algebra. Don't lie, James. [.snip.] >> So, I don't know if it is nonsensical in its face, but I do know >> that it is TRUE that when the value of m changes, the way in which the >> factors of 4 get distributed among g1, g2, and g3 changes as well. >That's your wish, but it doesn't make mathematical sense. So you claim. Yet there it is above: ALL g's have a non-unit common factor with 2. How come? [.snip.] Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Short Alg Int Argument >> >> [.snip.] >> >> >Then it follows from the constant terms that g_1 and g_2 each have a >factor that is f. > > There is no constant terms of g_1 and g_2. The constant term of > g_1(0) and the constant term of g_2(0) are multiples of f, yes: > g_1(0)=uf, and g_2(0)=uf. > > But you have NOT shown that g_1, g_2, or g_3, as functions of m that > they are, have a well defined constant term. For all we know, g_1(1) > is a polynomial with a DIFFERENT constant term from g_1(0); same for > g_2 and g_3. You have your big hunking gap here. >> >>So then the position is that the constant term might change with m, >>which is nonsensical on its face. >> >> Well, let's see. When u=1, f=2, we have >I answered this issue in another post where I used P(x) = 11x + 123, >and I'll copy from it to here. > Red herring. No, you are challenging my mathematical argument, and I clarified that your challenge depends on the assertion that the constant term might change with m. Rather than answer you went off on a tangent, and I presented an example using P(x) = 11x + 123 which considered that position hoping that a simpler example might help, but here I see you deleted it out. >> P(0) = 4(3x+2) = (0x+2)(0x+2)(3x+2), >> >> with a1=0, a2=0, a3=3. >> >> So g1=2, g2=2, g3=3x+2; the first two are multiples of 2, the last one >> is 'coprime to 2' when x is an algebraic integer coprime to 2. >> >> When m=1, we get >> >> P(1) = 4*(7x^3 - 9x + 2) >> = [ ((-7-sqrt(105))/2)x +2][ ((-7+sqrt(105))/2)x+2][-2x+2] >> >> with >> >> a1 = (-7-sqrt(105))/2 >> a2 = (-7+sqrt(105))/2 >> a3 = -2, >> >> obtained from your defining polynomial >> >> a^3 + 3(-1+mf^2)a^2 - f^2(m^3f^4-3m^2f^2 + 3m) = >> = a^3 +3(-1+4)a^2 - 4(16 - 12 + 3) >> = a^3 + 9a^2 - 28 >> = (a+2)(a^2 +7a - 14). >> >> a1 and a2 are the roots of a^2+7a-14. >> >> Then we can also write >> >> a1 = (-7-sqrt(105))/2 = sqrt(7) * sqrt[ (11+sqrt(105))/2] >> >> a2 = (-7+sqrt(105))/2 = sqrt(7) * sqrt[ (11-sqrt(105))/2]. >> >> Note that sqrt[ (11-sqrt(105))/2] and sqrt[ (11+sqrt(105))/2] >> are roots of the monic, irreducible, primitive polynomial with integer >> coefficients x^4 - 11x^2 + 4, and since the constant term is not 1 or >> -1 they are not units. Also, their product is: >> >> sqrt[ (11-sqrt(105))/2]*sqrt[ (11+sqrt(105))/2] = >> = sqrt[ (121-105)/4] >> = sqrt[ 16/4] >> = sqrt[4] >> = 2, >> >> so their are both factors of 2. >> >> So we have >> >> g1 = a1*x+2 >> = [(-7-sqrt(105))/2]*x + 2 >> = [sqrt(7) * sqrt[ (11+sqrt(105))/2]]*x + 2 >> = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] >> >> g2 = a2*x + 2 >> = [(-7+sqrt(105))/2]*x + 2 >> = [sqrt(7) * sqrt[ (11-sqrt(105))/2]]*x + 2 >> = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] >> >> >> g3 = a3*x + 2 >> = -2x + 2 >> = -2(x-1) >> >> So the factors of 2 now distribute as follows: >> >> sqrt[(11+sqrt(105))/2] into g1 >> sqrt[(11-sqrt(105))/2] into g2 >> -2 into g3. >And in fact one of the g's does not share non-unit factors with 2, but >how do you tell by physically looking? > No, they don't. ALL of them shre non-unit factors with 2: g3 shares > the non-unit factor -2. > g2 shares the non-unit factor sqrt[(11-sqrt(105))/2]. > And g1 shares the non-unit factor sqrt[(11+sqrt(105))/2]. > There they are. Explicitly stated, explicitly factored out. However, factors of 1 are typically excluded and here because of the error in core, though one of the factors *should* be a factor of 1, it is not *in the ring of algebraic integers*, which leads to the contradiction. Consider (2a)b = 2, where 'a' is NOT an algebraic integer, while 2a and 'b' both are, so then though ab=1 in a more inclusive ring, 'b' is not a unit *in the ring of algebraic integers* so trivially it is indeed a non-unit factor. The more inclusive ring is given by the the following definition: The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Notice it excludes a possibility like a=1/2 as then 2 would be a unit in the ring. My point to reader is that Arturo Magidin clearly knows what he is doing and he is relying on the very *error* in core that I've pointed out, which is not behavior that would be expected from a mathematician. >> I don't know what it is you think is the constant term of g1, g2, or >> g3. But as you can plainly see, the factors of 2 get distributed to >> ALL of a1, a2, a3 when m=1, and to ALL of g1, g2, g3; when m=0, only >> g1 and g2 are multiples of 2. >The constant terms of the g's are given by setting m=0. That's just >basic algebra as that way m is removed from the picture. > No, that's what happens with a polynomial. But g is NOT a > polynomial. Constant term has no meaning for arbitrary functions, so > it is not just basic algebra. Don't lie, James. > [.snip.] Setting m=0 gives those terms which do NOT vary as m varies, which IS basic algebra. >> So, I don't know if it is nonsensical in its face, but I do know >> that it is TRUE that when the value of m changes, the way in which the >> factors of 4 get distributed among g1, g2, and g3 changes as well. >That's your wish, but it doesn't make mathematical sense. > So you claim. Yet there it is above: ALL g's have a non-unit common > factor with 2. How come? > [.snip.] Because you have a situation like I showed where one factor is 2a while the other is 'b', but 'a' is not an algebraic integer. James Harris === Subject: Re: Short Alg Int Argument >>[...] > I don't know what it is you think is the constant term of g1, g2, or > g3. But as you can plainly see, the factors of 2 get distributed to > ALL of a1, a2, a3 when m=1, and to ALL of g1, g2, g3; when m=0, only > g1 and g2 are multiples of 2. >> >>The constant terms of the g's are given by setting m=0. That's just >>basic algebra as that way m is removed from the picture. >> No, that's what happens with a polynomial. But g is NOT a >> polynomial. Constant term has no meaning for arbitrary functions, so >> it is not just basic algebra. Don't lie, James. >> [.snip.] >Setting m=0 gives those terms which do NOT vary as m varies, which IS >basic algebra. Setting m = 42 _also_ gives something which does not vary as m varies. Saying the value at m = 0 is the constant term is _not_ basic algebra, for a function other than a polynomial. > So, I don't know if it is nonsensical in its face, but I do know > that it is TRUE that when the value of m changes, the way in which the > factors of 4 get distributed among g1, g2, and g3 changes as well. >> >>That's your wish, but it doesn't make mathematical sense. >> So you claim. Yet there it is above: ALL g's have a non-unit common >> factor with 2. How come? >> [.snip.] >Because you have a situation like I showed where one factor is 2a >while the other is 'b', but 'a' is not an algebraic integer. >James Harris ************************ David C. Ullrich === Subject: Re: Short Alg Int Argument Visiting Assistant Professor at the University of Montana. [.snip.] > P(0) = 4(3x+2) = (0x+2)(0x+2)(3x+2), > > with a1=0, a2=0, a3=3. > > So g1=2, g2=2, g3=3x+2; the first two are multiples of 2, the last one > is 'coprime to 2' when x is an algebraic integer coprime to 2. > > When m=1, we get > > P(1) = 4*(7x^3 - 9x + 2) > = [ ((-7-sqrt(105))/2)x +2][ ((-7+sqrt(105))/2)x+2][-2x+2] > > with > > a1 = (-7-sqrt(105))/2 > a2 = (-7+sqrt(105))/2 > a3 = -2, > > obtained from your defining polynomial > > a^3 + 3(-1+mf^2)a^2 - f^2(m^3f^4-3m^2f^2 + 3m) = > = a^3 +3(-1+4)a^2 - 4(16 - 12 + 3) > = a^3 + 9a^2 - 28 > = (a+2)(a^2 +7a - 14). > > a1 and a2 are the roots of a^2+7a-14. > > Then we can also write > > a1 = (-7-sqrt(105))/2 = sqrt(7) * sqrt[ (11+sqrt(105))/2] > > a2 = (-7+sqrt(105))/2 = sqrt(7) * sqrt[ (11-sqrt(105))/2]. > > Note that sqrt[ (11-sqrt(105))/2] and sqrt[ (11+sqrt(105))/2] > are roots of the monic, irreducible, primitive polynomial with integer > coefficients x^4 - 11x^2 + 4, and since the constant term is not 1 or > -1 they are not units. Also, their product is: > > sqrt[ (11-sqrt(105))/2]*sqrt[ (11+sqrt(105))/2] = > = sqrt[ (121-105)/4] > = sqrt[ 16/4] > = sqrt[4] > = 2, > > so their are both factors of 2. > > So we have > > g1 = a1*x+2 > = [(-7-sqrt(105))/2]*x + 2 > = [sqrt(7) * sqrt[ (11+sqrt(105))/2]]*x + 2 > = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] > > g2 = a2*x + 2 > = [(-7+sqrt(105))/2]*x + 2 > = [sqrt(7) * sqrt[ (11-sqrt(105))/2]]*x + 2 > = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] > > > g3 = a3*x + 2 > = -2x + 2 > = -2(x-1) > > So the factors of 2 now distribute as follows: > > sqrt[(11+sqrt(105))/2] into g1 > sqrt[(11-sqrt(105))/2] into g2 > -2 into g3. >> >>And in fact one of the g's does not share non-unit factors with 2, but >>how do you tell by physically looking? >> No, they don't. ALL of them shre non-unit factors with 2: g3 shares >> the non-unit factor -2. >> g2 shares the non-unit factor sqrt[(11-sqrt(105))/2]. >> And g1 shares the non-unit factor sqrt[(11+sqrt(105))/2]. >> There they are. Explicitly stated, explicitly factored out. >However, factors of 1 are typically excluded and here because of the >error in core, though one of the factors *should* be a factor of 1, it >is not *in the ring of algebraic integers*, which leads to the >contradiction. That's stupid. None of them are factors of 1. ALL OF THE g's HAVE COMMON FACTORS WITH 2. ALL OF THEM. You claimed one of them would necessarily be coprime to 2. NONE OF THEM ARE COPRIME TO 2, ACCORDING TO YOUR DEFINITION. They have common, non-unit factors. So your claim is false. Pure, and simple. Now you are saying that one of them should be a factor of 1. But if that is so, then you are still in trouble. Because either you will make 2 into a unit (which you do not want to), or else that means that you want to make sqrt[(11+sqrt(105))/2] into a unit. If you do that, then sqrt[(11-sqrt(105))/2] is ALSO a unit, which again makes 2 into a unit. And even if you don't believe that making one of them into a unit will make 2 into a unit, you are still in trouble. Because if you make sqrt[ (11+sqrt(105))/2] into a unit, then the best you can factor out from the right hand side would be sqrt[ (11-sqrt(105))/2 ] * 2, which does NOT equal 4. As to making a bigger ring where they are units, that's still nonsense: EVERY ELEMENT OF THE RING OF ALGEBRAIC INTEGERS IS A UNIT IN SOME LARGER RING. Just like in the integers. >Consider (2a)b = 2, where 'a' is NOT an algebraic integer, while 2a >and 'b' both are, so then though ab=1 in a more inclusive ring, 'b' is >not a unit *in the ring of algebraic integers* so trivially it is >indeed a non-unit factor. Red herring. >The more inclusive ring is given by the the following definition: >The Object Ring is a commutative ring that includes all numbers such >that -1 and 1 are the only members that are both a unit and an >integer, where no non-unit member is a factor of any two integers that >are coprime. This definition is nonsense as written, as we have pointed out numerous times. It has no referent. >Notice it excludes a possibility like a=1/2 as then 2 would be a unit >in the ring. Then none of 2, sqrt[ (11-sqrt(15))/2], and sqrt[ (11+sqrt(15))/2] are units in the Object ring, because if ANY of them are units, then 2 is a unit. So even if object ring made sense, you would still NOT be able to conclude that one the three factors is a unit in that ring. >My point to reader is that Arturo Magidin clearly knows what he is >doing and he is relying on the very *error* in core If there were an error in core (assuming you mean, if there is some contradiction in basic mathematics), then every proof is correct and valid. EVERYTHING can be proven. My proof would be just as correct as yours, so there is simply no way that you can claim I am lying because there is an error in core. But you have not pointed out any errors. Your argumetn has the gap of assuming that the behavior of complex functions of m at m=0 will yield information about their behavior at points other than m=0. When you say something should have a factor, what you mean is that you would ->like<- it to have a factor, but that it doesn't. That's not an error in mathematics, that you engaging in wishful thinking. [.snip.] > So, I don't know if it is nonsensical in its face, but I do know > that it is TRUE that when the value of m changes, the way in which the > factors of 4 get distributed among g1, g2, and g3 changes as well. >> >>That's your wish, but it doesn't make mathematical sense. >> So you claim. Yet there it is above: ALL g's have a non-unit common >> factor with 2. How come? >> [.snip.] >Because you have a situation like I showed where one factor is 2a >while the other is 'b', but 'a' is not an algebraic integer. No, I do not have that situation. ALL the numbers above are ALGEBRAIC INTEGERS. Your example relies on a not being an algebraic integer. But here they are again, James. Look CAREFULLY: > g1 = a1*x+2 > = [(-7-sqrt(105))/2]*x + 2 > = [sqrt(7) * sqrt[ (11+sqrt(105))/2]]*x + 2 > = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] sqrt[ (11+sqrt(105))/2 ] is an algebraic integer. It is the root of y^4 + 11y^2 + 4 = 0. sqrt(7) is an algebraic integer. x is an algebraic integer (by assumption). So sqrt(7)*x is an algebraic integer. sqrt[ (11-sqrt(105))/2] is an algebraic integer, root of y^4 + 11y^2 + 4 = 0. So (sqrt(7)*x) + sqrt [ (11-sqrt(105))/2] is an algebraic integer. So both factors of g1 above are algebraic integers. > g2 = a2*x + 2 > = [(-7+sqrt(105))/2]*x + 2 > = [sqrt(7) * sqrt[ (11-sqrt(105))/2]]*x + 2 > = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] Again, sqrt(7) is an algebraic integer. x is an algebraic integer. Therefore, sqrt(7)*x is an algebraic integer. sqrt[ (11+sqrt(105))/2] is an algebraic integer. So sqrt(7)*x + sqrt[ (11+sqrt(105))/2] is an algebraic integer. sqrt[(11-sqrt(105))/2] is an algebraic integer. So the factorization given is a factorization in which BOTH FACTORS are algebraic integers. > g3 = a3*x + 2 > = -2x + 2 > = -2(x-1) -2 is an algebraic integer. x is an algebraic integer. -1 is an algebraic integer. So x-1 is an algebraic integer. So this is a factorization INTO ALGEBRAIC INTEGERS. Your example does NOT apply. You are the one engaging in deceitful disinformation. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Short Alg Int Argument > [.snip.] > P(0) = 4(3x+2) = (0x+2)(0x+2)(3x+2), > > with a1=0, a2=0, a3=3. > > So g1=2, g2=2, g3=3x+2; the first two are multiples of 2, the last one > is 'coprime to 2' when x is an algebraic integer coprime to 2. > > When m=1, we get > > P(1) = 4*(7x^3 - 9x + 2) > = [ ((-7-sqrt(105))/2)x +2][ ((-7+sqrt(105))/2)x+2][-2x+2] > > with > > a1 = (-7-sqrt(105))/2 > a2 = (-7+sqrt(105))/2 > a3 = -2, > > obtained from your defining polynomial > > a^3 + 3(-1+mf^2)a^2 - f^2(m^3f^4-3m^2f^2 + 3m) = > = a^3 +3(-1+4)a^2 - 4(16 - 12 + 3) > = a^3 + 9a^2 - 28 > = (a+2)(a^2 +7a - 14). > > a1 and a2 are the roots of a^2+7a-14. > > Then we can also write > > a1 = (-7-sqrt(105))/2 = sqrt(7) * sqrt[ (11+sqrt(105))/2] > > a2 = (-7+sqrt(105))/2 = sqrt(7) * sqrt[ (11-sqrt(105))/2]. > > Note that sqrt[ (11-sqrt(105))/2] and sqrt[ (11+sqrt(105))/2] > are roots of the monic, irreducible, primitive polynomial with integer > coefficients x^4 - 11x^2 + 4, and since the constant term is not 1 or > -1 they are not units. Also, their product is: > > sqrt[ (11-sqrt(105))/2]*sqrt[ (11+sqrt(105))/2] = > = sqrt[ (121-105)/4] > = sqrt[ 16/4] > = sqrt[4] > = 2, > > so their are both factors of 2. > > So we have > > g1 = a1*x+2 > = [(-7-sqrt(105))/2]*x + 2 > = [sqrt(7) * sqrt[ (11+sqrt(105))/2]]*x + 2 > = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] > > g2 = a2*x + 2 > = [(-7+sqrt(105))/2]*x + 2 > = [sqrt(7) * sqrt[ (11-sqrt(105))/2]]*x + 2 > = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] > > > g3 = a3*x + 2 > = -2x + 2 > = -2(x-1) > > So the factors of 2 now distribute as follows: > > sqrt[(11+sqrt(105))/2] into g1 > sqrt[(11-sqrt(105))/2] into g2 > -2 into g3. >> >>And in fact one of the g's does not share non-unit factors with 2, but >>how do you tell by physically looking? >> >> No, they don't. ALL of them shre non-unit factors with 2: g3 shares >> the non-unit factor -2. >> >> g2 shares the non-unit factor sqrt[(11-sqrt(105))/2]. >> >> And g1 shares the non-unit factor sqrt[(11+sqrt(105))/2]. >> >> There they are. Explicitly stated, explicitly factored out. >However, factors of 1 are typically excluded and here because of the >error in core, though one of the factors *should* be a factor of 1, it >is not *in the ring of algebraic integers*, which leads to the >contradiction. > That's stupid. None of them are factors of 1. ALL OF THE g's HAVE > COMMON FACTORS WITH 2. ALL OF THEM. What's stupid is your objection to the argument at http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 where you keep challenging the conclusion that setting m=0 gives results that don't vary with m. As for your attempt at refuting that proof using an example, I've explained exactly how it fails as you have something like (2a)b = 2, where 'a' is NOT an algebraic integer, and for *that* reason as in a more inclusive ring ab=1, you have that 'b' is NOT a unit *in the ring of algebraic integers* as by definition if it's a unit in that ring then it has a factor in that ring, while 'a' is NOT an algebraic integer. > You claimed one of them would necessarily be coprime to 2. NONE OF > THEM ARE COPRIME TO 2, ACCORDING TO YOUR DEFINITION. They have common, > non-unit factors. > So your claim is false. Pure, and simple. A hallmark of an error in core mathematics is the ability for *two* sides to sit and argue that they're right, while luckily, the error here is simply one of arbitrary exclusion, as the definition for algebraic integers, with the focus on monic polynomials with integer coefficients, creates the problem. So consistency is restored by noting that that definition does not produce a complete ring. To believe you, readers must challenge algebra itself, and the necessary result that setting m=0 gives an expression that does not vary with m. However, otherwise, they need only accept that the definition does indeed exclude numbers that have to be included to avoid contradiction, and in this case that means that there are roots of non-monic polynomials with integer coefficients that have to be included. So readers can accept algebra, or accept an arbitrary definition. The algebra is quickly and easily gone over by considering it at the following link: http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 > Now you are saying that one of them should be a factor of 1. But if > that is so, then you are still in trouble. Because either you will > make 2 into a unit (which you do not want to), or else that means that > you want to make sqrt[(11+sqrt(105))/2] into a unit. If you do that, > then sqrt[(11-sqrt(105))/2] is ALSO a unit, which again makes 2 into a > unit. That last sentence does not follow logically from the previous ones. > And even if you don't believe that making one of them into a unit will > make 2 into a unit, you are still in trouble. Because if you make > sqrt[ (11+sqrt(105))/2] into a unit, then the best you can factor out > from the right hand side would be > sqrt[ (11-sqrt(105))/2 ] * 2, which does NOT equal 4. The factor is 2, and not 4. > As to making a bigger ring where they are units, that's still > nonsense: EVERY ELEMENT OF THE RING OF ALGEBRAIC INTEGERS IS A UNIT IN > SOME LARGER RING. Just like in the integers. Not by the definition which specifically precludes that which I give again. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. >Consider (2a)b = 2, where 'a' is NOT an algebraic integer, while 2a >and 'b' both are, so then though ab=1 in a more inclusive ring, 'b' is >not a unit *in the ring of algebraic integers* so trivially it is >indeed a non-unit factor. > Red herring. It's a specific explanation for what is happening that covers all the bases, including how 'b' can be a non-unit factor *in the ring of algebraic integers* yet still be a trivial factor because it's a unit in a proper ring. >The more inclusive ring is given by the the following definition: >The Object Ring is a commutative ring that includes all numbers such >that -1 and 1 are the only members that are both a unit and an >integer, where no non-unit member is a factor of any two integers that >are coprime. > This definition is nonsense as written, as we have pointed out > numerous times. It has no referent. Is that the royal we? The definition works quite well, and in fact it can immediately be seen that it includes integers, as integer are a set such that only -1 and 1 are units, so the Object Ring is not empty. >Notice it excludes a possibility like a=1/2 as then 2 would be a unit >in the ring. > Then none of 2, sqrt[ (11-sqrt(15))/2], and sqrt[ (11+sqrt(15))/2] are > units in the Object ring, because if ANY of them are units, then 2 is > a unit. That is not true. In the object ring, two of the three numbers given have a factor that is 2. > So even if object ring made sense, you would still NOT be able to > conclude that one the three factors is a unit in that ring. That is incorrect. >My point to reader is that Arturo Magidin clearly knows what he is >doing and he is relying on the very *error* in core > If there were an error in core (assuming you mean, if there is some > contradiction in basic mathematics), then every proof is correct and > valid. EVERYTHING can be proven. My proof would be just as correct as > yours, so there is simply no way that you can claim I am lying because > there is an error in core. Luckily the problem is a *definition* problem, and it is in core because the definition of the *ring* of algebraic integers as having as members roots of monic polynomials with integer coefficients is in core. > But you have not pointed out any errors. Your argumetn has the gap of > assuming that the behavior of complex functions of m at m=0 will yield > information about their behavior at points other than m=0. When you > say something should have a factor, what you mean is that you would > ->like<- it to have a factor, but that it doesn't. That's not an error > in mathematics, that you engaging in wishful thinking. > [.snip.] Which is an attack on algebra as necessarily in algebra, setting m=0 takes away the ability of the result to vary with m. So, there is no gap as in fact it IS true that setting m=0 removes the ability of what remains to vary with m. It's rather basic algebra. > So, I don't know if it is nonsensical in its face, but I do know > that it is TRUE that when the value of m changes, the way in which the > factors of 4 get distributed among g1, g2, and g3 changes as well. >> >>That's your wish, but it doesn't make mathematical sense. >> >> So you claim. Yet there it is above: ALL g's have a non-unit common >> factor with 2. How come? >> >> [.snip.] >Because you have a situation like I showed where one factor is 2a >while the other is 'b', but 'a' is not an algebraic integer. > No, I do not have that situation. ALL the numbers above are ALGEBRAIC > INTEGERS. Your example relies on a not being an algebraic > integer. But here they are again, James. Look CAREFULLY: > g1 = a1*x+2 > = [(-7-sqrt(105))/2]*x + 2 > = [sqrt(7) * sqrt[ (11+sqrt(105))/2]]*x + 2 > = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] > sqrt[ (11+sqrt(105))/2 ] is an algebraic integer. It is the root of > y^4 + 11y^2 + 4 = 0. > sqrt(7) is an algebraic integer. > x is an algebraic integer (by assumption). > So sqrt(7)*x is an algebraic integer. > sqrt[ (11-sqrt(105))/2] is an algebraic integer, root of > y^4 + 11y^2 + 4 = 0. > So (sqrt(7)*x) + sqrt [ (11-sqrt(105))/2] is an algebraic integer. > So both factors of g1 above are algebraic integers. Which does not preclude one of them being 2a where 'a' is NOT an algebraic integer, while 2a is. > g2 = a2*x + 2 > = [(-7+sqrt(105))/2]*x + 2 > = [sqrt(7) * sqrt[ (11-sqrt(105))/2]]*x + 2 > = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] > Again, sqrt(7) is an algebraic integer. x is an algebraic > integer. Therefore, sqrt(7)*x is an algebraic integer. > sqrt[ (11+sqrt(105))/2] is an algebraic integer. So > sqrt(7)*x + sqrt[ (11+sqrt(105))/2] is an algebraic integer. > sqrt[(11-sqrt(105))/2] is an algebraic integer. > So the factorization given is a factorization in which BOTH FACTORS > are algebraic integers. Which does not preclude one of them being 2a, where 'a' is NOT an algebraic integer. > g3 = a3*x + 2 > = -2x + 2 > = -2(x-1) > -2 is an algebraic integer. > x is an algebraic integer. -1 is an algebraic integer. So x-1 is an > algebraic integer. So this is a factorization INTO ALGEBRAIC INTEGERS. > Your example does NOT apply. You are the one engaging in deceitful > disinformation. Accepting algebra means that setting m=0 removes variability with respect to m from what results, and doing so takes away any doubts about the results that can be seen at the link: http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 Whereas, attacking algebra means trying to hold on to an *arbitrary* definition where roots of monic polynomials are accepted while all roots of non-monic polynomials with integer coefficients are rejected, which is actually an attack on mathematics. As who among you is willing to toss out algebra for an arbitrary definition? Apparently Arturo Magidin is. James Harris === Subject: Re: Short Alg Int Argument Visiting Assistant Professor at the University of Montana. >> [.snip.] >> P(0) = 4(3x+2) = (0x+2)(0x+2)(3x+2), >> >> with a1=0, a2=0, a3=3. >> >> So g1=2, g2=2, g3=3x+2; the first two are multiples of 2, the last one >> is 'coprime to 2' when x is an algebraic integer coprime to 2. >> >> When m=1, we get >> >> P(1) = 4*(7x^3 - 9x + 2) >> = [ ((-7-sqrt(105))/2)x +2][ ((-7+sqrt(105))/2)x+2][-2x+2] >> >> with >> >> a1 = (-7-sqrt(105))/2 >> a2 = (-7+sqrt(105))/2 >> a3 = -2, >> >> obtained from your defining polynomial >> >> a^3 + 3(-1+mf^2)a^2 - f^2(m^3f^4-3m^2f^2 + 3m) = >> = a^3 +3(-1+4)a^2 - 4(16 - 12 + 3) >> = a^3 + 9a^2 - 28 >> = (a+2)(a^2 +7a - 14). >> >> a1 and a2 are the roots of a^2+7a-14. >> >> Then we can also write >> >> a1 = (-7-sqrt(105))/2 = sqrt(7) * sqrt[ (11+sqrt(105))/2] >> >> a2 = (-7+sqrt(105))/2 = sqrt(7) * sqrt[ (11-sqrt(105))/2]. >> >> Note that sqrt[ (11-sqrt(105))/2] and sqrt[ (11+sqrt(105))/2] >> are roots of the monic, irreducible, primitive polynomial with integer >> coefficients x^4 - 11x^2 + 4, and since the constant term is not 1 or >> -1 they are not units. Also, their product is: >> >> sqrt[ (11-sqrt(105))/2]*sqrt[ (11+sqrt(105))/2] = >> = sqrt[ (121-105)/4] >> = sqrt[ 16/4] >> = sqrt[4] >> = 2, >> >> so their are both factors of 2. >> >> So we have >> >> g1 = a1*x+2 >> = [(-7-sqrt(105))/2]*x + 2 >> = [sqrt(7) * sqrt[ (11+sqrt(105))/2]]*x + 2 >> = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] >> >> g2 = a2*x + 2 >> = [(-7+sqrt(105))/2]*x + 2 >> = [sqrt(7) * sqrt[ (11-sqrt(105))/2]]*x + 2 >> = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] >> >> >> g3 = a3*x + 2 >> = -2x + 2 >> = -2(x-1) >> >> So the factors of 2 now distribute as follows: >> >> sqrt[(11+sqrt(105))/2] into g1 >> sqrt[(11-sqrt(105))/2] into g2 >> -2 into g3. > >And in fact one of the g's does not share non-unit factors with 2, but >how do you tell by physically looking? > > No, they don't. ALL of them shre non-unit factors with 2: g3 shares > the non-unit factor -2. > > g2 shares the non-unit factor sqrt[(11-sqrt(105))/2]. > > And g1 shares the non-unit factor sqrt[(11+sqrt(105))/2]. > > There they are. Explicitly stated, explicitly factored out. >> >>However, factors of 1 are typically excluded and here because of the >>error in core, though one of the factors *should* be a factor of 1, it >>is not *in the ring of algebraic integers*, which leads to the >>contradiction. >> That's stupid. None of them are factors of 1. ALL OF THE g's HAVE >> COMMON FACTORS WITH 2. ALL OF THEM. >What's stupid is your objection to the argument at I reject it because it is unfounded. >http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 >where you keep challenging the conclusion that setting m=0 gives >results that don't vary with m. It's right there, before your eyes. When m=0, one of the g's is coprime to 2. When m=1, NONE of the g's is coprime to 2. Your argument is false because there is NO reason to assume that a function will behave the same way with respect to divisibility when m=0 and when m<>0. Pure and simple. Your reply fails to address the OBSERVABLE FACT that all of g1, g2, and g3 have non-unit common factors with 2. Do you deny that all of them have non-unit common factors with 2 in the algebraic integers? Do you deny that you claimed that one of them would NOT have non-unit common factors with 2 in the algebraic integers? >As for your attempt at refuting that proof using an example, I've >explained exactly how it fails as you have something like (2a)b = 2, >where 'a' is NOT an algebraic integer, and for *that* reason as in a >more inclusive ring ab=1, you have that 'b' is NOT a unit *in the ring >of algebraic integers* as by definition if it's a unit in that ring >then it has a factor in that ring, while 'a' is NOT an algebraic >integer. No, this explanation is irrelevant. In your explanation, you have 2a, and a is not an algebraic integer. In the example above, ALL the factors are algebraic integers. So we are not in the situation you are claiming. sqrt[(11-sqrt(105))/2] is an algebraic integer, sqrt[(11+sqrt(105))/2] is an algebraic integer, sqrt(7) is an algebraic integer. So we are NOT in a situation like (2a)b = 2 where 'a' is NOT an algebraic integer. >> You claimed one of them would necessarily be coprime to 2. NONE OF >> THEM ARE COPRIME TO 2, ACCORDING TO YOUR DEFINITION. They have common, >> non-unit factors. >> So your claim is false. Pure, and simple. >A hallmark of an error in core mathematics is the ability for *two* >sides to sit and argue that they're right, IF there is an error in core mathematics, then EVERYTHING is provable, and my proof is just as correct as yours. There is nothing you can do to fix it. > while luckily, the error >here is simply one of arbitrary exclusion, as the definition for >algebraic integers, with the focus on monic polynomials with integer >coefficients, creates the problem. Do you agree that in the following factorizations g1 = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] g2 = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] g3 = -2(x-1) are valid? If not, which one is wrong? Why? Do you agree that, whenever x is an algebraic integer, each of sqrt[ (11+sqrt(105))/2 ] sqrt[ (11-sqrt(105))/2 ] sqrt(7) sqrt(7)*x sqrt(7)*x + sqrt[ (11+sqrt(105))/2 ] sqrt(7)*x + sqrt[ (11-sqrt(105))/2 ] -2 (x-1) are algebraic integers? If you do not agree, which one is not an algebraic integer? Do you agree that sqrt[ (11+sqrt(105))/2 ] * sqrt[ (11-sqrt(105))/2] = 2 and so that sqrt[ (11+sqrt(105))/2] and sqrt[ (11-sqrt(105))/2] are factors of 2 IN THE RING OF ALGEBRAIC INTEGERS? And if not, why not? Do you agree that sqrt[(11+sqrt(105))/2] is NOT a unit in the ring of algebraic integers? If not, why not? Note that it is a root of the monic irreducible polynomial y^4 + 11y^2 + 4. Do you agree that sqrt[(11-sqrt(105))/2] is NOT a unit in the ring of algebraic integers? If not, why not? It is a root of the same monic irreducible polynomial. So, do you agree that sqrt[(11+sqrt(105))/2] is a non-unit factor of 2, and a non-unit factor of g1, in the ring of algebraic integers? If not, why not? So, do you agree that sqrt[(11+sqrt(105))/2] is a non-unit common factor of g1 and 2, in the ring of algebraic integers? If not, why not? So, do you agree that g1 and 2 are not coprime in the ring of algebraic integers? Do you agree that sqrt[(11-sqrt(105))/2] is a non-unit factor of 2, and a non-unit factor of g2, in the ring of algebraic integers? Thus, that it is a non-unit common factor of g2 and 2 in the ring of algebraic integers? Thus, that g2 and 2 are not coprime in the ring of algebraic integers? Do you agree that 2 is a common factor of 2 and g3, and not a unit in the ring of algebraic integers? So do you agree that g3 and 2 are not coprime in the ring of algebraic integers? So, why do you not agree that none of g1, g2, and g3 are coprime to 2 in the ring of algebraic integers? And why do you not agree that this FLATLY CONTRADICTS your claim that one of them is? >So consistency is restored by noting that that definition does not >produce a complete ring. Because things that you ->want<- to happen don't happen? Sorry, the problem isn't with the algebraic integers, it's with your wishes. You want things which are impossible. This is nonsense. If the problem were that the algebraic integers are missing stuff, as you proclaim, and the stuff that is missing is the nontrivial common divisors, then how are you going to get rid of the nontrivial common divisors that ALREADY exist? Which one are you going to make a unit? Surely not 2. But if you make sqrt[(11-sqrt(105))/2] a unit, then you make (11-sqrt(105))/2 a unit as well. And if (11-sqrt(105))/2 is a unit, being the root of x^2+11x+4, should not that make ALL roots of that polynomial units? And if all roots of that polynomial are units, then their product, which is 4, is a unit, which would make 2 a unit again. >To believe you, readers must challenge algebra itself, and the >necessary result that setting m=0 gives an expression that does not >vary with m. No, there is no challenge to algebra. Your problem is that you think that if you write g(m) = (g(m)-g(0)) + g(0) then the expression g(m)-g(0) will always be non-coprime to m. This is not true for arbitrary functions (though it is true for polynomials and for polynomials in fractional powers of m). ->That<-, as far as you have seen fit to explain your thought process, is your mistake. As for challenging algebra itself, are you not challenging explicit algebraic expressions above? There's the factors, yet you claim they aren't factors. There's the calculations, yet you claim they aren't calculations. >> Now you are saying that one of them should be a factor of 1. But if >> that is so, then you are still in trouble. Because either you will >> make 2 into a unit (which you do not want to), or else that means that >> you want to make sqrt[(11+sqrt(105))/2] into a unit. If you do that, >> then sqrt[(11-sqrt(105))/2] is ALSO a unit, which again makes 2 into a >> unit. >That last sentence does not follow logically from the previous ones. If sqrt[(11+sqrt(105))/2] is a unit, then so is its square; because if v is such that sqrt[(11+sqrt(105))/2]*v = 1 then (sqrt[(11+sqrt(105))/2])^2*v^2 = (sqrt[(11+sqrt(105))/2]*v)^2 = 1^2 = 1. So (11+sqrt(105))/2 would have to be a unit. If (11+sqrt(105))/2 is a unit, then since it is the root of x^2 + 11x+ 4, a monic irreducible polynomial with integer coefficients, then surely all roots of that polynomial are units. You have stated in the past that your object ring will be closed under conjugation implicitly, e.g. on So if a root of 4x^2 + 11x + 1 is in the ring, namely the inverse of (11+sqrt(105))/2, why is the other root not in the ring? So obviously, once we know that (11+sqrt(105))/2 is a unit, we conclude that the other root of its minimal polynomial is also a unit. So (11-sqrt(105))/2, the other root, is also a unit. So let u be such that [(11-sqrt(105))/2]*u = 1. But then [(11+sqrt(105))/2]*[(11-sqrt(105))/2] is a unit, since [(11+sqrt(105))/2]*[(11-sqrt(105))/2]*(uv) = 1. (product of units is a unit). But the product is [(11+sqrt(105))/2]*[(11-sqrt(105))/2] = (121 -105)/4 = 16/4 = 4. So 4 is a unit. But if 4 is a unit, then 1/4 is an element of your ring. So then 2*(1/4) is an element of your ring. So then (1/2) is an element of your ring. So 2 is a unit in your ring. Ooops. >> And even if you don't believe that making one of them into a unit will >> make 2 into a unit, you are still in trouble. Because if you make >> sqrt[ (11+sqrt(105))/2] into a unit, then the best you can factor out >> from the right hand side would be >> sqrt[ (11-sqrt(105))/2 ] * 2, which does NOT equal 4. >The factor is 2, and not 4. No, the total factor is 4: f^2 = 4. One 2 you get from g3, the other 2 you get from sqrt[(11-sqrt(105))/2] * sqrt[(11+sqrt(105))/2] = 2, one factor each from g1 and g2. >> As to making a bigger ring where they are units, that's still >> nonsense: EVERY ELEMENT OF THE RING OF ALGEBRAIC INTEGERS IS A UNIT IN >> SOME LARGER RING. Just like in the integers. >Not by the definition which specifically precludes that which I give >again. Your definition has no referent as currently stated. It does not uniquely define anything. >The Object Ring is a commutative ring that includes all numbers such >that -1 and 1 are the only members that are both a unit and an >integer, where no non-unit member is a factor of any two integers that >are coprime. Your definition has no referent. And if you want your ring to contain all roots of an irreducible polynomial with integer coefficients once it contains one of them, then by the above you can see that if you make ANY of the three factors into a unit, then you will necessarily make 2 into a unit. Which means that you cannot make ANY of the three factors into a unit. Which is still a red herring, since you claimed that one of the g's would be coprime IN THE RING OF ALGEBRAIC INTEGERS to 2, but none of them are. >>Consider (2a)b = 2, where 'a' is NOT an algebraic integer, while 2a >>and 'b' both are, so then though ab=1 in a more inclusive ring, 'b' is >>not a unit *in the ring of algebraic integers* so trivially it is >>indeed a non-unit factor. >> Red herring. >It's a specific explanation for what is happening that covers all the >bases, No, it does not cover the case at hand. In the case at hand, NONE of the factors fail to be algebraic integers. This specific explanation may be a specific explanation for ->something<-, but not for what you are addressing. Hence, it is a red herring. [.snip.] >>The more inclusive ring is given by the the following definition: >> >>The Object Ring is a commutative ring that includes all numbers such >>that -1 and 1 are the only members that are both a unit and an >>integer, where no non-unit member is a factor of any two integers that >>are coprime. >> This definition is nonsense as written, as we have pointed out >> numerous times. It has no referent. >Is that the royal we? No. Many people, myself included, have pointed out that it is nonsese as written. You do not specify what a number is supposed to mean. That already renders it meaningless. >The definition works quite well, and in fact it can immediately be >seen that it includes integers, as integer are a set such that only -1 >and 1 are units, so the Object Ring is not empty. The definition works quite well? So, is 1/sqrt[(11-sqrt(105))/2] in the object ring? If the definition works quite well, you should be able to answer it and PROVE IT, using the definition. If the answer is yes, then surely 1/sqrt[(11-sqrt(105))/2] must also be in the ring, being the other root of 4x^2 + 11x + 1. If it is not, why not? If the answer is no, then you must have 1/sqrt[(11+sqrt(105))/2] in your ring. If so, why is one root of 4x^2 +11x + 1 in your ring, but not the other? And they cannot BOTH be in the ring. So which one is in it, and why? And why is the other excluded, if they are both roots of the same irreducible polynomial? >>Notice it excludes a possibility like a=1/2 as then 2 would be a unit >>in the ring. >> Then none of 2, sqrt[ (11-sqrt(15))/2], and sqrt[ (11+sqrt(15))/2] are >> units in the Object ring, because if ANY of them are units, then 2 is >> a unit. >That is not true. In the object ring, two of the three numbers given >have a factor that is 2. So, that means that in the object ring one of sqrt[(11-sqrt(105))/2] (typo above) or sqrt[(11+sqrt(105))/2] is a unit, and not the other. Why is one of them in and the other out? Which one is in and why? You seem to be the only one who thinks your definition works quite well, so surely you'll be able to PROVE which one is in and which one is out, and why. [.snip.] >>My point to reader is that Arturo Magidin clearly knows what he is >>doing and he is relying on the very *error* in core >> If there were an error in core (assuming you mean, if there is some >> contradiction in basic mathematics), then every proof is correct and >> valid. EVERYTHING can be proven. My proof would be just as correct as >> yours, so there is simply no way that you can claim I am lying because >> there is an error in core. >Luckily the problem is a *definition* problem, and it is in core >because the definition of the *ring* of algebraic integers as having >as members roots of monic polynomials with integer coefficients is in >core. That is nonsense. A definition cannot introduce a contradiction. I've explained this before: definitions are like macro expansions. They are nothing but shorthand for longer meanings. When you define algebraic integer to mean root of a monic polynomial with integer coefficients, you are not creating anything. You are just saying that instead of writing root of a monic polynomial with integer coefficients, you will write the shorter phrase algebraic integer. When YOU write x and y are coprime in the ring of algebraic integers, it is meant to be shorthand for x and y do not have any common factors which are algebraic integers and are not units in the ring of algebraic integers. Definitions CANNOT introduce contradictions. Claiming they do is to betray absolute ignorance about what you are talking about. [.snip.] >> But you have not pointed out any errors. Your argumetn has the gap of >> assuming that the behavior of complex functions of m at m=0 will yield >> information about their behavior at points other than m=0. When you >> say something should have a factor, what you mean is that you would >> ->like<- it to have a factor, but that it doesn't. That's not an error >> in mathematics, that you engaging in wishful thinking. >> [.snip.] >Which is an attack on algebra as necessarily in algebra, setting m=0 >takes away the ability of the result to vary with m. This is sophistry. Yes, fixing the value of m means that the value no longer changes. That does not mean that the information you get from a fixed value of m gives you information about what happens to any other value of m. And it is the latter that you do. >So, there is no gap as in fact it IS true that setting m=0 removes >the ability of what remains to vary with m. >It's rather basic algebra. No, it is nonsense. [.snip.] >>Because you have a situation like I showed where one factor is 2a >>while the other is 'b', but 'a' is not an algebraic integer. >> No, I do not have that situation. ALL the numbers above are ALGEBRAIC >> INTEGERS. Your example relies on a not being an algebraic >> integer. But here they are again, James. Look CAREFULLY: >> g1 = a1*x+2 >> = [(-7-sqrt(105))/2]*x + 2 >> = [sqrt(7) * sqrt[ (11+sqrt(105))/2]]*x + 2 >> = sqrt[(11+sqrt(105))/2] * [ sqrt(7)x + sqrt[(11-sqrt(105))/2] ] >> sqrt[ (11+sqrt(105))/2 ] is an algebraic integer. It is the root of >> y^4 + 11y^2 + 4 = 0. >> sqrt(7) is an algebraic integer. >> x is an algebraic integer (by assumption). >> So sqrt(7)*x is an algebraic integer. >> sqrt[ (11-sqrt(105))/2] is an algebraic integer, root of >> y^4 + 11y^2 + 4 = 0. >> So (sqrt(7)*x) + sqrt [ (11-sqrt(105))/2] is an algebraic integer. >> So both factors of g1 above are algebraic integers. >Which does not preclude one of them being 2a where 'a' is NOT an >algebraic integer, while 2a is. Which is irrelevant. That is true for ANY algebraic integer; most of them (anything which is not a multiple of 2) can be written as 2*a where a is not an algebraic integer. But that is just a red herring. I have produced EXPLICITLY a non-unit common factor of g1 and 2 which is an algebraic integer. Whether or not that factor is equal to 2 a non-algebraic integer is immaterial. Having produced an explicit common factor which is a factor in the algebraic integers, an algebraic integer, and not a unit, you MUST agree that g1 and 2 are NOT coprime in the ring of algebraic integers. >> g2 = a2*x + 2 >> = [(-7+sqrt(105))/2]*x + 2 >> = [sqrt(7) * sqrt[ (11-sqrt(105))/2]]*x + 2 >> = sqrt[(11-sqrt(105))/2] * [ sqrt(7)x + sqrt[(11+sqrt(105))/2] ] >> Again, sqrt(7) is an algebraic integer. x is an algebraic >> integer. Therefore, sqrt(7)*x is an algebraic integer. >> sqrt[ (11+sqrt(105))/2] is an algebraic integer. So >> sqrt(7)*x + sqrt[ (11+sqrt(105))/2] is an algebraic integer. >> sqrt[(11-sqrt(105))/2] is an algebraic integer. >> So the factorization given is a factorization in which BOTH FACTORS >> are algebraic integers. >Which does not preclude one of them being 2a, where 'a' is NOT an >algebraic integer. Which is a red herring. I have produced an explicit factorization, showing that there is an algebraic integer which is a common factor of g2 and 2 in the ring of algebraic integers, and which is not a unit. So g2 and 2 have a non-unit common algebraic integer factor. So you MUST agree that g2 and 2 are NOT coprime in the ring of algebraic integers. >> g3 = a3*x + 2 >> = -2x + 2 >> = -2(x-1) >> -2 is an algebraic integer. >> x is an algebraic integer. -1 is an algebraic integer. So x-1 is an >> algebraic integer. So this is a factorization INTO ALGEBRAIC INTEGERS. >> Your example does NOT apply. You are the one engaging in deceitful >> disinformation. >Accepting algebra means that setting m=0 removes variability with >respect to m from what results, and doing so takes away any doubts >about the results that can be seen at the link: >http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782 Irrelevant. I'm talking about this 2a nonsense. Apparently, you agree that the factorizations are correct and that each term is an algebraic integer. So you agree that g1 and 2 are not coprime ACCORDING TO YOUR DEFINITION, that g2 and 2 are not coprime ACCORDING TO YOUR DEFINITION, and that g3 and 2 are not coprime ACCORDING TO YOUR DEFINITION. So you must agree that it is false that one of g1, g2, g3 are coprime to 2. >Whereas, attacking algebra means trying to hold on to an *arbitrary* >definition where roots of monic polynomials are accepted while all >roots of non-monic polynomials with integer coefficients are rejected, >which is actually an attack on mathematics. >As who among you is willing to toss out algebra for an arbitrary >definition? It's not for an arbitrary definition. We are tossing out your faulty logic in favor of algebra. [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee fixe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a flash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not find it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Short Alg Int Argument >> An interesting question to me is, what is GCD(2,28/(7+sqrt(105))? >> I've made some attempts to find it, but had no luck. > Found it. It was reasonably easy, once I knew what I was looking for. > Note that > 28/(7+sqrt(105)) = 28*(7-sqrt(105))/(7-sqrt(105))*(7+sqrt(105)) > = 28(7-sqrt(105))/(-56) > = (-7+sqrt(105))/2 > Now consider > [(-7+sqrt(105))/2)]^2 = (49+105-14*sqrt(105))/4 > = (154-14*sqrt(105))/4 > = (77-7*sqrt(105))/2 > = 7* (11-sqrt(105))/2. > So (-7+sqrt(105))/2 = sqrt(7) * sqrt( (11-sqrt(105))/2 ). > Note that this is a decomposition as algebraic integers, since sqrt(7) > is a root of x^2 - 7, and (11-sqrt(105))/2 is a root of > x^2 - 11x + 4, so its square root is a root of x^4 - 11x^2 + 4. > Symmetrically, > 28/(7-sqrt(105)) = sqrt(7)*sqrt[(11+sqrt(105))/2]. > Also, sqrt[(11-sqrt(105))/2] and sqrt[(11+sqrt(105))/2] are factors of > 2; in fact, they multiply together to equal 2: > sqrt[ (11-sqrt(105))/2] * sqrt[ (11+sqrt(105))/2] > = sqrt[ (11-sqrt(105))(11+sqrt(105))/4 ] > = sqrt[ (121-105)/4] > = sqrt[ (16/4) ] > = sqrt(4) > = 2. Easier, and more generally: assume (b,c) = 1 (b = 7, c = 2 above) We have w^2 + bw = bc for w = (-b + sqrt(b(b+4c)))/2 with norm w w' = - bc, refining to (b,w)(b,w')(c,w)(c,w') But (b,w)^2 = (b,w)(b,w') via w' = -w-b = -w (mod b) = b(b,w,w',c) = (b) via (b,c) = 1 Thus (b,w) = sqrt(b) = (b,w') so (c,w) = (sqrt(b+4c)-sqrt(b))/2 [= w /sqrt(b) ] and (c,w') = (sqrt(b+4c)+sqrt(b))/2 [= -w'/sqrt(b) ] E.g. w = sqrt(7) (sqrt(15)-sqrt(7))/2 = (7,w) (2,w) = (sqrt(105)-7)/2 for b = 7, c = 2 See also my prior post on Schreier refinement -Bill Dubuque > All that remains is to show that neither sqrt[(11-sqrt(105))/2] nor > sqrt[(11+sqrt(105))/2] are units in the ring of algebraic > integers. One way would be to note that their squares are not units, > and using the following theorem that James has recently invoked and > accepted: > THEOREM. If an algebraic integer is a unit, then it is the root of a > monic polynomial with integer coefficients and last coefficient equal > to 1 or -1. > (The theorem is stronger, but that is what James said. Quoting from: > Here Magidin IS correct in that none of the roots can be units, > but that's because *in the ring of algebraic integers* it's > provably true that any unit is a root of a monic polynomial with > integer coefficients and a last coefficient of 1 or -1. > So we have the polynomial f(x) = x^4 - 11x^2 + 4. > CLAIM 1. f(x) is irreducible over Q. > Proof of claim: (a) It has no rational roots: the only possible rational > roots are 1, -1, 2, -2, 4, and -4, and plugging them in shows you do > not get zero. > (b) Since it has no rational roots, the only way it could be reducible > would be if it were a product of two quadratics, and by Gauss's Lemma > we may assume the quadratics have integer coefficients and therefore > are monic: > x^4 - 11x^2 + 4 = (x^2 + ax + b)(x^2 + cx + d) > = x^4 + (a+c)x^3 + (b+d+ac)x^2 + (ad+bc)x + bd > with a, b, c, d integers. > Therefore, > a+c = 0 > b+d+ac = -11 > ad+bc = 0 > bd = 4. > b+d-a^2 = -11 > a(d-b) = 0 > bd=4. > If a=0, then c=0, bd=4, b+d= -11; that's impossible with b, d > integers. So we must have d-b=0, or d=b, a=-c. We then reduce to > 2b-a^2 = -11 > b^2 = 4. > That means that b=2 or b=-2; so > 4-a^2 = -11, or a^2 = 15, impossible; or > -4-a^2 = -11, or a^2 = 7, also impossible. > Thus, the polynomial is irreducible over Q. > CLAIM 2: If g(x) is any polynomial with integer coefficients such that > g(sqrt[(11-sqrt(105))/2]) = 0, then g(x) is a multiple of f(x) by a > polynomial with integer coefficients. > Proof. This follows from Gauss's Lemma and the fact that f(x) is > irreducible: it is the minimal polynomial for sqrt[(11-sqrt(105))/2], > so it must divide any polynomial with integer coefficients which has > sqrt[(11-sqrt(105))/2] as a root. > CLAIM 3. If g(x) is any polynomial with integer coefficients such that > g(sqrt[(11+sqrt(105))/2])=0, then g(x) is a multiple of f(x) by a > polynomial with integer coefficients. > Proof: same as above, since f(x) is also the minimal polynomial for > this number. > CLAIM 4: If g(x) is any polynomial with integer coefficients that > has sqrt[(11+sqrt(105))/2] as a root, then its constant term is a > multiple of 4. If g(x) is any polynomial with integer coefficients > which has sqrt[(11-sqrt(105))/2] as a root, then its constant term is > a multiple of 4. > Proof. Let g(x) be a polynomial as given. By Claim 3 or 4, as > appropriate, there exists a polynomial h(x) with integer coefficients > such that g(x)=h(x)f(x). Write > h(x) = a_nx^n + ... + a_1x + a_0, with a_i an integer. > Then f(x)h(x) = 4a_0 + (higher terms) > so the constant term of g(x) is a multiple of 4. > CONCLUSION. sqrt[(11+sqrt(105))/2] is not a unit in the ring of > algebraic integers. sqrt[(11-sqrt(105))/2] is not a unit in the ring > of algebraic integers > Proof of conclusion. Suppose it's a unit. By James's statement, it > must be the root of a monic polynomial with integer coefficients and > last coefficient 1 or -1. But every polynomial with integer > coefficients that has it as a root has last coefficient equal to a > multiple of 4, so it cannot be a unit. (Contrapositive of the > statement). > So we have: > a_3 = -2 > a_2 = sqrt(7) * sqrt[(11-sqrt(105))/2] > a_1 = sqrt(7) * sqrt[(11+sqrt(105))/2] > where sqrt[(11-sqrt(105))/2] and sqrt[(11+sqrt(105))/2] are both > non-unit factors of 2. > So a_3 is not coprime to 2, since they have the non-unit common > factor 2. > a_2 is not coprime to 2, since they have the non-unit common factor > sqrt[(11-sqrt(105))/2]. > a_3 is not coprime to 2, since they have the non-unit common factor > sqrt[(11+sqrt(105))/2]. > Therefore, none of the a_i are coprime to 2. > James claims that one of them is coprime to 2, and the other 2 are > multiples of 2. > Therefore, James's claim is wrong. > Why do you take so much trouble to expose such a reasoner as > Mr. Smith? I answer as a deceased friend of mine used to answer > on like occasions - A man's capacity is no measure of his power > to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more; > and a long purse, which does most of all. He has made at least > ten publications, full of figures few readers can critize. A great > many people are staggered to this extend, that they imagine there > must be the indefinite something in the mysterious all this. > They are brought to the point of suspicion that the mathematicians > ought not to treat all this with such undisguised contempt, > at least. > -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan > Arturo Magidin > magidin@math.berkeley.edu === Subject: Re: Short Alg Int Argument Visiting Assistant Professor at the University of Montana. [.snip.] >Easier, and more generally: assume (b,c) = 1 (b = 7, c = 2 above) >We have w^2 + bw = bc for w = (-b + sqrt(b(b+4c)))/2 >with norm w w' = - bc, refining to (b,w)(b,w')(c,w)(c,w') >But (b,w)^2 = (b,w)(b,w') via w' = -w-b = -w (mod b) > = b(b,w,w',c) = (b) via (b,c) = 1 >Thus (b,w) = sqrt(b) = (b,w') >so (c,w) = (sqrt(b+4c)-sqrt(b))/2 [= w /sqrt(b) ] >and (c,w') = (sqrt(b+4c)+sqrt(b))/2 [= -w'/sqrt(b) ] >E.g. w = sqrt(7) (sqrt(15)-sqrt(7))/2 = (7,w) (2,w) > = (sqrt(105)-7)/2 for b = 7, c = 2 (figuring out the gcd) with cannonballs (the class number of the number field). It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: After Coxeter Suppose that, having read Coxeter's Introduction to Geometry the reader wished for something equally broad but more deep. What books might he try? Surely there is a GTM as thick as Lang's Algebra that he may curl up with during the coming autumnal evenings? Any suggestions? -- G.C. === Subject: Re: After Coxeter > Suppose that, having read Coxeter's Introduction to Geometry > the reader wished for something equally broad but more deep. > What books might he try? Surely there is a GTM as thick as > Lang's Algebra that he may curl up with during the coming > autumnal evenings? Any suggestions? geometry deeper and broader than coxeter may be hard to find in one volume. you may want to try a variety of sources. for example, foundations of geometry by david hilbert euclidean and non-euclidean geometries by marvin j. greenberg a survey of geometry by howard eves geometric constructions by george e. martin > -- > G.C. === Subject: Re: After Coxeter > Suppose that, having read Coxeter's Introduction to Geometry > the reader wished for something equally broad but more deep. > What books might he try? Surely there is a GTM as thick as > Lang's Algebra that he may curl up with during the coming > autumnal evenings? Any suggestions? > geometry deeper and broader than coxeter may be hard to find > in one volume. you may want to try a variety of sources. for > example, > foundations of geometry by david hilbert Yes, I've got that. > euclidean and non-euclidean geometries by marvin j. greenberg > a survey of geometry by howard eves > geometric constructions by george e. martin Was e. e. cummings a geometer :-) -- G.C. === Subject: Re: After Coxeter > Suppose that, having read Coxeter's Introduction to Geometry the reader > wished for something equally broad but more deep. What books might he > try? Surely there is a GTM as thick as Lang's Algebra that he may curl > up with during the coming autumnal evenings? Any suggestions? May I suggest you Marcel Berger : Geometrie (Nathan) in French === Subject: Re: After Coxeter > Suppose that, having read Coxeter's Introduction to Geometry the reader > wished for something equally broad but more deep. What books might he > try? Surely there is a GTM as thick as Lang's Algebra that he may curl > up with during the coming autumnal evenings? Any suggestions? > May I suggest you > Marcel Berger : Geometrie (Nathan) in French But I don't read French :-( Don't the insular English make you sick! -- G.C. === Subject: Re: After Coxeter charset=iso-8859-1 > > Suppose that, having read Coxeter's Introduction to Geometry the reader > wished for something equally broad but more deep. What books might he > try? Surely there is a GTM as thick as Lang's Algebra that he may curl > up with during the coming autumnal evenings? Any suggestions? > May I suggest you > Marcel Berger : Geometrie (Nathan) in French > But I don't read French :-( Don't the insular English make you sick! Marcel Berger's Geometry is available in English translation from Springer-Verlag. --Jim Buddenhagen === Subject: Re: After Coxeter > > Suppose that, having read Coxeter's Introduction to Geometry the reader > wished for something equally broad but more deep. What books might he > try? Surely there is a GTM as thick as Lang's Algebra that he may curl > up with during the coming autumnal evenings? Any suggestions? > May I suggest you > Marcel Berger : Geometrie (Nathan) in French > But I don't read French :-( Don't the insular English make you sick! > Marcel Berger's Geometry is available in English translation > from Springer-Verlag. > --Jim Buddenhagen -- G.C. === Subject: Good Old Springer (Was: Re: After Coxeter) >> Marcel Berger's Geometry is available in English translation >> from Springer-Verlag. A message from Joan Birman to another mailing list begins as follows: former Springer-Verlag (more recently BertelsmannSpringer) was merged with Kluwer Academic Publishers. As the press release says, the merger will create the second largest professional publisher in science, technology and medicine. What's the other one: Elsevier, of course! [...] (Birman goes on to describe why this is not especially good news for fans of Springer and more generally for those who look for reasonably-priced published mathematics; she then advocates some alternatives.) dave === Subject: Re: Good Old Springer (Was: Re: After Coxeter) >> Marcel Berger's Geometry is available in English translation >> from Springer-Verlag. > A message from Joan Birman to another mailing list begins as follows: > former Springer-Verlag (more recently BertelsmannSpringer) was merged with > Kluwer Academic Publishers. As the press release says, the merger will > create the second largest professional publisher in science, technology > and medicine. What's the other one: Elsevier, of course! [...] Are Reed Elsevier and Wolters Kluwer going to merge? I think that there was a plan once. > (Birman goes on to describe why this is not especially good news for > fans of Springer and more generally for those who look for reasonably-priced > published mathematics; she then advocates some alternatives.) > dave -- G.C. === Subject: Re: Good Old Springer > Marcel Berger's Geometry is available in English translation > from Springer-Verlag. > A message from Joan Birman to another mailing list begins as follows: > former Springer-Verlag (more recently BertelsmannSpringer) was merged with > Kluwer Academic Publishers. As the press release says, the merger will > create the second largest professional publisher in science, technology > and medicine. What's the other one: Elsevier, of course! [...] > (Birman goes on to describe why this is not especially good news for > fans of Springer and more generally for those who look for reasonably-priced > published mathematics; she then advocates some alternatives.) I agree, this is bad news indeed. Which particular alternatives does she mention? Marc === Subject: Chelsea Marie Sproule - August 16th 1984 S P R O U L E 19 16 18 15 21 12 5 = 106 In the evening I went to Crazy Cactus at 2404 Melrose Ave (phone 975-1266) and met Chelsea, she is in town for the day and from Calgary, and works there as a waitress. A bit of a space cadet, Chelsea says that she thinks that maybe dad was born on February 29th, but I think that if he was indeed born on such an unusual day then most people could remember that without any problem. 106+ Dad 29 2 60/306 106+ Mom 17 1 17/ 205 Chelsea 16 8 84 229/137 10042 Chelsea 53 Marie 46 Sproule 106 Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 <-16th-> 25 --- --- 381 225 Chelsea was born on the 16th, her first names adds to 53 (16th prime), her last name adds to 106 (twice the 16th prime and is Leviticus 16). Her given initials add together for 16 while her last two initials average 16. Chelsea's middle name adds to 46, it is the 32nd (16+16th) non-prime. Her square valued letters add to 25 (16th non-prime). The vowels in her given names are in positions adding to 48 (16+16+16). The even valued letters in her given names are in positions adding to 16. In her given names, her odd valued letters plus their positions together exceed her vowels plus their positions by 69 (16 plus the 16th prime). Only her P (16) is in proper alphabetical order. Chelsea was born with 137 days remaining in the year, I dreamt of a 16mm film projector being sold at a garage sale at 137 Forester Road (the film projector was mottled gray and had a plate rivited onto it identifying it as being model #4561). We think that maybe dad and mom were born on days 29 and 17, these Bible Books differ in length by 94 verses, pretty as 94 is 16 plus the 16th prime (53) plus the 16th non-prime (25) while Chelsea (16th prime) was born on the 16th. Non-Primes 1 27 4 28 6 30 8 32 9 33 10 34 12 35 14 36 15 38 16 39 18 40 20 42 21 44 22 45 24 46 <-32nd 25 48 26 49 Perhaps dad was born on the 29th day of the month, while Chelsea was born on the 229th day of the year. The consonants in her given names add to the 73 verses of Bible Book 29 (and is the Lucas numbers up to 29). In her given names, her consonants plus tehir positions together exceed her vowels plus their positions by 29. Her initials add to 35, pretty as Genesis 29 contains 35 verses, Genesis 35 contains 29 verses, and there are 35 chapters in the Bible that contain the length of 29 verses. Deuteronomy 29 with 29 verses is chapter 182 (the 140th non-prime), it exceeds it's 140th non-prime position by 42 (29th non-prime), while Chelsea's represented letters add to 140. And I am 10042 days older than Chelsea, it ends in 42 (29th non-prime), Bible Book 29 is Joel (42, the 29th non-prime). I meet Chelsea on the 29th day of the month. Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Perhaps the parents were born on days of the month adding to 46 while Chelsea's middle name adds to 46. Mom and Chelsea were born on days of the year adding to 246. Chelsea was born on the 16th (Nehemiah with 406 verses). Chelsea was born 92 (46+46) days closer to the end of the year than to the beginning of the year. Perhaps they were born on days 29, 17 and 16, together these Bible Books contain 646 verses. Perhaps the parents were born on days 17 and 29, these Bible Books contain 167 and 73 verses, these are 39th and 21st primes, together for 60, pretty as dad may have been born on the 60th day of the year, while Chelsea's last two names differ in value by 60. Bible Books 17 and 29 contain an average of 120 verses while Chelsea's repeating letters add to 120. I meet Chelsea at Crazy (73) Cactus (67), the name adds to 140 while Chelsea's 12 different letters add to 140. Chelsea was born on day 229 (50th prime), while the prime valued letters in her given names add to 50. Chelsea has 19 letters. Her vowels add to 67 (19th prime), her Fibonacci valued letter add to 67 (19th prime) and she is 67 inches tall (19th prime). The even valued letters in her given names add to 38 (19+19). Her letters in proper alphabetical order and her letters in reverse alphabetical order add together for 38 (19+19). Old Testament Book 19 is Psalms with 2460 verses, it is 7x19x19 minus the 19th prime (67). Old Testament Book 19 and New Testament Book 19 together contain 163 chapters (the 19+19th prime). Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations Chelsea was born in 84, her even valued letters add to 84 (the first 13 primes minus the first 13 non-primes). Her day, month and year of birth adds to 108 (the primes in prime position up to the 13th prime). She was born on the 16th (Nehemiah with 13 chapters). Perhaps Chelsea and her parents were born on days of the month adding to 62, or the 13th prime (41) plus the 13th non-prime (21). The Fibonacci valued letters in her given names add to 41 (13th prime). The Lucas valued letters in her full name add to 41 (13th prime). The unrepeated letters in her full name add to 82 (twice the 13th prime). Her consonants add to 135. The vowels in her given names add to 26 (13+13). She has 13 odd valued letters in her full name. We meet on the 29th, it's the number of chapters in Bible Book 13. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 Mom was born on the 17th (7th prime), Chelsea was born 212 days after mom's birthday (1 through 17 plus the 17th prime, it's the opening chapter of Bible Book 7). Chelsea's common names adds to 159, it ends in the 17th prime (59), it is the 7x7th prime (227) minus the 7x7th non-prime (68) and is the 618 verses of Bible Book 7 short of 777. She has 7 lettered first and last names while her given names differ in value by 7. The vowels in her given names add to 26 (17th non-prime). Her prime, square and cube valued letters add together for 107 (Leviticus 17, it's the 7+7+7+7th prime), while her letters that have neither prime, square nor cube values add together for 98 (7x7+7x7). Chelsea was born in 84 and her even valued letters add to 84, or 7 times the 7th non-prime (12). Her vowels add with their positions for the 167 verses of Bible Book 17 (it's the 7 primes in prime positions up to the 17th prime). Her consonants add with their positions for 228 (Judges 17, or Book 7 chapter 17). Her unrepeated letters add to 85, it is 5x17 and is a comination of the 17th prime (59) plus the 17th non-prime (26), Ruth with 85 verses is the 17th shortest Book in the Bible. Chelsea's day, month and year of birth adds together for the 108 verses of James, Bible Book 59 (the 17th prime), it is the 17th prime short of the 167 verses of Book 17. Perhaps the parents were born on days of the year adding to 77. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 I met Chelsea at 2404 Melrose Ave., there are 404 verses in Bible Book 66, Revelation, prettier as there are 24 chapters in Old Testament Books 6 and 10 (6th non-prime). The phone number here is 975-1266, it ends in the 66 Books of the Bible. We meet at Crazy (73) Cactus (67), the first name exceeds the 6 lettered last name by 6, the initials add to 6. Chelsea had her last birthday 317 days ago while I was born with 317 days remaining in the year (the 66th prime). We meet on the 29th day of the 6th month, pretty as there are 29 chapters in Bible Book 13 (6th prime) and 29 verses in chapter 666 (Ecclesiastes 7, the chapter refers to counting), for 29 is a combination of 6 plus the 6th prime (13) plus the 6th non-prime (10). If Marcia marries me and then Chelsea marries Marcia and me, then Chelsea's name could add to 216 (6x6x6). Isaiah is the Book with 66 chapters while Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia and me, and six other women. If my abusive parents, Protestants, Catholics and Hindus think that they should have the right to have me arrested and tortured for not being married (so that they can shut me up about about the Egyptian penises on the roofs of their churches and other of their false traditions), then I should have the right to ask women to marry me, or even to marry both Marcia and me. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 216 Chelsea 16 8 84 229/137 10042 Chelsea 53 Marie 46 Acevedo-Kabatoff 55-62 Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! If you take my name (or Marcia's and my name), then you should consider it as a gift from God. === Subject: Kelly Anne Yanko - June 6th 1982 Y A N K O 25 1 14 11 15 = 66 In the evening I went to The Yard and Flagon Pub at 718 Broadway Ave. (Penis Street, it's the street with the penis poster poles) in Saskatoon and met Kelly, she was my cute nubile bartenderette. A bit of a space cadet, Kelly said her dad's first name is Tomas, but dad's first name is really Thomas, and she dothn't know her mom's year of birth, only said that mom was younger than dad. 221 Tom 25 8 52 238/128 +1637 Thomas 76 William 79 Yanko 66 139 Barb 18 1 18/ Barbara 43 Jean 30 Yanko 66 186 Robyn 21 11 79 325/40 8312 Robyn 74 Alise 46 Yanko 66 165 Kelly 6 6 82 157/208 9240 Kelly 65 Anne 34 Yanko 66 The females were likely together born 95 days closer to the beginning of their years than to the end of their years (the number of verses in Bible Book 51). Kelly was born 51 days closer to the beginning of the year than to the end of the year. The family was born on days of the month adding to 70 (51st non-prime). The kids have names adding together for 351. Dad and the kids were born on days of the century adding to 78516. Dad is 51 years old. Non-Primes 1 27 50 4 28 51 6 30 52 8 32 54 9 33 55 10 34 56 12 35 57 14 36 58 15 38 60 16 39 62 18 40 63 20 42 64 21 44 65 22 45 66 24 46 68 25 48 69 26 49 70 <-51st Dad was born in 52, Kelly was born with 208 (4x52) days remaining in the year. Kelly's vowels add to 52. The even valued letters in Kelly's given names add to 52. Kelly's middle name adds to 52.3% of her first name. Primes 2 61 3 67 5 71 7 73 11 79 13 83 17 89 19 97 23 101 29 103 31 107 37 109 41 113 43 127 47 131 53 137 59 139 <-34th Mom's name adds to 139 and Kelly was born 139 days after mom's birthday. The kids have first names adding together for 139. This 139 is the 34th prime while Kelly's middle name adds to 34. Bible Book 34 contain 47 verses while her first name adds to 65 (47th non-prime). The parents have names differing in value by 82, Kelly was born in 82. Kelly's prime and square valued letters add together for 82. 22 Song 117 23 Isaiah 1292 24 Jeremiah 1364 25 Lamentations 154 26 Ezekiel 1273 27 Daniel 357 <-the 6th in the set 28 Hosea 197 29 Joel 73 30 Amos 146 ---- 4973 <- the 666th prime I met Kelly's dad on May 6th 2001, he has a little Yanko (66) brother that was born in 66 (see Ted below). Kelly Yanko (66) was born on the 6th day of the 6th month, she is 66 inches tall, says her flavorite number is 6. Kelly has 6 different letters in her given names. The vowels in Kelly's given names add to 36 (6x6). Kelly's even valued letters add to 66, this is 66.666...% of the value of her odd valued letters. Kelly's last name adds to 66, this is 66.666...% of the combined value of her given names. Kelly's given names add to 65 and 34, these Bible Books contain an average of 36 (6x6) verses. Sister's first name is Robyn, the name begins with the 6+6+6th letter of the alphabet and adds to 74 (a factor of 666). Robyn and mom have given names adding together for 193 (Bible Book 6 chapter 6). Robyn and her parents have first names adding together for 193 (Bible Book 6 chapter 6). Together the kids have 29 (6+6p+6np) letters, there are 29 chapters in Bible Book 13 (6th prime) and 29 verses in chapter 666. Mom and the kids were born on days of the year averaging 166.666... Mom was born on the 18th (6+6+6th) day of the month. The parents were born on the 238th and on the 18th days of the year, these are the 187th and the 11th non-primes, together for 198 (66+66+66). The Yanko (66) parents have names adding together for 360 (6x60). The Yanko (66) family was born on days of the year adding to 738, or 6x123 (6 times Numbers 6). The kids were born on days 21 and 6, together these Bible Books contain 36 (6x6) chapters. Mom and the kids were born on days 18, 21 and 6, together these Bible Books contain 78 chapters (6 times the 6th prime). Dad and the kids were born in years adding to 213 (166th non-prime). Dad and Kelly were born in years adding to 134, corresponding to Numbers 17 with 13 verses (the 6th prime). 1-50 - Genesis 51-90 - Exodus 91-117 - Leviticus 118-153 - Numbers 154-187 - Deuteronomy 188-211 - Joshua 930-957 - Matthew 958-973 - Mark 974-997 - Luke 998-1018 - John 1019-1046 - Acts 1047-1062 - Romans 188 <-the opening chapter of Book 6 is 6x6x6 short of the 404 verses of Bible Book 66, it is the 6th prime squared (13x13) short of the 357 verses of Daniel (also in part about 666) 193 <-Book 6 chapter 6 is the 44th prime, while 44 is in turn 66.666...% of 66 211 <-the terminating chapter of Book 6 is approximately 66.6% of the 66th prime (317) 357 <-the opening chapter of Book 6 plus the 6th prime squared is the 357 verses of Daniel (in part about 666) 404 <-the 6th prime squared (13x13) plus the 6th prime squared (13x13) plus 66 adds to the 404 verses of Bible Book 66 1062 <-666 plus 6x66 is a combination of the 658 verses of Bible Book 6 plus the 404 verses of Bible Book 66, and is the terminating chapter of New Testament Book 6 1070 <-666 plus the 404 verses of Book 66 is the 1070 verses of Job (Book 6+6+6) 1213 <-Exodus terminates at chapter 90 (66th non- prime) with 1213 verses (the 198th or the 66+66+66th prime) 1292 <-the 658 verses of Book 6 plus twice the 66th prime (317) is the 1292 verses of Isaiah (the Book contains 66 chapters) Kelly's names add to 65, 34 and 66, together these Bible Books contain 476 (238+238) verses, pretty as dad was born on day 238 (the first 13 primes). Kelly's dad, and dad's brother and sister were born in years adding to 165 while Kelly's name adds to 165. Dad was an average age of 28.5 years old when the kids were born, and Kelly was born 285 days after dad's birthday. Kelly was born 285 (5x57) days after dad's birthday. The parents have given names adding together for 228 (4x57). Kelly was born on day 157 and with 104+104 days remaining in the year, chapter 104 is Leviticus 14 with 57 verses. The first three family members were born on days 238, 18 and 325, these are the 187th, 11th and 259th non-primes, together for 457. The parents have given names adding to 155 and 73, these are the lengths of Bible Books 49 and 29, together for 78 (57th non-prime). Mom and the kids were born on days 18, 21 and 6, together these Bible Books contain 78 chapters (57th non-prime). I meet Kelly 57+57 days after her birthday. Mom is Barbara (43), she and dad were born on days of the month adding to 43 (14th prime). Kelly has 14 letters. Kelly adds to 65, or the 14th prime (43) plus the 14th non-prime (22). Kelly's common name adds to 131 (Numbers 14). Probababbly mom was born in a non-leap year and so the parents were together born 219 days closer to the beginning of their years than to the end of their years, while the kids have given names adding together for 219. Mom's given names add together for 73 while the kids have given names adding together for 73+73+73. There are 73 verses in Bible Book 29, it is the Lucas numbers up to 29, while the 29th element is Copper (73). Dad's middle name adds to 79 (Exodus 29) while the parents have middle names adding together for 109 (29th prime). Their first kid arrived in 79 (Exodus 29). Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 18 1 19 19 = 59 Kelly's names have an average value of 55. Her 7 different letters add to the 83 verses of Bible Book 55. Kelly's dad is Thomas, his first name adds to 76 (55th non-prime). Tom's given names add together for 155. Tom's last name adds to 83% of his middle name (the number of verses in Bible Book 55). Tom was born 110 (55+55) days closer to the end of the year than to the beginning of the year while his mom (Kelly's grandma) was born on the 110th (55+55th) day of the year. 66+ Dad 30 7 20 212/154 +13351 66+ Mom 19 4 24 110/256 +11992 173 Louanne 24 7 47 205/160 +3496 Louanne 82 Gale 25 Yanko 66 221 Tom 25 8 52 238/128 +1637 Thomas 76 William 79 Yanko 66 197 Ted 4 9 66 247/118 3486 Edward 55 Louis 76 Yanko 66 I met Tom in the afternoon of May 6th 2001 at The Broadway Roaster at 2318 8th Street (phone 652-8882), we sat at the counter together. His little brother is Edward, this first name adds to 55, and his middle name adds to 76 (the 55th non-prime), pretty as he is the 5th of 5 family members. Dad was 9855 days old when Louanne was born. Mom was 10355 days old when she gave birth to Tom. The kids were born in years averaging 55. Tom's names add to the 55th non-prime, 22nd prime and 48th non-prime, together for 125 (5x5x5). Tom was born on the 25th (5x5 or 5+5p+5np), he is coming out of a family of 5. Dad was born with 154 days remaining in the year, it is the length of Bible Book 25, the first of the kids has a middle name adding to 25 while Tom was born on the 25th. Born on the 25th, Tom's names add to the 55th non-prime, 22nd prime and to the 48th non-prime, together for 125 (5x25). Dad was born with 154 days remaining in the year, while the kids have first names adding to the 60th, 55th and 39th non-primes, together for 154 (77+77). Dad was born with 154 days remaining in the year, it is a combination of the first 13 non-primes. Tom was born on day 238, it is a combination of the first 13 primes. His middle name adds to 79 (Psalm 79 is the 13th chapter to contain the length of 13 verses). His full name adds to 221 (13x17) and he was born in '52 (4x13). Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations Mom was born in 24 and first gave birth on the 24th, while the brothers have names differing in value by 24. And the first and last kids have names differing in value by 24. Ted was born on day 247. Then I meet Kelly when I am 17024 days old. Bible Book 24 is Jeremiah with 52 chapters, and mom gave birth in year '52. The family was born on days 30, 19, 24, 25 and 4, these Bible Books together contain 252 chapters. They were born on days 212, 110, 205, 238 and 247, corresponding to chapter numbers adding to 52. The kids are commonly known as Louanne, Tom and Ted, the first initials of these names add together for 52. The parents were born on days and in months and years adding to 104 (52+52). Together the family had their birthdays 1081 days ago, corresponding to the 152nd chapter of the New Testament (Second Corinthians 3). Primes Non-Primes Lucas 2 1 1 3 4 3 5 6 4 7 8 7 11 9 11 13 10 18 17 12 29 19 14 47 23 15 76 29 16 123 31 18 199 37 20 322 41 21 521 43 22 843 47 <-15th-> 24 <-15th-> 1364 <-Jeremiah is Book 24 with 1364 verses Mom was born with 16x16 days remaining in the year and married Yanko, the name adds to 66, corresponding to Exodus 16, pretty as it is the 48th (16+16+16th) non-prime. The first of the kids was born with 160 days remaining in the year, the last Yanko (66) was born in 66. Together the kids were born with 406 days remaining in their years (406 verses in Bible Book 16). Dad's combined age at the births of the kids amounts to 38406 days. Tom was born on the 25th (16th non-prime), the kids were born on days of the month adding to 53 (the 16th prime). The males were born on days of the month adding to 59, the females on days of the month adding to 43, it is a difference of 16. The kids have first names adding to the 60th, 55th and 39th non-primes... and the kids have middle names adding to the 16th non-prime, 22nd prime and to the 55th non-primes, all together for Ted's 247th day of birth. Louanne was born in 47 while Ted was born on day 247. This 47 is the 15th prime while the females were born on days of the year adding to 315. Yanko adds to 66 (the 48th non-prime), the kids have names differing in value by 48. The parents were born on days of the month adding to 49 while the first two kids were born on days of the month adding to 49. Tom's given names add to the 155 verses of Bible Book 49. The first two kids have names averaging the 197 verses of Bible Book 28 while the last of the kids has a name adding to 197. Kelly was born 197 days after her sister's birthday. Tom was born with 128 days remaining in the year. The males were born in years '20, '52 and '66, these Bible Books together contain 1408 verses, it is a multiple of 128 (2 to the 7th). The represented letters in Tom's given names add to 120 and also the consonants in his given names add to 120. The represented letters in Tom's full name add to 170 and also the consonants in his given names add to 170. 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter Dad's day, month and year of birth adds to 57. Louanne's middle name adds to the 25 verses of chapter 57 and Book 57, her given names differ in value by 57, her common name adds to 148 (the 57+57th non-prime). The first two kids were born on days 24 and 25, together these Bible Books contain 57 chapters. The last two kids were born on days 25 and 4, these Bible Books together contain 41 chapters (57 is the 41st non-prime, Genesis 41 contains 57 verses). The kids had common name Louanne (82) Yanko (66), Tom (48) Yanko (66) and Ted (29) Yanko (66), their common names add together for the 357 verses of Daniel. Non-Primes 1 57 110 158 207 4 58 111 159 208 6 60 112 160 209 8 62 114 161 210 9 63 115 162 212 <-165th 10 64 116 164 213 12 65 117 165 214 14 66 118 166 215 15 68 119 168 216 16 69 120 169 217 18 70 121 170 218 20 72 122 171 219 21 74 123 172 220 22 75 124 174 221 24 76 125 175 222 25 77 126 176 224 26 78 128 177 225 27 80 129 178 226 28 81 130 180 228 30 82 132 182 230 32 84 133 183 231 33 85 134 184 232 34 86 135 185 234 35 87 136 186 235 36 88 138 187 236 38 90 140 188 237 39 91 141 189 238 40 92 142 190 240 42 93 143 192 242 44 94 144 194 243 45 95 145 195 244 46 96 146 196 245 48 98 147 198 246 49 99 148 200 247 50 100 150 201 248 51 102 152 202 249 52 104 153 203 250 54 105 154 204 252 55 106 155 205 253 56 108 156 206 254 Grandpa's 212th day of birth is the 165th non-prime while the kids were born in years adding to 165, while granddaughter Kelly has a name adding to 165. I met Tom at The Broadway Roaster at 2318 8th Street (phone 652-8882), now I meet his daughter Kelly at The Yard and Flagon at 718 Broadway Ave. (phone 653-8883). 165 Kelly 6 6 82 157/208 9240 Kelly 65 Anne 34 Yanko 66 66+ Grandpa 30 7 20 212/154 +13351 66+ Grandma 19 4 24 110/256 +11992 173 Louanne 24 7 47 205/160 +3496 Louanne 82 Gale 25 Yanko 66 221 Tom 25 8 52 238/128 +1637 Thomas 76 William 79 Yanko 66 139 Barb 18 1 18/ Barbara 43 Jean 30 Yanko 66 186 Robyn 21 11 79 325/40 8312 Robyn 74 Alise 46 Yanko 66 197 Ted 4 9 66 247/118 3486 Edward 55 Louis 76 Yanko 66 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 Shown above is Kelly with her family standing between Marcia and me. Tom's and Barb's kids were born in years adding to 161, and that could be the value of Kelly's name if she took my last name (and she would become a Queen, like Esther in Book 17). Or if Kelly marries Marcia and me, then Kelly could become Kelly Anne Acevedo-Kabatoff, resulting in her name adding to 216 (6x6x6). Kelly and Marcia were born on days of the year averaging 188 (the opening chapter of Bible Book 6), great as 6 is Kelly's flavorite number. Isaiah is the Book with 66 chapters while Isaiah 4, 12 and 20 (adds to 6x6) each contains 6 verses. Isaiah 4:1 is about Marcia and me, and six other women, all are probababbly begging me to take away their clothes (or some damn thing like that). If my abusive parents, Protestants, Catholics and Hindus think that they should have the right to have me arrested and tortured for not being married, then I should have the right to ask women to marry me, or even to marry both Marcia and me. If Kelly marries me, great, but if Marcia marries me then Tom will win a new Cadillac!!! And if BOTH Kelly and Marcia marries me, then Robyn and Barb will win new Cadillac's too, it's all now included in my suit against the filthy powers that be. Good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Subject: Jennifer Lynn Lemeurn - May 7th 1981 L E M E U R N 12 5 13 5 21 18 14 = 88 In the evening I went to The Yard and Flagon Pub at 718 Broadway Ave. (Penis Street, it's the street with the penis poster poles) in Saskatoon, a nubile sweety provided stats when she ventured near. 88+ Dad 24 9 /98 88+ Mom 28 3 /278 88+ Sis 26 4 76 117/249 7008 234 Jennifer 7 5 81 127/238 8845 Jennifer 81 Lynn 65 Lemeurn 88 Dad generally has his birthday on day 267 (First Samuel 31). The kids were born on days 26 and 7, together these Bible Books contain 1891 (31x61) verses. Jennifer was born in 81 and her first name adds to 81 (Exodus 31). She was born on the 127th day of the year (31st prime). Her day, month and year of birth averages 31. Primes 2 61 149 3 67 151 5 71 157 <-37th 7 73 163 11 79 167 13 83 173 17 89 179 19 97 181 23 101 191 29 103 193 31 107 197 37 109 199 41 113 211 43 127 223 47 131 227 53 137 229 59 139 233 Dad was born on the 24th, corresponding to Jeremiah with 52 chapters (37th non-prime). The parents were born on days of the month adding to 52 (37th non-prime). Mom generally has her birthday on day 87 (Exodus 37). The kids were born on days 117 and 127, corresponding to Leviticus 27 and Numbers 10, together for 37. The kids were born in years adding to 157 (37th prime). Jennifer was born 111 (37+37+37) days closer to the beginning of the year than to the end of the year. The kids are separated by 1837 days. Generally dad has his birthday 169 (13x13) days closer to the end of the year than to the beginning of the year. The first of the kids was born on the 26th (13+13th) day of the month and on the 117th (3x3x13th) day of the year. Jennifer's first and last names add together for 169 (13x13), her middle name adds to 65 (5x13), her full name adds to 234 (2x3x3x13). The family was born in months adding to 21 (13th non-prime). Jennifer was born with 238 days remaining in the year (the first 13 primes). Primes Non-Primes Fibonacci Lucas 2 1 0 1 3 4 1 3 5 6 1 4 7 8 2 7 11 9 3 11 13 10 5 18 17 12 8 29 19 14 13 47 23 15 21 76 29 16 34 123 31 18 55 199 37 20 89 322 41 <-13th-> 21 <-13th-> 144 <-13th-> 521 --- --- 238 154 <-Lamentations The parents were born on days 24 and 28, together these Bible Books contain 66 chapters. The sister was born 132 (66+66) days closer to the beginning of the year than to the end of the year. The 6th letter of Jennifer's name is the 6th letter of the alphabet. The repeating letters in her given names add to 66. In her full name, her repeating letters exceed her unrepeated letters by 66. Mom first gave birth 29 days after her birthday, it's 6 plus the 6th prime (13) plus the 6th non-prime (10), there are 29 (6+6p+6np) chapters in Book 13 (6th prime) and 29 verses in chapter 666. Primes Non-Primes 2 1 3 4 5 6 7 <-4th-> 8 <-Bible Book 8 contains 11 4 chapters, pretty as 13 8 is the 4th non-prime 17 19 -- 77 <-the first 8 primes plus 8 more adds to the 85 verses of Bible Book 8 Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 18 61 19 67 <- 67 <-the 8th prime in prime position R U T H <-Book 8 18 21 20 8 = 67 The family name adds to 88. The Lemeurn (88) family was born on days of the month adding to the 85 verses of Bible Book 8, it's a combination of the first 8 primes plus 8 more. Jennifer has an 8 lettered first name, and her middle name is 4 lettered (8 is the 4th non-prime, Bible Book 8 contains 4 chapters). Jennifer has 8 different letters in her given names. Jennifer is missing 16 (8+8) letters, these 8+8 missing letters add to 218 and exceed the letters that she doths have by the 85 verses of Bible Book 8 (the first 8 primes plus 8 more). Jennifer has 19 letters (the 8th prime). Dad was born with 98 days remaining in the year (Leviticus 8). Mom generally has her birthday 185 days after dad has his, pretty as Bible Book 8 terminates at chapter 236 (the 185th non-prime), prettier as there are 85 verses in Book 8, and prettier still as the terminating chapter of Book 8 is 118+118 (twice the 88th non-prime). The kids were born on days of the year adding to 244 (First Samuel 8). Dad was born 8 months and 8+8+8 days into the year. The parents were born on days 24 and 28, the former is 85.71% of the latter (85 verses of Bible Book 8, it's the first 8 primes plus 8 more). The Lemeurn (88) kids were born on days 117 and 127, these are the 87th non-prime and the 31st prime, together for 118 (88th non-prime). Jennifer's names add to the 59th, 47th and 65th non-primes, together for 171, pretty as Old Testament Book 8 and New Testament Book 8 contain an average of 171 verses, it is the numbers 1 through to 18, prettier as chapter 171 is Deuteronomy 18. Jennifer's given names differ in value by 16 (8+8). Jennifer's given names add to the 59th and 47th non-primes, for an average position of 53 (8+8th prime). Her first 8 and last 8 letters add together for 183 while her full name adds to 234 (the 183rd non-prime). Her 10 different letters add to 133, corresponding to Numbers 16 (8+8). The First 8 18's ---------------- 13 <-Genesis 13 20 <-Genesis 20 74 <-Exodus 24 81 <-Exodus 31 142 <-Numbers 25 166 <-Deuteronomy 13 188 <-Joshua 1 204 <-Joshua 17 --- 888 The parents generally have their birthdays on days of the year averaging 177 (59+59+59). The kids were together born 243 (81+81+81) days closer to the beginning of their years than to the end of their years, Jennifer adds to 81, she was born in 81 (59th non-prime). Jennifer and her parents were born on days of the month adding to 59 (17th prime) and in months adding to 17. The sister was born on the 26th day of the month (17th non-prime) and in year 76 (17 plus the 17th prime), corresponding to Exodus 26 (17th non-prime), and on the 117th day of the year. The family was born on days of the month adding to the 85 verses of Bible Book 8, it is 5x17 and is the 17th prime (59) plus the 17th non-prime (26), with 85 verses Ruth is the 17th shortest Book in the Bible. The kids were together born 440 days after dad's birthdays (the first 17 primes up to 59). In Jennifer's given names, her consonants add to 102 (6x17 and is 17+17p+17np, it is the 76th non-prime while 76 is 17 plus the 17th prime), while all of her consonants add to 159 (ends in the 17th prime, it is the 7x7th prime minus the 7x7th non-prime and is the 618 verses of Bible Book 7 short of 777). In Jennifer's given names, her consonants exceed her vowels by 58 (the 7 primes up to 17). Jennifer's last two names add together for 153 (1 through 17 and is the 117th non-prime, it's the terminating chapter of Numbers and is the number of fish in the net in John 21). The kids were together born 509 days after their parent's birthdays, it plays upon the 17th prime (59). Perhaps the family was born on days of the year adding to 598, keeping in mind that dad was born with 98 days remaining in the year. The 1837 days separating the kids is 11 times the 167 verses of Bible Book 17. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 The sister was born on the 26th (Iron) while the parents were born on days 24 (Chromium) and 28 (Nickel), pretty as Chromium and Nickel are used to strengthen Iron, and to make it resistant to corrosion. See that Aluminum is primarily strengthened with Copper, these are elements 13 and 29 so it is pretty that Bible Book 13 contains 29 chapters. The God who made the Bible also made (and named) the elements, and I think there is a mathematical structure in the Bible that can give instruction on making alloys. Likely the parents were born on days of the year adding to 354, the kids were born on days of the year adding to 244 (a difference of 55+55). The sister was born on the 26th (Iron with an atomic mass of 55.845). The parents were born on days 24 and 28, these elements have atomic masses averaging 55.34475. The parents and the first were born on days 24, 28 and 26, these elements have atomic masses averaging 55.5115. The sister was born in 76 (55th non-prime). The kids were together born 440 (8x55) days after dad's birthdays. The parents were born on days 24 and 28, the kids on days 26 and 7, Bible Books 26 and 7 together contain 330 (6x55) more verses than Bible Books 24 and 28. Perhaps mom Lemeurn (88) was born on the 88th day of the year, and so the parents were born on days of the year adding to 355. Jennifer and I were together born with 555 days remaining in our years. Bible Book 10 brings The Samuels up to 55 chapters, pretty as 1 through 10 adds to 55. And New Testament Book 10 contains 155 verses, again pretty as 1 through 10 adds to 55. Lucas 1 3 4 7 11 18 29 -- 73 <-the Lucas numbers up to 29 add to the 73 verses of Bible Book 29 J O E L <-Bible Book 29 10 15 5 12 = 42 <-29th non-prime C O P P E R <-29th element 3 15 16 16 5 18 = 73 <-Book 29 and is the Lucas numbers up to 29, there is a copper riding a horse on the 1973 Canadian 25 cent piece C E N T <-made out of 29th element 3 5 14 20 = 42 <-29th non-prime Copper and Zinc are elements 29 and 30 (together for 59), and together they make Brass (59): B R A S S 2 18 1 19 19 = 59 Mom and Jennifer were born on days of the month adding to 35, they were born on days 28 and 7, together these Bible Books contain 35 chapters. Jennifer's 234 valued name is 35 plus the 35th prime (149) plus the 35th non-prime (50), or simply 35+35p+35np. The represented letters in her given names add to 99, or the 35th prime minus the 35th non-prime. Likely the family was born on days 267, 87, 117 and 127, corresponding to First Samuel 31, Exodus 37, Leviticus 27 and Numbers 10, together for 105 (35+35+35). The parents were born on days of the year adding to 354, 355 or 356. If Jennifer was born in the evening, then she was born 35% into the year. Jennifer and I were born on days of the year adding to 175 (5x35). Genesis 29 contains 35 verses, Genesis 35 contains 29 verses and there are 35 chapters in the Bible that contain the length of 29 verses while Jennifer was born 265 days after her parent's birthdays (First Samuel 29). Perhaps the family was together born 265 days closer to the beginning of their years than to the end of their years (First Samuel 29). Jennifer's given names average the 73 verses of Bible Book 29 (and is the Lucas numbers up to 29). Jennifer's prime, square and perfect valued letters add together for the 73 verses of Bible Book 29 (and is the Lucas numbers up to 29). In Jennifer's given names, her even valued letters exceed her odd valued letters by 58 (29+29 and is the 29th non-prime in non-prime position). Mom first gave birth 29 days after her birthday. 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter The kids were together born on days of the year adding to 244, mom was likely born on day 87 (a difference of 157). The kids were born in years adding to 157. The parents and the kids were born on days of the month adding to 52 and 33, the former is 157.5757...% of the latter. Mom was likely born 191 days closer to the beginning of the year than to the end of the year, it's the 57th non-prime (78) short of the 57th prime (269). The family was born on days 24, 28, 26 and 7, these Bible Books contain an average of 25.57 verses per chapter, and these Bible Books together contain 135 chapters (57 plus the 57th non-prime). Jennifer's names add to the 59th, 47th and 65th non-primes (an average of 57). I was seated at the bar and working on our bartenderette's stats, stats when Jennifer came to me, Kelly Anne Yanko was born on the 157th day of the year. I was born in 57, you find your storms of destruction in Psalm 57. 187 Dar 17 2 57 48/317 00 Daryl 60 Shawn 65 Kabatoff 62 187 Marcia 6 8 80 219/147 8571 Marcia 45 Veronica 87 Acevedo 55 234 Jennifer 7 5 81 127/238 8845 Jennifer 81 Lynn 65 Lemeurn 88 165 Kelly 6 6 82 157/208 9240 Kelly 65 Anne 34 Yanko 66 Marcia, Jennifer and Kelly were born on days of the year averaging 167.666... The 167 verses of Bible Book 17 is a combination of the primes in prime positions up to the 17th prime. Esther becomes Queen in Book 17 and Q is the 17th letter of the alphabet. Jennifer's name adds to 234 (117+117) while Acevedo-Kabatoff adds to the 117 verses of Song of Solomon, Jennifer could marry Marcia and me and become a Queen like Esther (in Book 17). Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 And so if Marcia marries me, then Jennifer will win a new Cadillac. And if Jennifer marries Marcia and me, then Kelly will win a new Cadillac, it is all now included in my suit against the filthy powers that be (the Protestants, Catholics, Hindus, and the political parties that allowed them to torture me at the U of S and then later at Saskatoon City Hospital). If my abusive parents, Protestants, Catholics and Hindus think that they should have the right to have me arrested and tortured for not being married, then I should have the right to ask women to marry me, or even to marry both Marcia and me. They tortured me year after year, and year after year I begged people in the community for assistance to flee the country, and you people are so cheap and ignorant that you can't even offer to buy me a measly cookie for my work when I show you evidence that your very name is a gift from God, you won't even pay me minimum wage for my work so that I could attempt to get out of this country on my own dime, you people are so cheap and ignorant that not a single one of you has the ability to spend 48 cents for a stamp so that you can send me a cheap letter and express thanx for showing you evidence that the true God of the Bible provided you with your very names. You people are compassionless turds, the of the earth, I show you people sevens and seventeens and all you want to do is turn a seven foot tall tree into a decorated idol with 17 blinkin' bulbs, call me a stalker and a pedophile and then fly off to Europe on holidays, may He blow your trees (and children) over, and burn and smite the remainder with diseases. You people used my interest in mathematics as a reason to have me repeatedly arrested and tortured (at a cost of many millions of dollars), my work is worth a little more than a needle up my ass!!! Anyway, good luck and may God bless you!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Subject: Fermat's Last Theorem/Andrew Wiles Hi All, I'm a sincere math amateur, and I saw a series on PBS about Andrew Wiles' attempt to solve Fermat's Last Theorem. If I recall, he was close, but that an associate pointed out a fundamental flaw in his theory. Does anyone know if he fixed the theorem -- and is it now considered proved? Also, can anyone recommend a good book with regard to Wiles' attempt to solve the problem over 7 years? === Subject: Re: Fermat's Last Theorem/Andrew Wiles You might find Algebraic Number Theory and FLT by Ian Stewart and David Tall (3rd edition A K Peters 2002) helpful. Guy Corrigall > Hi All, > I'm a sincere math amateur, and I saw a series on PBS about Andrew Wiles' > attempt to solve Fermat's Last Theorem. If I recall, he was close, but that > an associate pointed out a fundamental flaw in his theory. > Does anyone know if he fixed the theorem -- and is it now considered > proved? Also, can anyone recommend a good book with regard to Wiles' > attempt to solve the problem over 7 years? === Subject: Re: Fermat's Last Theorem/Andrew Wiles >I'm a sincere math amateur, and I saw a series on PBS about Andrew Wiles' >attempt to solve Fermat's Last Theorem. If I recall, he was close, but that >an associate pointed out a fundamental flaw in his theory. >Does anyone know if he fixed the theorem -- and is it now considered >proved? Also, can anyone recommend a good book with regard to Wiles' >attempt to solve the problem over 7 years? For the sincere and dedicated amateurs: Fermat's Last Theorem for Amateurs (currently on sale in the Springer Yellow sale) and Invitation to the Mathematics of Fermat-Wiles (written by the individual who first realized that elliptic curves were an essential step towards the solution) Both these are by what I believe to be respected individuals writing serious mathematics and are not just popularized near-fiction. === Subject: Re: Fermat's Last Theorem/Andrew Wiles >Also, can anyone recommend a good book with regard to Wiles' >attempt to solve the problem over 7 years? Aczel's Fermat's Last Theorem is a florid account of the history and basic concepts of the proof. It is O.K. Perhaps others here have objections to it. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Fermat's Last Theorem/Andrew Wiles >Also, can anyone recommend a good book with regard to Wiles' >attempt to solve the problem over 7 years? Aczel's Fermat's Last Theorem is a florid account of the history and basic concepts of the proof. It is O.K. Perhaps others here have objections to it. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Fermat's Last Theorem/Andrew Wiles > Hi All, > I'm a sincere math amateur, and I saw a series on PBS about Andrew Wiles' > attempt to solve Fermat's Last Theorem. If I recall, he was close, but that > an associate pointed out a fundamental flaw in his theory. > Does anyone know if he fixed the theorem -- and is it now considered > proved? Also, can anyone recommend a good book with regard to Wiles' > attempt to solve the problem over 7 years? If you are referring to the Nova documentary, The Proof, then the answer to your question is contained in the same program. Wiles explained how he tried one line of reasoning, then rejected it in favor of another, which turned out to contain the flaw. He eventually fixed it (with some help from a colleague) by returning to the first line of reasoning that he had previously rejected. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Fermat's Last Theorem/Andrew Wiles Visiting Assistant Professor at the University of Montana. >I'm a sincere math amateur, and I saw a series on PBS about Andrew Wiles' >attempt to solve Fermat's Last Theorem. If I recall, he was close, but that >an associate pointed out a fundamental flaw in his theory. >Does anyone know if he fixed the theorem -- and is it now considered >proved? Yes, and yes. The error was fixed with joint work with Richard Taylor. >Also, can anyone recommend a good book with regard to Wiles' >attempt to solve the problem over 7 years? Simon Singh's book, _Fermat's Enigma_ is usually well spoken of, although I have not read it myself. Stay away from Marilyn vos Savant's, though. Really, really, far away. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Xiaoxia Zhao - May 30thn 197_ Z H A O 26 8 1 15 = 50 In the evening Xiaoxia (pronounced Sosya) provided stats, she is Chinese and from Inner Mongolia. She has been in Canada for the last couple of years and taking a biology program at the U of S, where she is focusing her studies upon fish. The nubile sweety hopes to become an ictyologist. 133 Xiaoxia 30 5 150/215 Xiaoxia 83 Zhao 50 Xiaoxia says she was born in a non-leap year, I guess some times during the 1970's. Zhao (50) was born on the 150 (50+50+50th) day of the year. Her initials add to 50. Zhao adds to 50 (35th non-prime), the first three letters in Zhao add to 35. The vowels in her first name add to 35, the odd valued letters in her first name add to 35. Xiaoxia was born on day 150, there are 150 chapters in Bible Book 19. Her name adds to 133 (7x19). Her first name adds to 83 while all 6 different letters in her full name add to 83. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 <-7th-> 12 -- -- 58 50 Xiaoxia has a 7 lettered first name and a 4 lettered last name (7 is the 4th prime). Her 7 vowels add to 51 (3 times the 7th prime). Her two unrepeated letters average 17 (7th prime). The 4 different letters in her first name add to 49 (7x7). Her last name adds to 50 and her initials add to 50 (the first 7 non-primes). Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Subject: Xiaoxia Zhao - May 30th 197_ Z H A O 26 8 1 15 = 50 In the evening Xiaoxia (pronounced Sosya) provided stats, she is Chinese and from Inner Mongolia. She has been in Canada for the last couple of years and taking a biology program at the U of S, where she is focusing her studies upon fish. The nubile sweety hopes to become an ictyologist. 133 Xiaoxia 30 5 150/215 Xiaoxia 83 Zhao 50 Xiaoxia says she was born in a non-leap year, I guess some times during the 1970's. Zhao (50) was born on the 150 (50+50+50th) day of the year. Her initials add to 50. Zhao adds to 50 (35th non-prime), the first three letters in Zhao add to 35. The vowels in her first name add to 35, the odd valued letters in her first name add to 35. Xiaoxia was born on day 150, there are 150 chapters in Bible Book 19. Her name adds to 133 (7x19). Her first name adds to 83 while all 6 different letters in her full name add to 83. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 <-7th-> 12 -- -- 58 50 Xiaoxia has a 7 lettered first name and a 4 lettered last name (7 is the 4th prime). Her 7 vowels add to 51 (3 times the 7th prime). Her two unrepeated letters average 17 (7th prime). The 4 different letters in her first name add to 49 (7x7). Her last name adds to 50 and her initials add to 50 (the first 7 non-primes). Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Moderator: You. Marcia and Me: Me? Moderator: Yes, you. Get off of my newsgroup, and out of my newsreader. Off, off! You're washed up, you're finished! I'll see to it that you never make another post again! Marcia and Me: Does that include alt.math.idiots too, sir? === Subject: Re: Brownian motion approximation > A while back I posted a question about whether or not > P[sup_{0<=t<=1}|B_t - f(t)| < d] > 0 for all d > 0 and f(t) continuous > on [0,1] with f(0) = 0. > In other words, does Brownian motion uniformly approximate any > continuous function(with f(0)=0) with positive probability? Someone > replied that it does, and this follows from first proving it for f(t) > = 0 for all t and then applying the Cameron-Martin Theorem. I can do > it for f(t)=0, but I don't seem to be able to find a reference for the > Cameron-Martin Theorem, though it seems to be related to Girsanov's > Theorem, and maybe even follows from it. Can someone give me some help > or lead me to a reference? The paper of Cameron & Martin is: Transformations of Wiener integrals under translations, Annals of Math., vol. 45, (1944), pp. 386-396. The paper gives conditions on a (non-random!) continuous function f(t), 0<= t<= 1, under which the distribution of {B_t - f(t), 0<= t <= 1} is absolutely continuous with respect to Wiener measure. Girsanov's theorem extends C&M by allowing the translation function f to be random (but adapted to the filtration of the Brownian motion). [Cameron & Martin imposed a condition on f that was later shown to be superfluous by G. Maruyama. For this reason the theorem is often referrred to as the Cameron-Martin-Maruyama theorem.] -- A. === Subject: Re: Brownian motion approximation > A while back I posted a question about whether or not > > P[sup_{0<=t<=1}|B_t - f(t)| < d] > 0 for all d > 0 and f(t) continuous > on [0,1] with f(0) = 0. > > In other words, does Brownian motion uniformly approximate any > continuous function(with f(0)=0) with positive probability? Someone > replied that it does, and this follows from first proving it for f(t) > = 0 for all t and then applying the Cameron-Martin Theorem. I can do > it for f(t)=0, but I don't seem to be able to find a reference for the > Cameron-Martin Theorem, though it seems to be related to Girsanov's > Theorem, and maybe even follows from it. Can someone give me some help > or lead me to a reference? > The paper of Cameron & Martin is: > Transformations of Wiener integrals under translations, > Annals of Math., vol. 45, (1944), pp. 386-396. > The paper gives conditions on a (non-random!) continuous function f(t), > 0<= t<= 1, under which the distribution of > {B_t - f(t), 0<= t <= 1} is absolutely continuous with respect to Wiener > measure. Girsanov's theorem extends C&M by allowing the translation > function f to be random (but adapted to the filtration of the Brownian > motion). > [Cameron & Martin imposed a condition on f that was later shown to be > superfluous by G. Maruyama. For this reason the theorem is often > referrred to as the Cameron-Martin-Maruyama theorem.] === Subject: Re: Brownian motion approximation > A while back I posted a question about whether or not > P[sup_{0<=t<=1}|B_t - f(t)| < d] > 0 for all d > 0 and f(t) continuous > on [0,1] with f(0) = 0. > In other words, does Brownian motion uniformly approximate any > continuous function(with f(0)=0) with positive probability? Someone > replied that it does, and this follows from first proving it for f(t) > = 0 for all t and then applying the Cameron-Martin Theorem. I can do > it for f(t)=0, but I don't seem to be able to find a reference for the > Cameron-Martin Theorem, though it seems to be related to Girsanov's > Theorem, and maybe even follows from it. Can someone give me some help > or lead me to a reference? I don't believe there is any significant difference between the Cameron-Martin-Girsanov theorem, the Girsanov theorem and the Cameron-Martin theorem. As I understand it the same theorem was discovered independently and is now attributed to all three. Rather like the Green-Gauss-Ostrogradsky theorem. === Subject: Please help with convex sets! Hi everyone I've been struggling with this problem for a few days, I would greatly appreciate any help. Question: Prove that the sum and difference of two convex sets in Rn are convex. === Subject: Re: Please help with convex sets! >Question: Prove that the sum and difference of two convex sets in Rn >are convex. The sum of sets A and B is A+B = {a + b : a in A, b in B}. Consider points a1 + b1 and a2 + b2 in A+B, and 0 <= t <= 1. How do you think you could write t (a1 + b1) + (1-t) (a2 + b2) as the sum of a member of A and a member of B? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Newby : A question on decideability I have just finished re-reading Fermats Last Theorem by Simon Singh and on Page 165 a statement made me stop and think. It said that if the Theorem were undecideable then it must be true. The argument is written below and explored further in my random ramblings on the issue. Surely this means that it is indeed decideable? Indeed, it seems to me that a whole class of problems must be decideable, at least in principle, and that this class is very broad indeed. Could someone please give a little more information on this (or a web reference? I have looked and failed.) I may just have misunderstood the terminology (but the term 'fundamentally undecideable' seems clear enough to me.) I appologise in advance for my lack of rigour (and possible abuse of the symbology.) I am an electronics/software engineer with a good knowledge of mathematical tools (3D calculus / Fourier transforms etc.) but a laymans knowledge of the higher structures of mathematics. Finding digestible information on these issues is not easy. I have read and understood Godel, Escher, Bach (an astonishingly well written book) and an explaination pitched at that level would be perfect. Also, I have the desire to extend my maths. I know enough to have seen the inherent beauty and I would like to see more (it is 20 years since I did my degree...) Any suggestions for an accessible area to begin with? I have started looking at Groups because this seems accessible and yet it is very different from the maths that I am used to. Is this a good place to start? And where next? I have an interest in parsers and compilers on the software side so anything that might complement that interest would be good. Andy PS If this is not an appropriate place to post this, I appologise. Please advise. I have only just subscribed to this group (I know, I'm impatient) but I guess I'll be sticking around for a while. I certainly seems very active. Oh! and I repeat, I am not a mathematician (yet), please also be gentle!! Conjecture C is ~$x | f(x), where x is an integer and f(x) is any predicate function of x. (if you can't read the font this is 'There does not exist x such that f(x)...) C is then stated to be mathematically undecideable. Therefore, there cannot be a counter-example, x, since the very existence of such a counter-example would make C decideable (and indeed, false.) If there is no counter-example then C must be true (because that is precisely what C asserts.) Therefore C is, again, decideable. The only unproven assertion is that C is undecideable. This must be invalid and therefore all conjectures of the form of C must be decideable. The flip-side Conjecture D is $x | f(x), where f(x) is any predicate function of x. (...and this is 'There exists x such that f(x)) D is then stated to be mathematically undecideable. Therefore, there cannot be an example, x, since the very existence of such an example would make D decideable (and indeed, true.) If there is no example then D must be false (because that is precisely what D refutes.) Therefore D is, again, decideable. The only unproven assertion is that D is undecideable. This must be invalid and therefore all conjectures of the form of D must be decideable. So my best guess is... There cannot be a fundamentally undecideable proposition that posits the existence or non-existence of a member (or members) of an enumerable set. Simply enumerating the set will eventually locate one such member and make the proposition decideable. It is irrelevant that such a search might be impractical provided that it is possible. [This is, I guess, where Turing steps in with his computability theory?] OK, release the hounds.... ; ) === Subject: Re: Newby : A question on decideability > I have just finished re-reading Fermats Last Theorem by Simon Singh and on > Page 165 a statement made > me stop and think. It said that if the Theorem were undecideable then it > must be true. The argument is Yes. If it were false there would be a counterexample. However in the case of FLT the discussion is moot, since we now know that FLT is true. The same reasoning applies to any statement that is falsifiable by a counterexample, such as the Goldbach conjecture. === Subject: Re: Newby : A question on decideability > I have just finished re-reading Fermats Last Theorem by Simon Singh and on > Page 165 a statement made > me stop and think. It said that if the Theorem were undecideable then it > must be true. ... Perversely, if U is Fermat's theorem is undecidable and T is Fermat's theorem is true, then if U then T is true just because _any_ statement if P then Q is true if Q is true. I wonder what the force of must is? Syntactic sugar? -- G.C. === Subject: Re: Newby : A question on decideability > .... > an electronics/software engineer with a good knowledge of mathematical tools > (3D calculus / Fourier transforms etc.) ... You could look at real and complex analysis to make it rigorous (if it isn't already). > I have started looking at Groups because this seems accessible > and yet it is very different from the maths that I am used to. Is this a > good place to start? And where next? Read books backwards for motivation. For example near the end of a UG algebra text you might find a proof of the unsolvability of the quintic in radicals which makes all the boring stuff about groups at the beginning seem worthwhile. > I have an interest in parsers and compilers on the software side so anything > that might complement that interest would be good. Formal grammars and finite automata. With a background in software, various logical topics might interest you: lambda calculus (connects to Lisp), modal/tense logic (connects to formal proofs of correctness), decidable fragments of first order logic (connects to Prolog), recursion theory, etc. -- G.C. === Subject: least square of y = 1 + bx was wondering if someone might go through a least squares derivation of y = 1 + bx, a = 1, so theres only one unknown I guess.. I don't know where to start. === Subject: Re: least square of y = 1 + bx forget it, I got it....... > was wondering if someone might go through a least squares derivation of y = > 1 + bx, > a = 1, so theres only one unknown I guess.. > I don't know where to start. === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind > 1. Mathematics is the science in which we make something out of > nothing. Was it Quine who said that everything could be built out of the empty set? Not true of course, one needs the axiom of infinity (in the presence of which the assumption of the existence of the empty set becomes redundant). > 2. All of man-made Mathematics consists of an abstraction from > physical processes. There are famous examples of pure maths preceding the physicists' need for it: matrices & Heisenberg, Riemannian geometry & Einstein. > 3. Since Mathematics in general needs nothing to be created, then the > human mind seems to be limited in that it can only consider > possibilities that are analogous to physical processes. analogous is a bit vague. A religious person might say that God (and related concepts) are not analogous to physical processes, but you might reply that with analogous taken broadly enough, they are. Each case would need considering. -- G.C. === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind > > 1. Mathematics is the science in which we make something out of > nothing. > Was it Quine who said that everything could be built out of the empty > set? Of course we all know about the empty set and its importance in the foundation of Mathematics is almost indisputable. What I am saying is to note that physical objects and our 5 senses (sight, hearing, smell, taste, feeling) are used (needed) in all other branches of science, but not in Mathematics. Did Quine say anything about that? I would be very interested in hearing his opinion on that idea. > 2. All of man-made Mathematics consists of an abstraction from > physical processes. > There are famous examples of pure maths preceding the physicists' need > for it: matrices & Heisenberg, Riemannian geometry & Einstein. With all due respect, my statement has nothing to do with the chronological order in which certain events occurred on our planet. It has to do with the timeless structure of Mathematics and the subset of it which man has explored. In fact, you are only citing more examples of how man-made Mathematics does in fact consist only of abstractions from physical processes. > 3. Since Mathematics in general needs nothing to be created, then the > human mind seems to be limited in that it can only consider > possibilities that are analogous to physical processes. > analogous is a bit vague. How about models? > A religious person might say that God (and > related concepts) are not analogous to physical processes, but you might > reply that with analogous taken broadly enough, they are. You certainly might! :) Charlie Volkstorf Cambridge, MA http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1 http://www.arxiv.org/html/cs.lo/0003071 === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind >Yes. Abstracting from physical reality produces Mathematics. >Abstracting from Mathematics produces metamathematics. And >abstracting from metamathematics produces metametamathematics. etc. >Now, we all know what Mathematics and metamathematics look like. But >what does metametamathematics consist of? >>What is (or was, at least) referred to as metamathematics is a part of >>mathematics, so metametamathematics is just a particular portion of >>metamathematics. > The proposition undecidable in the system PM is thus decided by > metamathematical arguments. - Kurt Godel, 1931 The undecidable propostition is seen to be true in the natural numbers by an ordinary mathematical argument. The argument is only metamathematical in the sense that it treats formal systems, which are taken to axiomatise portions of arithmetical truths (and by Rosser's generalisation, also some that provably don't), but not in any sense which would imply that it's not part of mathematics. > A Turing Machine calculates Mathematics, but does not decide the > halting predicate, which lies within Metamathematics. The halting predicate lies well within ordinary mathematics. > Aside from where Metametamathematics happens to lie, I ask, what does > it look like? It looks like mathematics. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind >>What is (or was, at least) referred to as metamathematics is a part of >>mathematics, so metametamathematics is just a particular portion of >>metamathematics. > > The proposition undecidable in the system PM is thus decided by > metamathematical arguments. - Kurt Godel, 1931 > The undecidable propostition is seen to be true in the natural numbers > by an ordinary mathematical argument. If it is ordinary mathematics, then why is it not decidable in the system PM, or is Godel wrong? > A Turing Machine calculates Mathematics, but does not decide the > halting predicate, which lies within Metamathematics. > The halting predicate lies well within ordinary mathematics. Doesn't a computer implement ordinary mathematics (yet a computer cannot decide the halting predicate)? > Aside from where Metametamathematics happens to lie, I ask, what does > it look like? > It looks like mathematics. You haven't made any distinction concerning the nature of Metametamathematics. I think that Metamathematics concerns Mathematical problems that cannot be solved by Mathematics. My question is still, what is Metametamathematics like? Charlie Volkstorf Cambridge, MA http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1 http://www.arxiv.org/html/cs.lo/0003071 === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind >>What is (or was, at least) referred to as metamathematics is a part of >>mathematics, so metametamathematics is just a particular portion of >>metamathematics. > > The proposition undecidable in the system PM is thus decided by > metamathematical arguments. - Kurt Godel, 1931 > The undecidable propostition is seen to be true in the natural numbers > by an ordinary mathematical argument. > If it is ordinary mathematics, then why is it not decidable in the > system PM, or is Godel wrong? > A Turing Machine calculates Mathematics, but does not decide the > halting predicate, which lies within Metamathematics. > The halting predicate lies well within ordinary mathematics. > Doesn't a computer implement ordinary mathematics (yet a computer > cannot decide the halting predicate)? > Aside from where Metametamathematics happens to lie, I ask, what does > it look like? > It looks like mathematics. > You haven't made any distinction concerning the nature of > Metametamathematics. I think that Metamathematics concerns > Mathematical problems that cannot be solved by Mathematics. My > question is still, what is Metametamathematics like? meta languages are languages *about* the language, not what is *outside* the language. a formal specification language could be termed a meta language, as it refers to constructions of the language for development, more typical are languages pertaining mainly to the grammar. this opposes your interpretation that meta means outside of a formal system. I don't think its an active area though so the terminology is still open. I've centered on a language for my formal distributed computer design. I call it F.M.L. Formal Markup Language. The basis is that the mechanism our mind uses to give illusion of tangible objects is paralleled to a hypertext document, that from a text string (html file) a graphical interactive page 'appears'. i.e. from sequences of neural firing patterns our imagery appears, as a web page appears from a sequential markup language. So the base domain is a grahical file editor, I would initially need to formalise the operations of programs such as dreamweaver that interact on the editable web page and calculate where to insert the code into the html file. The formal explictness of the domain is lost when the collection of html objects on the screen are referenced by association. e.g. even though a computer screen is purely digital, we don't refence objects by saying pixels 546,333 to 634,333, the domain becomes associative. There were programs like this using blocks that interpreted english like put the red block on the blue block and the program would manipulate a virtual crane. But that's not the goal its a parallel system, some way all the major algorithms must be incorporated into the dynamics of its operations. Primitive Turing machine like processes are the building primitives. Herc === Subject: Re: Deep Thoughts # 1: A new limitation to the human mind > Deep thoughts indeed ... pulled out from the depth of your ass, I > presume. > Another fine example of thoughtful intellectual analysis from the > dissenting side. :) > Charlie Volkstorf > Actually, Texas Jokes aside, your deep thoughts are interesting, but it > looks like you really don't want to defend the argument you present. Point # 1 (Definition of Mathematics): When asked for examples, I described in detail how other branches of science (physics and biology) fail my definition of Mathematics, and how branches of Mathematics (Set Theory and Recursion Theory) meet my definition. (I also pointed out that this was not so of other popular definitions of Mathematics, which instead apply to only subsets of Mathematics or science in general.) Point # 2 (Man-made Mathematics is limited to physical abstractions.): When asked how it applied to the natural numbers, real numbers, and the infinitude of the prime numbers, I explained each in detail. (I also applied it to the irrational numbers.) Point # 3 (Conclusion): When presented with a counterexample, I expressed logic to the contrary. Be careful when you use the old trick of making unverifiable (and irrefutable) claims such as one's mental state, as actions can in fact suggest the truth. (It is also irrelevant.) Also note another response: You see Ghosty, Charlie supports his claims! > So far , it seems, you are discarding any dissention as moronic ravings. When did I say (quote, please) anything like that? What would be an appropriate response to a story about hillbillies driving off a bridge or off-color references to human anatomy? > Actually, I judge the dissention here to be a plea for something generally > deemed cogent and supported by documentation. How do you go from hillbillies and human anatomy to appeals for cognizance and documentation? (I did respond to actual appeals for details, as I describe above.) > Your definition is unacceptable to many, the minor premise is a vague > generality, and seems to be a fiat of one school of thought, widely > disputed. Which school of thought is that and where is it disputed (citation, please)? > By declaring all dissent as moronic ( if that's which you have done?) you > give a peculiar brand of hubris and self-flagellation. Again, where did I degrade any response? I responded to coherent questions in detail, and politely ignored those which had no Mathematical content (and I ask above what would be a more appropriate response.) > There won't be much exchange > or broadening if you insist ( in a fundamentalist manner) that everyone you > invite into your world must accept your beliefs. explanations as well. in one post, and Is there an alternate explanation? in another, thus inviting alternate lines of reasoning. > The best I can do for you, seeing that I am not a distinguished philosopher > or a published expert on Predicate Calculus, but a net gadfly of moderate > to low exposure,is to deign the SLACKrament of Chillout by my Authority.. I think you've lost me on that one. Charlie Volkstorf Cambridge, MA > Pope Bobby II > 69th Clench of the Stark Fist of Removal > Reformed Church of the Subgenius === Subject: Re: Two coin flip/ clarification for C Bond > I'll toss in my 2 cents. > We have prior agreement, even with Dr. Ullrich that the following two > statements are the same: > > Two coins were flipped and at least one is a head. > Two coins were flipped and at least one is a tail. > Obviously they are different, but if you are talking about > probabilities, you can substitute head for tail and tail for > head everywhere in a problem and generate an equivalent problem. We can be assured that with HH, the at least one is a heads statement was made, and with TT, the at least one is a tails statement was made. > > If you agree to here, then, flip two identical coins. You can make > neither statement without some way inspecting at least one coin. > > Inspect one and only one, if it's a head, make the heads statement, if > it's a tail, make the tails statement. > Neither of your statements refers to inspecting one and only one coin. We're talking about re iterating one, two coin toss. Two coins were tossed, one time. We don't do probability on a one time event, we have to iterate the event thousands, or millions of times. We have to iterate the event which has already happened. We have to be sure not to alter the event. All we know about the event we learned from the problem statement. All we know is what it told us. We can't go by what it suggests. With HH, the at least one is a head stateement is true. With TT, the at least one is a tail statement is true. With HT, or TH, either statement would be true. We know that our statement is true, or we are answering a different question. > The statement above corresponds to: > Two coins were flipped and the first one is a head. or > Two coins were flipped and the second one is a head. > Of course, you can substitute tail for head without affecting the > nature of the problem. Therefore we can iterate jillions of times. With HH, the heads statement was made, with TT, the tails statement was made. With HT, and TH, one statement, or the other was made. As our number of iterations gets large, the ratio of heads statements to tails statements should approach equality. > > The answer to the corresponding question is 1/2. > Only after changing what you mean by at least one is a head to a > specific coin (1st or 2nd) is a head. When our heads statements to tails statements approach equality, the the answer to any individual question is 1/2. You can let them suggest whatever you wish. A bettor betting for one of each will win half the time, a bettor betting for two of a kind will win half the time. That's a 1/2 probability. > > Define a repeatable coin flip sequence which has correct answer 1/3. > You will have to inspect both coins. > Which is what at least one suggests. You can let it suggest whatever you wish, you will have to designate a 'color' prior to the flip and flip around it. You have to 'look' at both coins to flip around it, but you must do it, and can't just 'suggest it. Someway, somehow, prior to the toss, tell the bettor's that you'll reflip on the designated outcome. You can reflip anytime, prior to the 'look'. That won't change the odds. Toss the coins, say, two coins were tossed, at least one is a, you can go no farther, prior to the 'look'. Look....Then, it's one of four, you can't change it. With HH, say heads, with TT, say tails, With HT, or TH, say one, or the other. No matter what it 'suggests', the answer is 1/2. > > There is no evidence in our question that the writer of the question > knew the outcome of both coins. Imagine that there is always two > bettors. One bets for two of the kind, the other covers the bet. On > HH, for the losing bettor to pay two to one, there must be some > assurance, in the statement, to the losing bettor, that there would > have been some kind of demur at TT. > Here the distinction seems to come down to whether you are looking at a > sequence of flips, or simultaneous flips of distinguishable coins. Is > it not reasonable for me to take a dime and nickel, flip them (concealed > from you) and announce that at least one is a head after inspecting > both? Can you reasonably assume that I will be consistent as to whether > I am informing you about the dime or the nickel? I can reasonably assume that the problem statement is true. Maybe the flipper lied to us is conjecture, and assumes a different question. Our question, as written, is a true statement, or it's asking one question, but answering another. Eldon > > example: Two coins were flipped until at least one is a head. The > 'until' tells the losing bettor to pay double on HH. It protects her > from also having to pay double on TT. > > Eldon:) > This is a completely different type of scenario. How you work out the > unmentioned payments is up to the gamblers, but I would certainly hope > they can work out a system of payments for 0, 1, or 2 heads that > satisfies both.