mm-4349 === Subject: JSH: Really weird story, not your faults There really is NO reason at this point for surrogate factoring to not be taken seriously as potentially a powerful factoring technique. And I don't rely on Usenet alone to get my ideas out. I HAVE contacted the NSA repeatedly and now am too scared to keep bugging them so I haven't sent them the latest, but still... I have contacted mathematicians around the world and some people that I have reason to believe are connected with the intelligence community. others. Sometimes I think that there are people who really, really, really want me to post factorizations of RSA Challenge numbers--which I have not yet achieved--and collapse confidence overnight in the current system, but why? Or do they really think I can't do it? Where are the people who are supposed to protect us? Why would world leaders sit back and let this happen? Why? Is our world really this stupid? James Harris === Subject: Re: Really weird story, not your faults > There really is NO reason at this point for surrogate factoring to not > be taken seriously as potentially a powerful factoring technique. Reason #1, it is * TOO SLOW *, Reason #2 *worser than random guessing*. Reason #3 *no way to figure out n, k, T, x, y, z, p_1, p_2...etc* Reason #4 you have admitted before, it will never work And I don't rely on Usenet alone to get my ideas out. But WE control your ideas to the world. > I HAVE contacted the NSA repeatedly and now am too scared to keep > bugging them so I haven't sent them the latest, but still... NSA dosen't need to factor anything, they have other ways of encryption/decryption > I have contacted mathematicians around the world and some people that > I have reason to believe are connected with the intelligence > community. BS. They would never tell you that, or anybody else. others. and they rejected it, you. > Sometimes I think that there are people who really, really, really > want me to post factorizations of RSA Challenge numbers--which I have > not yet achieved It is the only way you will ever prove you have achieved something. >--and collapse confidence overnight in the current > system, but why? WRONG - they do not use factoring in the current system Or do they really think I can't do it? Show that you can't ! Where are the people who are supposed to protect us? Right here, We are protecting the world from your disfunctional lies. > Why would world leaders sit back and let this happen? Why? Is our > world really this stupid? No, just you, JSH. You think you have a high IQ, but your math indicates you are stuck in 7th grade with a math IQ of 70. Sorry, but you just suck so badly at math. And it is your fault, you haven't taken a math course. > James Harris > === Subject: Re: JSH: Really weird story, not your faults > I have contacted [...] some people that I have reason to believe are > connected with the intelligence community. This is the most depressing sentence you have ever written. You really do need to seek help. -- Jesse F. Hughes The sole cause of all human misery is the inability of people to sit quietly in their rooms. -- Blaise Pascal === Subject: Re: JSH: Really weird story, not your faults > There really is NO reason at this point for surrogate factoring to not > be taken seriously as potentially a powerful factoring technique. And I don't rely on Usenet alone to get my ideas out. I HAVE contacted the NSA repeatedly and now am too scared to keep > bugging them so I haven't sent them the latest, but still... That's okay. I'll be going to a mathematics meeting, and they will contact me there. What message do you want me to pass on? > I have contacted mathematicians around the world and some people that > I have reason to believe are connected with the intelligence > community. Anyone take you seriously? > others. Sometimes I think that there are people who really, really, really > want me to post factorizations of RSA Challenge numbers--which I have > not yet achieved--and collapse confidence overnight in the current > system, but why? That alone won't do it. They'd simply move to larger numbers. > Or do they really think I can't do it? Where are the people who are supposed to protect us? The politicians? They're working for corporations. > Why would world leaders sit back and let this happen? Why? Is our > world really this stupid? Portions of it. === Subject: Re: JSH: Really weird story, not your faults > There really is NO reason at this point for surrogate factoring to not > be taken seriously as potentially a powerful factoring technique. And I don't rely on Usenet alone to get my ideas out. I HAVE contacted the NSA repeatedly and now am too scared to keep > bugging them so I haven't sent them the latest, but still... I have contacted mathematicians around the world and some people that > I have reason to believe are connected with the intelligence > community. others. Sometimes I think that there are people who really, really, really > want me to post factorizations of RSA Challenge numbers--which I have > not yet achieved--and collapse confidence overnight in the current > system, but why? Or do they really think I can't do it? Where are the people who are supposed to protect us? Why would world leaders sit back and let this happen? Why? Is our > world really this stupid? James Harris I *know* you can't do it. There's a personal check here for $10k waiting if you factor (an as-yet unfactored) challenge number. === Subject: Re: JSH: Really weird story, not your faults [JSH ] > ... > Sometimes I think that there are people who really, really, really > want me to post factorizations of RSA Challenge numbers--which I have > not yet achieved--and collapse confidence overnight in the current > system, but why? Because people genuinely interested in factoring would /genuinely love/ to see a method that factored RSA-class composites quickly. How stupid can you be not to realize this? The idea that financial ruin would follow is just a fantasy, held by you and you alone, and used by use as a transparently lame excuse for not producing any results. > Or do they really think I can't do it? Duh. > ... === Subject: Re: JSH: Really weird story, not your faults > There really is NO reason at this point for surrogate factoring to not > be taken seriously as potentially a powerful factoring technique. And I don't rely on Usenet alone to get my ideas out. I HAVE contacted the NSA repeatedly and now am too scared to keep > bugging them so I haven't sent them the latest, but still... I have contacted mathematicians around the world and some people that > I have reason to believe are connected with the intelligence > community. others. Sometimes I think that there are people who really, really, really > want me to post factorizations of RSA Challenge numbers--which I have > not yet achieved--and collapse confidence overnight in the current > system, but why? Or do they really think I can't do it? Yeah, we really think you can't do it. Where are the people who are supposed to protect us? There aren't any. The role of the police is to investigate crime, not prevent it. Why would world leaders sit back and let this happen? Why? Is our > world really this stupid? You are. That's a start. James Harris === Subject: Fundamentals of differential equations does anyone have the solution manual for fundamentals of differential equations forth edition Paul === Subject: Re: Coding of ordered pairs Nntp-Posting-Host: hera.cwi.nl ... > I've been sloppy, rather thinking about sqrt(4) = 1.999999999999.. and > sqrt(25) = 4.999999999999.. in a computer. > > OK, if you want to have your computer calculate the true value of > trunc(sqrt(integer)), simply use trunc(sqrt(x + .5)) > This is safer than trunc(sqrt(x) + eps) -- your sqrt should not > be so bad that it can produce sqrt(n^2+.5) < n for reasonable values > of n This is actually a ridiculous discussion. If your computer computes sqrt(4) = 1.999999999.., you better upgrade the software. Getting the correct answer is trivial. And what a computer does in computing is part of mathematics, but it is in the field of numerical mathematics. So if you want to allow such discrepancies, you have to look what that field has to say about it. In mathematics, sqrt(4) = 2, period. Or is Han also considering computers that have no representation for 0.0? (Yes, they *did* exist, so on those computers 2 * 0.0 != 0.0.) Or computers where a + b is not necessarily b + a? And I can go on. On the other hand, trunc(sqrt(n^2 + .5)) does not necessarily work... Especially if .5 is much smaller than n^2. But let me give some actual facts (long ago I did analyse them with truncating arithmetic). Given floating point numbers with k bits of precision, and assume truncating arithmetic and N-R for the calculation of the square root. In the range (1 - 2^(-2k), 1) there are exactly *two* numbers where N-R does not converge. In all other cases N-R converges to the properly truncated square root. In those two cases N-R oscillates between two numbers, one too small and one too large. But neither of the two cases is the square of a number. I have good reasons to suspect that something similar happens with rounding arithmetic. And beware a bit about Cody and Waite. Especially for the square root their algorithm is slightly wrong. That explains why IBM's software got better results than their algorithm when implemented on that machine. -- 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: Coding of ordered pairs > ... > > I've been sloppy, rather thinking about sqrt(4) = 1.999999999999.. and > > sqrt(25) = 4.999999999999.. in a computer. > > > > OK, if you want to have your computer calculate the true value of > > trunc(sqrt(integer)), simply use trunc(sqrt(x + .5)) > > This is safer than trunc(sqrt(x) + eps) -- your sqrt should not > > be so bad that it can produce sqrt(n^2+.5) < n for reasonable values > > of n > > This is actually a ridiculous discussion. If your computer computes > sqrt(4) = 1.999999999.., you better upgrade the software. Getting the > correct answer is trivial. And what a computer does in computing is > part of mathematics, but it is in the field of numerical mathematics. So > if you want to allow such discrepancies, you have to look what that field > has to say about it. In mathematics, sqrt(4) = 2, period. Or is Han > also considering computers that have no representation for 0.0? (Yes, > they *did* exist, so on those computers 2 * 0.0 != 0.0.) Or computers > where a + b is not necessarily b + a? And I can go on. There is _no_ problem with the floating point values as such - they are accurate enough for most purposes. But a problem could occur with the _truncation_ which is required in order to implement Daryl McCullough's algorithm. If the (numerical) logarithm would give 1.99999999, instead of 2.000000001, then Trunc(ln()) would give 1 instead of 2, which is a significantly different outcome. I didn't actually check out if it can happen on my computer, with my favorite programming language. It's just a matter of precaution. Any suggestion better than this is appreciated. > On the other hand, trunc(sqrt(n^2 + .5)) does not necessarily work... > Especially if .5 is much smaller than n^2. But let me give some > actual facts (long ago I did analyse them with truncating arithmetic). > Given floating point numbers with k bits of precision, and assume > truncating arithmetic and N-R for the calculation of the square root. > In the range (1 - 2^(-2k), 1) there are exactly *two* numbers where > N-R does not converge. In all other cases N-R converges to the > properly truncated square root. In those two cases N-R oscillates > between two numbers, one too small and one too large. But neither of > the two cases is the square of a number. I have good reasons to suspect > that something similar happens with rounding arithmetic. > > And beware a bit about Cody and Waite. Especially for the square root > their algorithm is slightly wrong. That explains why IBM's software got > better results than their algorithm when implemented on that machine. Han de Bruijn === Subject: Solution Manual Interested in Engineering Mechanics Dynamics (11ed) R.C. Hibbeler === Subject: Re: TOMMY1729 STRIKES AGAIN !!! Nntp-Posting-Host: hera.cwi.nl ... > The rate of growing has nothing to do with it. Try > x^(1/4). > > > > easy g(x)= x^(1/2) > > Indeed. The rate of growing has nothing to do with > the difficulty. > > of course not in this case ... > > however > > x^23 - x^5 + x^3 -3*x^2 + x = g(g(x)) seems a lot harder than the cases > i investigated ?? The rate of growing has nothing to do with it. x^144 grows much faster and *has* an easy solution. > so wheiter or not the rate of growing is involved in the difficulty is > in the realms of the conjectures rather than the known. No. It is easy to prove. Given a function f(x) with a particular fast grow that is not easy to decompose as g(g(x)) = f(x). I would state that f(f(x)) is growing much faster, but has an easy solution. > however the growing rate is probably not the dominant factor in the > level of difficulty... It is no factor at all. > length of expression for instance is another parameter... I do not think so. It is easy to formulate polynomials of arbitrary length for which it is simple. Did you look at the reference I gave? There is was stated that for the quadratic polynomial a.x^2 + b.x + c it was proven that under some conditions on a, b and c, the problem could be solved and under other conditions it could not be solved. (After a week I disremember the condition, but it was something like b^2 - 4ac > 1.) -- 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: Two results of set geometry Nntp-Posting-Host: hera.cwi.nl ... > But a stronger result is that > > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. Yup. Because |{2, 4, 6, ..., 2n}| = n. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, and hence, there is a greater natural number > than > |{2, 4, 6, ...}|. Eh? Why? The first is about sets with a last element, the second not. -- 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: Two results of set geometry > But a stronger result is that > > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. Yup. Because |{2, 4, 6, ..., 2n}| = n. > So there is always, i.e. for any n, a greater natural number than > > |{2, 4, 6, ..., 2n}|, and hence, there is a greater natural number > > than > > |{2, 4, 6, ...}|. Eh? Why? The first is about sets with a last element, the second not. All are about sets with only natural numbers as elements. And lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1 is about an infinite set. === Subject: Re: Two results of set geometry Nntp-Posting-Host: hera.cwi.nl > > I only wanted to have you confess that there are countable sets which > > can not be put in a constructable bijection with N. > > Darn. I thought you *did* read what I write? Apparently not. When did > I write that Turings computable numbers where countable, but the list > of them could not be computed? > > We are not talking about Turings numbers but about finitely definable > numbers, i.e., numbers which can be defined. No, indeed, you do not read what I write, not even what you write yourself. Read your quote above: I only wanted to have you confess that there are countable sets which can not be put in a constructable bijection with N. And I have stated again and again, that you do *not* mean constructable but finitely definable, and that the latter term is pretty ill-defined, and that I have shown that such was indeed not possible. So what confession did you want? > You seem to have misunderstood. There is an injection from the set of > paths to the set of nodes and hence to the natural numbers. You state so. Show it. > Take any > path, pi (or what ever you like). Map it on a node which is crossed by > the path at level n (any natural number n can be chosen to start > with). Those paths which down to level n run the same way, will > subsequently get mapped on nodes below level n. The existence of such > nodes is guaranteed by the fact that those paths must distinguish > themselves by infinitely many nodes from pi and from each other. This does *not* show an injection. The simple reason being that you do not give a method to map a particular path to a particular node. Suppose I map pi to the root node. To what nodes are the other paths mapped? Your method lacks precision. When you define an injection you should be able to tell (given a particular path) to which node it maps without reference to other paths. > So the proof of countability of the set of paths (the proof of an > injection from the paths to N, not a construction) is complete. Not at all, because no injection is shown. > > Choose any paths P1. Map it on one of its nodes, K1, at the n-th leven > > (n e N). Choose any other path P2. Map it on a node, K2, which belongs > > to level n+1. Choose any other path P3. Map it on a node, K3, which > > belongs to level n+2. Continue. In this way every path is mapped on a > > node which no other path is mapped on. (If you believe that a path is > > missing in the bijection, then map it on a node at a level not yet > > used.) > > Ok. I map 0.1000... to its node at the first level. I map 0.11000... to > its node at the second level, etc. Where should I map 0.0111111... to? > All levels are used. So your methodology does not work. > > I assume that you meant 0.111... No, I meant 0.0111... > Of course you *can* choose a mapping which does not work. But it is *your* task to show that there is a mapping that *does* work. > To make it > simple map 1/pi and 1/sqrt(2) and 1/e all on the root node. Yes, that does (trivially) not give an injection. > But you > need not choose such a mapping. Therefore your argument does not prove > anything. This is proven by rational mapping. You remember? A simple > mathematical proof that every path has a node of its own (1/2 + 1/4 + > 1/8 + ... = 1). *What* proof? Please show a mapping that *does* work. > > As the tree contains at least one path for each real number, > > none of them is missing --- in particular, there is no chance to > > construct a diagonal number. > > No, as the construction does not work. It is based (in principle) on > circular reasoning. > > The argument that two existing numbers must be distinct by two > different nodes remains. And there is no possibility to disprove the > formula of the geometrical sum. That has nothing to do with it. > Obviously this is not a working procedure. It is clear nonsense. > Suppose I have the paths 0.000..., 0.1000..., 0.11000... (p1, p2, p3 > respectively). When I consider the pair p1 and p2, I should map > p1 to the node 0.0 and p2 to the node 0.1; when I consider the pair > p1 and p3, I should map p1 to the node 0.0 and p3 to the node 0.1, but > when I consider the pair p2 and p3 I should map p2 to the node 0.10 > and p3 to the node 0.11. So your construction is ambiguous. > > You can start with any desired node at any desired level. Two > different paths go through a last common node and then go through > infinitely many different nodes. These can be used. Do you really > think that your arbitrary constructions prove anything? Do you really think that your ramblings prove anything? I used the mapping your stated and did show that it was ambiguous. Pray, for once, show a *working* procedure. > Cantor's proof employed the diagonal number. This cannot be done in > the tree. Everything else is insufficient - and wrong as the simple > mathematical proof shows that every path has a node of its own, (1/2 + > 1/4 + 1/8 + ... = 1). So you can't succeed. What kind of proof is this? > > If *in the infinite tree* the real numbers cannot be > > individualized and distinguished from any other real number, then this > > would also be impossible in Cantor's list. > > But they can. > > Therefore every path can be mapped on a node of its own. Proof, please. > > Don't forget that the tree is infinite. It is as infinite as Cantor's > > list. It is assumed that Cantor's diagonal number differs from every > > number of the list. In this same way every path of my tree differs > > from every other paths. And for each of these paths there is a pair of > > nodes where they differ. > > The last statement is clear nonsense. Do you see the nonsense? > > You are in error. Oh. > The > correct statement is that for each pair of paths there is a node where > they differ. > > Each pair of paths has a last node in common. After that there is a > pair of nodes where they differ for the first time. of nodes where they differ? How can you use singular each of these paths and later on plural they? > > But clearly it is not possible that only one path goes in > > because of necessity at least two paths come out. At each > > node the number of incoming paths is the same > > as the number of outgoing paths. That is *your* construction. > > > > That is your wrong interpretation of an infinite tree. > > O. So some of your paths start at intermediate nodes? How is it otherwise > possible that there are more outgoing than incoming paths? > > At every level and at every node there are more outgoing *separated* > path bunches than incoming *separated* path bunches. So, pray, what was wrong about what I did write? Now you change back again to path bunch. What is your definition of a path bunch? Please take one of the two conflicting definitions from your book. > > It becomes not separated at any finite level. But if the reals do not > > even differ in their infinite sequence of digits, then they are not > > different numbers. > > You misread. There is no node where a real number is separated from all > other real numbers. But for each pair of real numbers there is a node > where they do separate. > > And exactly that one can be used for that path which up to then has > not yet got its node. But at each node there are infinitely many paths that separate from the other paths. So which of these paths uses this node? -- 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: Two results of set geometry but finitely definable, and that the latter term is pretty ill-defined, > and that I have shown that such was indeed not possible. So what > confession did you want? The confession that for the set of decimal representations of definable numbers no bijection with N can be defined. > You seem to have misunderstood. There is an injection from the set of > > paths to the set of nodes and hence to the natural numbers. You state so. Show it. I have done it so many times. If you can't understand, it is a waste of time to repeat it. ... But for the lurkers: Lev. 0 0. / 1 0 1 / / 2 0 1 0 1 / / / / ........ Here you see the first levels (number 0 to 2) of the infinite binary tree which contains all real numbers of the interval [0, 1] as infinite path bunches, i.e., as sets of infinite paths which are separated from any path outside of this set. As there are no two infinite path with exactly the same set of nodes, an infinite path bunch is a single path. Down to level n there are 2^(n+1) -1 nodes and 2^(n+1) separated path bunches. Therefore the ratio of path bunches to nodes is 1/(1 -1/2^(n +1)) at level n. This keeps on without end, which is the definition of infinity. Therefore, in the infinite binary tree, for n --> oo, the ratio of infinite path bunches to nodes is 1. *What* proof? Please show a mapping that *does* work. Show a mapping of the decimal representations of all definable numbers on N. > Cantor's proof employed the diagonal number. This cannot be done in > > the tree. Everything else is insufficient - and wrong as the simple > > mathematical proof shows that every path has a node of its own, (1/2 + > > 1/4 + 1/8 + ... = 1). So you can't succeed. What kind of proof is this? Every path bunch in the tree with two levels 1/2 node from level 2, 1/4 node from level 1 and 1/8 node from level 0, that makes 1/2 + 1/4 + 1/8 = 7/8 nodes. You can easily calculate the same for a tree with 3, 4 or 5 levels, respectively, 1/2 + 1/4 + 1/8 + 1/16 1/2 + 1/4 + 1/8 + 1/16 + 1/32 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64. For the infinite tree we get for every infinite path bunch 1/2 + 1/4 + 1/8 + ... = 1 node of its own. > > If *in the infinite tree* the real numbers cannot be > > > individualized and distinguished from any other real number, then this > > > would also be impossible in Cantor's list. > > But they can. > > Therefore every path can be mapped on a node of its own. Proof, please. See above. > Now you change back > again to path bunch. What is your definition of a path bunch? See above: A path bunche is a set of infinite paths which are separated from any path outside of this set. How many paths belong to a path bunch depends on the level n which you refer to. In the limit n --> oo every path bunch is a single path, because paths, which have not yet separated after infinitely many nodes, will not separate at all and, hence, are the same path. > Please take one of the two conflicting definitions from your book. In the limit n --> oo every path bunch is a single path. Therefore there is no conflict. PS: Here I have a riddle for you and those readers who can understand German. I cannot translate the letter because when translating it further mistakes could be introduced unintentionally. The following German letter from Cantor to Dedekind of 7. Dec. 1873 contains three mistakes (introduced by me in the verbal text, errors in the formulae are not intended). Who can find them? Who is able to state why these mistakes are no mistakes in fact? (The letter has been printed, e.g., by Meschkowski, Nilson: Georg Cantor Briefe, Springer, 1991) In den letzten Tagen habe ich die Zeit gehabt, etwas nachhaltiger meine Ihnen gegen?ber ausgesprochene Vermutung zu verfolgen; erst heute glaube ich mit der Sache fertig geworden zu sein; sollte ich mich jedoch t?uschen, so finde ich gewiss keinen nachsichtigeren Beurtheiler, als Sie. Ich nehme mir also die Freiheit, Ihrem Urtheile zu unterbreiten, was soeben in der Unvollkommenheit des ersten Conceptes zu Papier gebracht ist. Man nehme an, es k?nnten alle positiven definierbaren Zahlen w < 1 in die Reihe gebracht werden: (I) w 1, w 2, w 3, .. , w n, ... Auf w 1 folgend sei w a das n?chst gr?ssere Glied, auf dieses folgend w b das n?chst gr?ssere, u. s. f. Man setze: w 1 = w 11, w a = w 12, w b = w 13 u. s. f. und hebe aus (I) die unendliche Reihe aus: w 11, w 12, w 13, ..., w 1n, ... In der ?brig bleibenden Reihe werde das erste Glied mit w 21, das n?chst folgende gr?ssere mit w 22 bezeichnet, u. s. f. so hebe man die zweite Reihe aus: w 21, w 22, w 23, ..., w 2n, ... Wird diese Betrachtung fortgesetzt, so erkennt man dass die Reihe (I) sich in die unendlich vielen zerlegen l??t: (1) w 11, w 12, w 13, ..., w 1n, ... (2) w 21, w 22, w 23, ..., w 2n, ... (3) w 31, w 32, w 33, ..., w 3n, ... ... .......... in jeder von ihnen wachsen aber die Glieder fortw?hrend von links nach rechts zu; es ist: w kl < w kl+1 Man nehme nun ein Intervall (p... q) so an, dass kein Glied der Reihe (1) in ihm liegt; also etwa innerhalb (w 11, ..., w 12); nun k?nnten auch etwa s?mtliche Glieder der zweiten Reihe, oder der dritten ausserhalb (p... q) liegen; es muss jedoch einmal eine Reihe kommen, ich will sagen die kte, bei welcher nicht alle Glieder ausserhalb (p... q) liegen; (denn sonst w?rden die innerhalb (p... q) liegenden Zahlen nicht in (I) enthalten sein, gegen die Voraussetzung); dann kann man ein Intervall (p'... q') innerhalb (p... q) fixieren, so dass die Glieder der kten Reihe alle au?erhalb desselben liegen; von selbst verh?lt sich dann (p'... q') in gleicher Weise in Bezug auf die vorhergehenden Reihen; im weiteren Verlaufe muss jedoch eine k' te Reihe erscheinen, deren Glieder nicht s?mmtlich ausserhalb (p'... q') liegen und man nehme dann innerhalb (p'... q') ein drittes Intervall (p''... q'') an, so dass alle Glieder der k' ten Reihe ausserhalb desselben liegen. So sieht man, dass es m?glich ist eine unendliche Reihe von Intervallen zu bilden: (p... q), (p'... q'), (p''... q''), ... von denen jedes die folgenden einschliesst und die zu unsern Reihen (1), (2), (3), ... sich wie folgt verhalten: Die Glieder der 1ten, 2ten, k - 1ten Reihe liegen ausserhalb (p... q) Die Glieder der kten, ..., k' - 1ten Reihe liegen ausserhalb (p'... q') Die Glieder der k' ten, ..., k'' - 1ten Reihe liegen ausserhalb (p''... q'') .......... Es l??t sich nun stets wenigstens eine definierbare Zahl, ich will sie h nennen, denken, welche im Innern eines jeden dieser Intervalle liegt; von dieser Zahl h, welche offenbar >0, < 1, sieht man rasch, da? sie in keiner unserer Reihen (1), (2), ..., (n), enthalten sein kann. So w?rde man, von der Voraussetzung ausgehend, da? alle definierbaren Zahlen >0, < 1, in (I) enthalten seien, zu dem entgegengesetzten Resultate gelangt sein, da? eine bestimmte Zahl h nicht unter (1) zu finden sei; folglich ist die Voraussetzung eine unrichtige gewesen. So glaube ich schlie?lich zum Grund gekommen zu sein, weshalb sich der in meinen fr?heren Briefen mit (x) bezeichnete Inbegriff nicht dem mit (n) bezeichneten eindeutig zuordnen l??t. Mit den besten Gr??en Ihr ergebenster G. Cantor. === Subject: Re: Two results of set geometry If this method works, then so does the method of saying, at the nth step, write down the nth decimal place of the number. This is a bijection between the segment (0,1) and the countable direct product of the set {0,1,...,9}. Instead of trying to come up with new ways that are actually equivalent to the old (failed) ways, wouldn't it be simpler to actually understand the material? === Subject: Re: Two results of set geometry <20244873.1189694690403.JavaMail.jakarta@nitrogen.mathforum.org If this method works, then so does the method of saying, at the nth step, write down the nth decimal place of the number. This is a bijection between the segment (0,1) and the countable direct product of the set {0,1,...,9}. I don't know what you are refering to (you should not delete the topic), but, yes, real numbers are defined such that a decimal is attached to the n-th decimal place. Instead of trying to come up with new ways that are actually equivalent to the old (failed) ways, wouldn't it be simpler to actually understand the material? The new material is useless, because it has nothing to do with mathematics. If infinity is completely decoupled from finite sets then it is completely without value. === Subject: Re: Two results of set geometry Nntp-Posting-Host: hera.cwi.nl > Above you talk about the number of first indices of matrix elements, > where are the initial segments? > > Here they are: Darn. Can you give a clear answer to my questions? You talk about the number of first indices of matrix elements. Where in that sentence are the initial segments? > The contradiction evident in set theory is that the full column is > said to contain omega 1's and that there shall be omega natural > numbers but it is not so that there was one 1 for every natural > number. There is one. > What prevents the order-preserving bijection between omega and > omega??? Nothing. But that is *not* what I did state. I stated that there is no order-preserving bijection between the initial segments and N. That is something completely different. > But whatever, there is *no* order-preserving > bijection between the complete set of initial segments and the natural > numbers. > > What prevents the order-preserving bijection between omega and > omega??? The order of the complete set of initial segments is *not* omega. It is omega+1. The order of the set of *finite* initial segments is omega. There is one initial segment (the complete initial segment) that follows all finite initial segments. > And why is that impossible? I have not yet seen a coherent argument about > that. You argue (without proof) that there is some specific bijection, > but the bijection you insist that does exist does not exist. > > Of course, it does not exist. But an order preserving bijection > between omega and omega does exist. > Take the following matrix: > > 111... > 11 > 111 > 1111 > ... > There is an order preserving bijection between all inital segments of > first row and first columnm including the complete segments. There is > even a trijection including the initial segments of the diagonal. Yes, so what? But that is not an order-preserving bijection between omega and omega. It is an order-preserving bijection between omega+1 and omega+1. The order type of the ordered set of all initial segments is omega+1, not omega. > It is. But when talking about cardinalities I prefer to use aleph-0. > > Cantor, in his later years, used omega too for expressing > cardinalities. Further my argument is based upon order preservation > (which is not a problem for identical sets like omega and omega). But it is for non-identical sets, like one with order type omega (N) and one of order type omega+1 (the initial segments). > > No? There are not omega natural numbers? Cantor had this opinion. > > Where do I state that that is not the case? Moreover, there is a trivial > order-preserving bijection between the set of *finite* (non-empty) initial > segments with the natural numbers: {1, ..., n} <-> n. > > I see. The set of natural numbers, however, has no finite number of > elements, but omega elements: {1, 2, 3, ...}. Why again is omega not > in order preserving bijection with itself? It *is*. Why do you think it is not? The set of initial segments of natural numbers of N (when we include N itself) is omega+1, not omega. > > The ordinal number of the set of (all indices of) the post comma > > digits of pi is omega + 1? > > Where do I state such? Are the digits of pi initial segments? > > They are in order preserving bijection with the finite initial > segments. There are omega of them. But pi is not yet among them. Indeed. If you do not include the *complete* initial segment, pi is not among them, and there is a bijection. > But > we can consider each digit in bijection with the initial segment that > terminates with that digit. If we look at the set of initial segments > that we do construct this way, we get a set of *finite* initial segments. > pi itself is *not* in that set. > > You are confusing sets of indices with sets of initial segments. > > Because both are sadi to represent omega. No. The set of *finite* initial segments has order type omega. > > That you are confusing. You are actually talking about indices while > > every time only writing '1'. > > > > The 1's indicate which indices are considered. > > > I consider the typical pairs of indices of a matrix as used in > mathematics: > 11, > 21, 22 > 31, 32, 33 > ... > > *Not* considered are indices like 12, 13, ... In that case, *write* it that way. Not '1' to represent either '11', '21', '22', '31, '32', '33', or whatever. > > Moreover it is not clear. Is it a column > > of natural numbers, or a triangular matrix? > > > > Both does not exclude each other. > > Well, in my opinion there is quite some difference. > > It depends on the interpretation. There one must have some > flexibility. We had the same case with 0.111... which can stand for > 1/9 as well as 1/1 as well as omega, depending on the point of view. Yes, but your point of view should be consistent during a discussion, otherwise it becomes profoundly confusing. > > Of course. Empty segments are not considered (because they are no > > segments --- like empty sets are no sets). > > Your very personal form of set theory. > > A suspicion which I share with Zermelo, Fraenkel and Cantor (ZFC), > some people which have worked a lot in set theory. Ah, going back some seventy years again. We are talking *current* set theory. Moreover, I thought that one of the axioms of ZFC was that an empty set did exist? > > That is the case because every natural number n remains a natural > > number after a 1 has been added. If in the following order preserving > > bijection > > > > 1 > > 11 > > 111 > > ... > > > > every initial segments of the first column is mapped on a natural > > number, > > To what natural number is the complete first column mapped? Do you mean > an order-preserving mapping or not? > > Yes. So, to what natural number is the complete first column mapped? > Why are you always so extremely > unspecific? The ordered set of initial segments of the first column > has order type omega+1. > > It has as many elements as the number pi has digits behind the decimal > point. When you consider *as many*, yes, the cardinal number is the same, the ordinal number is not. > It is *not* in order-preserving bijection > with the set of natural numbers. > > I observed this about four years ago. And I asked myself for the > reason of omega not mapping on omega. Do you know a reason? Because the order type of the ordered set of initial segments is not omega but omega+1. So there is *no* order preserving bijection. The set of initial segments has a last element, which N does not have. But there *is* a bijection (the cardinal numbers are the same). > No, it does not. But that can be defined. > > As you see, the result is a contradiction. There are *wrong* > definitions like the set N does exist. What contradictions? What is *wrong* about that definition? > > And if I do not add an infinite line, then the matrix nevertheless > > has, according to you, all second indexes. This, however, is > > impossible: A matrix with all second indices has them in one row, > > because all second indicess of a triangular matrix can be found in one > > row. > > Proof, please. > > Proof: Those indices which exist in the rows 1 to n are all contained > in row n+1. That is no proof. > > There is an order preserving bijection between the second indices > > and the sequences of 1's in the rows of the matrix. Hence, if there > > are actually all second indices, then there must be a row with omega > > 1's. > > Proof, please. If all second indices do exist (and they do > > In fact? Even omega/2 does exist, and omega - 1 as well as > floor(sqrt(omega))? How do you *define* omega/2 and omega-1 and floor(sqrt(omega))? As long as I have not seen a definition, I do not know about the existence. > Yes, and yes. So what of it? That does *still* not show that there is > a order-preserving bijection between *all* initial segments and the > natural numbers. Only that there is a bijection between all *finite* > initial segments and the natural numbers. > > Look at my matrix above. There is an order preserving trijection > between omega and omega and omega. What should be the reason to > prevent an oreder preserving bijection between two identical sets? Because the sets are not identical. > > If you think there is, what natural number > > does the complete first column map to? > > > > None. Therefore, if the complete first column exists, i.e., if > > there are omega numbers, then one of them must be non-natural. > > Proof, please. > > There are omega 1's but not each of them has a natural number in its > row. I cannot follow. Each of the omega 1's corresponds with a natural number. > You are assuming a mapping does exist which does not > exist. And based on that assumption you claim the above. But your > assumption is simply wrong. There does *not* exist an order-preserving > bijection between the set of initial segments and the natural numbers. > Just as there is *no* order-preserving bijection between all initial > segments of N (if we include N) and N itself. The order-preserving > bijections only exist between N and the subset of *finite* initial > segments. > > Therefore it is simply wrong to state that there are omega natural > numbers. As I have been stating for several years now: in order to > have an actually infinite set of numbers, there must be an infinite > number. An actually infinite set of natural numbers does not exist. Yes, you are stating that again and again, but only stating, not proving. > > Hence, > > there is no bijection of N with omega > > There is. In Von Neumann terminology, omega = N. Because they are > identical, there is a trivial bijection. > > As trivial as my matrix? The complete set of initial segments has *not* order type omega. > Consider the equivalence classes of (finite and infinite sets) under > order-preserving bijections. That relation is an equivalence relation, > so there are equivalence classes. We call the equivalence classes > ordinal numbers. And moreover, why is omega not in trichotomy with > the natural number if, by its very definition, omega > each and every > natural number? > > Omega is greater than each and every natual number. That is obvious. > But omega is said to be the number of natural numbers. That is > impossible, because, you know, then for every 1 in the first column of > 1 > 11 > 111 > ... > there must be a natural number. There is. Show an 1 for which there is *not* a natural number. -- 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: Two results of set geometry So, to what natural number is the complete first column mapped? That question should be answered by those who insist that omega exists. > > And if I do not add an infinite line, then the matrix nevertheless > > > has, according to you, all second indexes. This, however, is > > > impossible: A matrix with all second indices has them in one row, > > > because all second indicess of a triangular matrix can be found in one > > > row. > > Proof, please. > > Proof: Those indices which exist in the rows 1 to n are all contained > > in row n+1. That is no proof. No? Then give a counter proof: Which indices cannot be found in one row? (Of course all is not an answer, because the existence of all is just in question.) > > There is an order preserving bijection between the second indices > > > and the sequences of 1's in the rows of the matrix. Hence, if there > > > are actually all second indices, then there must be a row with omega > > > 1's. > > Proof, please. If all second indices do exist (and they do) > > In fact? Even omega/2 does exist, and omega - 1 as well as > > floor(sqrt(omega))? How do you *define* omega/2 and omega-1 and floor(sqrt(omega))? As > long as I have not seen a definition, I do not know about the existence. An indication that not all do exist. The complete set of initial segments has *not* order type omega. The complete set of natural numbers has order type omega. The first column has order type omega. > > Omega is greater than each and every natual number. That is obvious. > > But omega is said to be the number of natural numbers. That is > > impossible, because, you know, then for every 1 in the first column of > > 1 > > 11 > > 111 > > ... > > there must be a natural number. There is. Show an 1 for which there is *not* a natural number. If we start with the mapping 1 <-> 1 1 1 <--> 1,1 ... then an order preserving bijection would alsways continue and also end with same number of elements. But there is is not w <--> w. === Subject: Re: Two results of set geometry has, according to you, all second indexes. Yes. This is clear to see. Just don't surpress the 0's M 1000 ... 11000 ... 111000 ... ... Each line has all second indexes. The only thing that changes is which indexes contain 1 and which indexes contain 0. M' 111... 11000... 111000... ... Note that M' has exactly the same second indexes as M. The only change is that some indexes that contain 0 in M contain 1 in M'. However, there is no second index added or taken away. - William Hughes === Subject: Re: Two results of set geometry ...if I do not add an infinite line, then the matrix nevertheless >has, according to you, all second indexes. Yes. This is clear to see. Just don't surpress the 0's M 1000 ... > 11000 ... > 111000 ... > ... Each line has all second indexes. The only thing that changes > is which indexes contain 1 and which indexes contain 0. M' 111... > 11000... > 111000... > ... Note that M' has exactly the same second indexes as M. > The only change is that some indexes that contain 0 in M > contain 1 in M'. However, there is no second index added > or taken away. You are right. But what is of interest here is the non-completed matrix, namely the attempted bijection of the *finite* natural numbers with omega (in form of the first column). Every finite sequence of 1's (representing a natural number) is followed by infinitely many 0's which consume some indices which are not available for 1's. Hence, the diagonal, consisting exclusively of 1's, cannot use all second indices. If I speak of the matrix I obviously mean that part which is occupied by 1's only. === Subject: Re: Two results of set geometry 1 11 111 ... > > What prevents the order-preserving bijection between omega and > > omega??? Nothing. Then look at the matrix. There are omega elements in the first colum. And there are omega natural numbers represented as finite seuquences of 1's in the rows. Nevertheless there is no order-preserving bijection because, as it starts with 1 <--> 1, ..., it must end with omega <--> omega which is not the case as there is no natural number omega (while the complete first column is omega). > > But whatever, there is *no* order-preserving > > bijection between the complete set of initial segments and the natural > > numbers. > > What prevents the order-preserving bijection between omega and > > omega??? The order of the complete set of initial segments is *not* omega. It > is omega+1. The order of the set of *finite* initial segments is omega. > There is one initial segment (the complete initial segment) that follows > all finite initial segments. So the order of the completed infinity, for instance of the digits in 0,123... is omega + 1? Strange. > And why is that impossible? I have not yet seen a coherent argument about > > that. You argue (without proof) that there is some specific bijection, > > but the bijection you insist that does exist does not exist. > > Of course, it does not exist. But an order preserving bijection > > between omega and omega does exist. > > Take the following matrix: > > 111... > > 11 > > 111 > > 1111 > > ... > > There is an order preserving bijection between all inital segments of > > first row and first columnm including the complete segments. There is > > even a trijection including the initial segments of the diagonal. Yes, so what? But that is not an order-preserving bijection between > omega and omega. It is an order-preserving bijection between omega+1 > and omega+1. The order type of the ordered set of all initial segments > is omega+1, not omega. In fact? The order type of the ordered set of all natural numbers {1,2,3,...} is omega + 1? Please note that the first column has not a last element. > It is. But when talking about cardinalities I prefer to use aleph-0. > > Cantor, in his later years, used omega too for expressing > > cardinalities. Further my argument is based upon order preservation > > (which is not a problem for identical sets like omega and omega). But it is for non-identical sets, like one with order type omega (N) > and one of order type omega+1 (the initial segments). So you would say that the complete initial segment 111... has order type omega + 1? I was of the opinion that order type omega belongs to sets with a last element. The first column, however, has not a last element. > No. The set of *finite* initial segments has order type omega. And the infinite column has order type omega too. Perhaps you will reflect somewhat longer on this question? === Subject: Re: Two results of set geometry Yes. However, the fact that an *infinite set* of steps > has infinite height, does not imply that any single step > has an infinite height. What is it then that causes infinite height? If a step is not the last step, the set of steps has a height greater > than that step. The set of steps has no height at all. It has a number of elements. No step is the last step, so the set of steps has greater > height than any step. The set of steps has no height at all. A set of green elements is not green! A set of large elements need not be large. For any natural number there is a step of that > height. The set of steps has height greater than any natural number. The set of natural numbers is not a natural number. The set N has omega elements but it has no numerical seize. The set of steps has infinite height. It has no height. > There is no last step. But all steps which are there, are finite. Therefore they do not cause > infinite height. This stands unrefuted. > Back to the subject. > It is not contradictory to say > The set A is complete. > There is no point at which the set A is completed. It is a contradiction. In order to say that A is complete, you must > know all elements of A. No. To say that A is a complete set of things of type X, you > have to know that given a thing of type X you can show that it > is an element of A. That is a tautology. The statement given a thing of type X implies that that the thing belongs to A. There is nothing to be shown. But even what you mean is not enough. You have to know that you are able to consider all elements of the set A simultaneously (simultaneous being together of elements is essential for sets and is impossible for some non-sets, as already Cantor knew). This is only possible if you know that there is no element left out. > You don't have to know all things of type X, > it is enough to know something that must be true of any thing of type > X. > Knowing all the elements of A is neither necessary nor sufficient. Why then is the set of all sets a problem? It is not contradictory to say The set A is complete. > There is no point at which the set A is completed. It is contradictory. But in order to reduce the argument to a rational level take my recent argument: Does the sequence S = 111... represent omega? Does it contain an infinite segment? Or do the of finite segments of S, namely 1, 11, 111, ..., represent omega? Or does counting stumble somewhat when counting over into the infinite? 1, 2, 3, ..., omega, omega, omega + 1, ... === Subject: Re: Two results of set geometry > Yes. However, the fact that an *infinite set* of steps > has infinite height, does not imply that any single step > has an infinite height. > What is it then that causes infinite height? If a step is not the last step, the set of steps has a height greater > than that step. The set of steps has no height at all. The height is the supremum of the natural numbers. If one assumes that this supremum does not exist, then the set of steps has no height and the fact that each step has finite height does not lead to a contradiction. If one assumes that this supremum, the ordinal which is the union of all ordinals that are elements of some natural number, exists, then the set of steps has a height, but the fact that each step has a finite height does not lead to a contradiction, as the set of steps has a height greater than any step. You choose the first assumption. No contradiction. I choose the second assuption. No contradiction. You claim that the second assumption leads to a contradiction. No. >It has a number of elements. > ITYM a number of steps. Yes, this number is omega. For each step there is an initial semgment that ends at that step. There are omega initial segments of steps that have a last step. There is one initial segment of steps that does not have a last step. There are omega+1 initial segments of steps. It is not contradictory to say > The set A is complete. > There is no point at which the set A is completed. > It is a contradiction. In order to say that A is complete, you must > know all elements of A. No. To say that A is a complete set of things of type X, you > have to know that given a thing of type X you can show that it > is an element of A. That is a tautology. The statement given a thing of type X implies > that that the thing belongs to A. Yes, the statement that The set of all natural numbers is the complete set of natural numbers is immediate. However, statements like the set of even natural numbers is not the complete set of natural numbers or the set of natural numbers that are prime or composite is the complete set of natural numbers are not. >There is nothing to be shown. But > even what you mean is not enough. You have to know that you are able > to consider all elements of the set A simultaneously. No. You can know that there is no element left out without considering all elements of the set A simultaneously. To show that the set of even natural numbers is not the complete set of natural numbers you need to find a natural number that is not even. To show that the set of natural numbers that are prime or composite is the complete set of natural numbers you need to show that: given a natural number x, x is either prime or composite. In neither case do you have to consider all elements of the set of natural numbers simultaneously. It is not contradictory to say The set A is complete. There is no point at which the set A is completed. Note that there are other ways to show that A is complete than finding a point at which the set A is completed. - William Hughes === Subject: Re: Two results of set geometry > Yes. However, the fact that an *infinite set* of steps > has infinite height, does not imply that any single step > has an infinite height. > What is it then that causes infinite height? > If a step is not the last step, the set of steps has a height greater > than that step. The set of steps has no height at all. The height is the > supremum of the natural numbers. If one assumes that this supremum does not exist, > then the set of steps has no height and the > fact that each step has finite height does not lead to > a contradiction. If one assumes that this supremum, > the ordinal which is the union of all ordinals that > are elements of some natural number, exists, then the > set of steps has a height, but the fact that each step > has a finite height does not lead to a contradiction, > as the set of steps has a height greater than any > step. The set of steps has steps but no height. A set of large elements need not be large. You choose the first assumption. > No contradiction. I choose the second assuption. > No contradiction. Wrong. The contradiction is 1, 2, 3, ..., omega, omega, omega + 1, ... It has a number of elements. ITYM a number of steps. Yes, this number is omega. > For each step there is an initial semgment that ends at > that step. There are omega initial segments of steps that > have a last step. There is one initial segment of steps that > does not have a last step. There are omega+1 initial segments > of steps. So 1,2,3, ... has ordinal number omega + 1? You have to know that you are able > to consider all elements of the set A simultaneously. No. You can know that there is no element left out > without considering all elements of the set A simultaneously. Only by assuming that. > To show that the set of even natural numbers is not the complete > set of natural numbers you need to find a natural number that is > not even. That does neither prove that you had a complete set of even numbers nor that you have a complete set of natural numbers. > To show that the set of natural numbers that are prime > or composite is the complete set of natural numbers you need to > show that: given a natural number x, x is either prime or composite. > In neither case do you have to consider all elements of the > set of natural numbers simultaneously. given a natural number x does not imply that you can consider all natural numbers. In fact, as you know, it is impossible to do so. It is not contradictory to say The set A is complete. > There is no point at which the set A is completed. The set A is complete means, there is some point at which we can say now we have all elements because all contrary opportunities are exhausted. There is no point at which the set A is completed is just the opposite. Note that there are other ways to show that A is complete than > finding a point at which the set A is completed. I would be interested to see the other ways. And I remain being interested in your answer to my question: Does the sequence S = 111... represent omega? Does it contain an infinite segment? Or do the finite segments of S, namely 1, 11, 111, ..., represent omega? Or does counting stumble somewhat when counting over into the infinite? 1, 2, 3, ..., omega, omega, omega + 1, ... === Subject: Re: Two results of set geometry > Yes. However, the fact that an *infinite set* of steps > has infinite height, does not imply that any single step > has an infinite height. > What is it then that causes infinite height? > If a step is not the last step, the set of steps has a height greater > than that step. > The set of steps has no height at all. The height is the > supremum of the natural numbers. If one assumes that this supremum does not exist, > then the set of steps has no height and the > fact that each step has finite height does not lead to > a contradiction. If one assumes that this supremum, > the ordinal which is the union of all ordinals that > are elements of some natural number, exists, then the > set of steps has a height, but the fact that each step > has a finite height does not lead to a contradiction, > as the set of steps has a height greater than any > step. The set of steps has steps but no height. No. The height of the set of steps is the supremum of set of heights of the steps. If one assumes that this supremum does not exist, then the set of steps has no height and the fact that each step has finite height does not lead to a contradiction. If one assumes that this supremum, the ordinal which is the union of all ordinals that are elements of some natural number, exists, then the set of steps has a height, but the fact that each step has a finite height does not lead to a contradiction, as the set of steps has a height greater than any step. > You choose the first assumption. > No contradiction. I choose the second assuption. > No contradiction. Wrong. The contradiction is 1, 2, 3, ..., omega, omega, omega + 1, ... No the set of heights is 1,2,3,.. It does not contain omega. Recall, a set may or may not contain its supremem. The fact that the supremum of the set of heights is omega, does not that the set of heights contains omega. >It has a number of elements. > ITYM a number of steps. Yes, this number is omega. > For each step there is an initial semgment that ends at > that step. There are omega initial segments of steps that > have a last step. There is one initial segment of steps that > does not have a last step. There are omega+1 initial segments > of steps. > So 1,2,3, ... has ordinal number omega + 1? No the ordinal number of 1,2,3,... is the number of finite initial segments in 1,2,3... It is not the number of initial segements in 1,2,3,... There is no point at which the set A is completed. The set A is complete means, there is some point at which we can say > now we have all elements because all contrary opportunities are > exhausted. No, this is the Volkensmuekenheim definition of complete. Ouside of Volkensmuekenheim saying that A is a complete set of X, means that there is no X that is not in A. Outside of Volkensmuekenheim it is not contradictory to say The set A is complete. There is no point at which the set A is completed. - William Hughes === Subject: Re: Two results of set geometry > The set of steps has steps but no height. No. The height of the set of steps is the supremum > of set of heights of the steps. What is the colour of a set which consists of infinitely many coloured elements? What is the age of a set which consists of infinitely many elements of different age? What is the melody of a set of infinitely many different melodies? If one assumes that this supremum does not exist, > then the set of steps has no height and the > fact that each step has finite height does not lead to > a contradiction. It shows that there is no infinite height. And by the diagonal argument (not Cantor's but mine) there is also no actually infinite width. If one assumes that this supremum, > the ordinal which is the union of all ordinals that > are elements of some natural number, exists, then the > set of steps has a height, Assuming it or not does not make the steps larger than before. > but the fact that each step > has a finite height does not lead to a contradiction, > as the set of steps has a height greater than any > step. Only one of both cases can be true. > You choose the first assumption. > No contradiction. > I choose the second assuption. > No contradiction. Wrong. The contradiction is 1, 2, 3, ..., omega, omega, omega + 1, ... No the set of heights is 1,2,3,.. > It does not contain omega. It contains omega number and omega finite intial segments. And it is an infinite initial segment. > Recall, a set may or may not contain its supremem. > The fact that the supremum of the set of heights is omega, > does not that the set of heights contains omega. And it does not mean that it is infinitely heigh. >It has a number of elements. > ITYM a number of steps. Yes, this number is omega. > For each step there is an initial semgment that ends at > that step. There are omega initial segments of steps that > have a last step. There is one initial segment of steps that > does not have a last step. There are omega+1 initial segments > of steps. > So 1,2,3, ... has ordinal number omega + 1? No the ordinal number of 1,2,3,... is the number > of finite initial segments in 1,2,3... > It is not the number > of initial segements in 1,2,3,... I am interested of in the number of elements in the sequences (of the rows) 1, 11, 111, ... and in the column 111... and why both, if being the same, do not biject (order preserving, of course). > The set A is complete means, there is some point at which we can say > now we have all elements because all contrary opportunities are > exhausted. A is a complete set of > X, means that there is no X that is not in A. That saying is not a proof. And it does not establish that omega X's exist. === Subject: Re: Two results of set geometry > The set of steps has steps but no height. No. The height of the set of steps is the supremum > of set of heights of the steps. What is the colour of a set which consists of infinitely many coloured > elements? not defined > What is the age of a set which consists of infinitely many elements of > different age? not defined > What is the melody of a set of infinitely many different melodies? not defined What is the height of a set of infinitely many steps? Defined to be the supremum of the set of heights of the steps. Let the set of heights of the steps be A. The supremum of the set A is omega. Note that the set A does not contain omega. A set need not contain its supremum. - William Hughes === Subject: Re: Two results of set geometry > The set of steps has steps but no height. > No. The height of the set of steps is the supremum > of set of heights of the steps. What is the colour of a set which consists of infinitely many coloured > elements? not defined What is the age of a set which consists of infinitely many elements of > different age? not defined What is the melody of a set of infinitely many different melodies? not defined What is the height of a set of infinitely many steps? Defined to be the supremum of the set of heights > of the steps. But this definition is wrong, namely leading to a contradiction. Let the set of heights of the steps be A. The supremum of the set A is omega. > Note that the set A does not contain omega. > A set need not contain its supremum. The steps of the staircase do not contain their supremum. But the set of differences is said to contain its supremum, namely there are omega differences of 1. And just this is the contradiction - the same as with my matrix: 1 11 111 ... The infinite first column is (said to be) the supremum 111... of its finite initial segments (as pi is said to have omega digits), but the set of rows does not contain the supremum of its finite sequences, namely an infinite sequence 111.... Hence the diagonal does not know whether it should contain the supremum or not. The set of first indices contains it, the set of second indices does not. Therefore the diagonal is the contradiction of the assumption that omega can be assumed. === Subject: Re: Two results of set geometry > The set of steps has steps but no height. > No. The height of the set of steps is the supremum > of set of heights of the steps. > What is the colour of a set which consists of infinitely many coloured > elements? not defined > What is the age of a set which consists of infinitely many elements of > different age? not defined > What is the melody of a set of infinitely many different melodies? not defined What is the height of a set of infinitely many steps? Defined to be the supremum of the set of heights > of the steps. But this definition is wrong, namely leading to a contradiction. No, this does not lead to a contradiction Let the set of heights of the steps be A. The supremum of the set A is omega. > Note that the set A does not contain omega. > A set need not contain its supremum. The steps of the staircase do not contain their supremum. But the set > of differences is said to contain its supremum, namely there are omega > differences of 1. Let the set of differences be D. The maximum of D is 1. The supremum of D is 1. The fact that D contains 1 is not a contradiction. The sum of all the elements of D is an infinite sum. It is the supremum of the set of partial sums. It is equal to omega. Neither D, nor the set of partial sums contains omega. >And just this is the contradiction - the same as > with my matrix: 1 > 11 > 111 > ... The infinite first column is (said to be) the supremum 111... of its > finite initial segments (as pi is said to have omega digits), Yes. Note that the first indices correspond to the finite segments. The supremem 111... is not a finite segment. So no first index corresponds to the supremum >but the > set of rows does not contain the supremum of its finite sequences, So no second index correspond to the supremum > namely an infinite sequence 111.... Hence the diagonal does not know > whether it should contain the supremum or not. No. The diagonal does not contain an index (first or second) that corresponds to the supremum, > The set of first > indices contains it, No it does not. The sequence 111... is neither an index, nor is it a finite sequence. > the set of second indices does not. There is a bijection between the set of first indices and the set of second indices, - William Hughes === Subject: Re: Two results of set geometry > The set of natural numbers is not a natural number. The set N has > omega elements but it has no numerical seize. The set of natural numbers, if it could exist, would be equivalent to another natural number. Therefore it cannot exist. (Tip, hint: there is a bijection between sets and naturals). Han de Bruijn === Subject: Re: Two results of set geometry omega elements but it has no numerical seize. The set of natural numbers, if it could exist, would be equivalent to > another natural number. Therefore it cannot exist. (Tip, hint: there > is a bijection between sets and naturals). > Note: this bijection exists in Bruijnland and perhaps in Wolkensmuekenheim. It does not exist in Orlovia. - William Hughes === Subject: Re: Two results of set geometry > The set of natural numbers, if it could exist, would be equivalent to > another natural number. Therefore it cannot exist. (Tip, hint: there > is a bijection between sets and naturals). Where might one find this bijection described? -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Two results of set geometry > >>The set of natural numbers, if it could exist, would be equivalent to >>another natural number. Therefore it cannot exist. (Tip, hint: there >>is a bijection between sets and naturals). > > Where might one find this bijection described? Here: Relevant quote: > There are two main ways to represent a set in a programming language: > > - as a sorted array > - as a bit map > > For example the set {2,7,5,1} = [1,2,5,7] = 10100110 , where the bits > are numbered from the right to the left. Let's consider the bimapped > way to represent sets more closely. > > Inside a binary computer, the natural number 0 corresponds with a bit > map consisting of nothing but zeroes: 00000000 . But this is also the > bitmap of the empty set {} . Therefore 0 = {} . The natural number 1 > corresponds with 00000001. Counting from the right, this is bit zero > set up, which corresponds to the set {0} . But 0 = {} hence 1 = {{}} . > The number 2 is 00000010 , which is the set {1} = {{{}}} = 2 . Number > 3 is 00000011 , which is the set {0,1} = {{},{{}}} = 3 . Number 4 is > 00000100 , which is the set {2} = {{{{}}}} = 4 . Number 5 is the set > 00000101 , which is the set {0,2} = {{},{{{}}}} = 5 . > > 10100110 = 2^7 + 2^5 + 2^2 + 2^1 = 128 + 32 + 4 + 2 = 166 . Where in > turn: 7 = 2^2 + 2^1 + 2^0 , 5 = 2^2 + 2^0 , 2 = 2^1 , 1 = 2^0 . Hence: > 166 = {1,2,5,7} = { {{}} , {{{}}} , {{},{{{}}}} , {{{{}}},{{}},{}} } . > > It is noted that the commas do not add anything to the unambiguity of > the formulas and can be deleted: {{{}}{{{}}}{{}{{{}}}}{{{{}}}{{}}{}}} Han de Bruijn === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi> On 12 Sep., 12:55, Aatu Koskensilta The set of natural numbers, if it could exist, would be equivalent to > another natural number. Therefore it cannot exist. (Tip, hint: there > is a bijection between sets and naturals). Where might one find this bijection described? > Here: 1 <--> {1} 2 <--> {1, 2} 3 <--> {1, 2, 3} ... and even the points ... in {1, 2, 3, ...} describe nothing but natural numbers. But a stronger result is that lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. So there is always, i.e. for any n, a greater natural number than |{2, 4, 6, ..., 2n}|, and hence, there is a greater natural number than |{2, 4, 6, ...}|. As this cannot be true {2, 4, 6, ...} cannot exist. === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi On 12 Sep., 12:55, Aatu Koskensilta The set of natural numbers, if it could exist, would be equivalent to > another natural number. Therefore it cannot exist. (Tip, hint: there > is a bijection between sets and naturals). Where might one find this bijection described? Here: 1 <--> {1} > 2 <--> {1, 2} > 3 <--> {1, 2, 3} > ... No one questions that { | n in the positive naturals} is a bijection. What would be more interesting is, given the existence of the set of natural numbers, a bijection between between w and some member of w (as mentioned by de Bruijn). > and even the points ... in {1, 2, 3, ...} describe nothing but > natural numbers. So? > But a stronger result is that lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. A stronger result? It's trivial that for n>0, card({x | x is an even number & 0 < x <=2n})/2n = 1/2. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, Yes. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}|. You give no rule of inference nor principle of logic or mathematics to justify any such hence. I am truly fascinated what goes on in your mind when think you have such a hence. I mean, do you EVER stop to think by what prinicples of logic whatsoever you arrive at such a hence? > As this cannot be true {2, 4, 6, ...} cannot exist. Since your hence above is without justification, you have not refuted that the set of even natural numbers greater than 0 exists. MoeBlee === Subject: Re: Two results of set geometry You are on the hook again? :-) F. -- E-mail: infosimple-linede === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > and even the points ... in {1, 2, 3, ...} describe nothing but > natural numbers. So? But a stronger result is that lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. A stronger result? It's trivial that for n>0, card({x | x is an even number & 0 < x > <=2n})/2n = 1/2. Discovery consists of seeing what everybody has seen and thinking what nobody has thought. (Szent Gy?rgi) So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, Yes. and hence, there is a greater natural number > than > |{2, 4, 6, ...}|. You give no rule of inference nor principle of logic or mathematics to > justify any such hence. Very simple: Every element of 2N is a natural number 2n. Every 2n is the last element of an ordered set of even natural numbers which obeys |{2, 4, 6, ..., 2n}| / 2n < 1 or There exists a natural number larger than |{2, 4, 6, ..., 2n}|. The set 2N of all even natural numbers is tied to these finite sets by being the supremum of the set of finite sets. Therefore it cannot be totally different from every finite set, even if the supremum is not taken by the set of finite sets. As this cannot be true {2, 4, 6, ...} cannot exist. Since your hence above is without justification, That should rather read: You cannot understand the proof. > you have not > refuted that the set of even natural numbers greater than 0 exists. It is impossible that 2N has a cardinal number larger than every natural number. === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > and even the points ... in {1, 2, 3, ...} describe nothing but > natural numbers. So? > But a stronger result is that > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. A stronger result? It's trivial that for n>0, card({x | x is an even number & 0 < x > <=2n})/2n = 1/2. Discovery consists of seeing what everybody has seen and thinking > what > nobody has thought. (Szent Gy?rgi) And just muttering any number of trivialities is not discovery. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, Yes. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}|. You give no rule of inference nor principle of logic or mathematics to > justify any such hence. Very simple: Every element of 2N is a natural number 2n. What is 2N in this context? Does 'N' stand for the set of natural numbers and 2N stand for cardinal multipiclation of 2*N? But then it is not the case that every member of 2N is 2n for some n. Oh, I see below that you mean 2N to mean the set of even numbers. Why don't you just say that initially then? Or do you mean every member of each of the sets {2 4 6 ... 2n} is 2n for some n? Then okay, again, obvious. > Every 2n is > the last element of an ordered set of even natural numbers which obeys > |{2, 4, 6, ..., 2n}| / 2n < 1 or There exists a natural number larger > than |{2, 4, 6, ..., 2n}|. Yes. So? > The set 2N of all even natural numbers is tied to these finite sets by > being the supremum of the set of finite sets. Not unless you defined supremum in this instance. A supremum is a least upper bound per a particular ORDERING. Say what ordering you have in mind. > Therefore it cannot be > totally different from every finite set, If by totally different you mean disjoint, then, yes, of course, the set of even numbers is not disjoint from any nonempty set of even numbers. We don't need your roundabout argument about whatever supremum to see that! > even if the supremum is not > taken by the set of finite sets. Whatever that means, so far all you've added is that the set of even numbers is not disjoint from any non-empty set of even numberes. Yeah, we know that. It doesn't entail that there exists a natural number greater than the cardinality of the set of even numbers. > As this cannot be true {2, 4, 6, ...} cannot exist. Since your hence above is without justification, That should rather read: You cannot understand the proof. No, it means you've not given any axiom, theorem, or rule of inference that justifies that hence. > you have not > refuted that the set of even natural numbers greater than 0 exists. It is impossible that 2N has a cardinal number larger than every > natural number. Just repeating your conclusion is not a proof of it. (Unless you are using your conclusion as a premise itself in a proof of it.) The following statement is correct: For every non-empty FINITE set of even numbers (each of which is greater than 0), there is a member of that set that is greater than the cardinality of that set. You've not given any correct argument that the above statement entails: For ANY (finite OR infinite) set of even numbers, there is an even number greater than the cardinality of that set. And you even resorted to some kind of principle of yours about unions. But you've not shown anywehre in mathematics that there is general principle that if property P holds for every fi then P holds for Ui fi. MoeBlee === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi But a stronger result is that lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, Yes. Note that every sequence of this form has a last element. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}| No. Note that this sequence does not have a last element. You cannot assume that something that is true for sequence with a last element must be true for sequences without a last element. - William Hughes === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > But a stronger result is that lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, Yes. Note that every sequence of this form has a last element. and hence, there is a greater natural number > than > |{2, 4, 6, ...}| No. Note that this sequence does not have a last element. Note, however, that it has only natural numbers as elements - and only that is important. > You cannot assume that something that is true for sequence > with a last element must be true for sequences without > a last element. But you can assume that every element of N is a natural number and nothing else. For any set consisting only of even natural numbers I proved that there is a nunber contained which is greater than the cardinal number of the set. === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > But a stronger result is that > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, Yes. Note that every sequence of this form has a last element. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}| No. Note that this sequence does not have a last element. Note, however, that it has only natural numbers as elements - and only > that is important. No, you can't pretend that the set does not have a greatest member. > You cannot assume that something that is true for sequence > with a last element must be true for sequences without > a last element. But you can assume that every element of N is a natural number and > nothing else. For any set consisting only of even natural numbers I > proved that there is a nunber contained which is greater than the > cardinal number of the set. No, for any FINITE set of even numbers (greater than 0), there is an even number in the set and greater than the cardinality of the set. MoeBlee === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > But a stronger result is that > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, > Yes. Note that every sequence of this form has a last element. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}| > No. Note that this sequence does not have a last element. Note, however, that it has only natural numbers as elements - and only > that is important. No, you can't pretend that the set does not have a greatest member. I do not pretend that the set has a greatest member. But important is that the set has only natural numbers as members. And for each of them my statement holds. > You cannot assume that something that is true for sequence > with a last element must be true for sequences without > a last element. But you can assume that every element of N is a natural number and > nothing else. For any set consisting only of even natural numbers I > proved that there is a nunber contained which is greater than the > cardinal number of the set. No, for any FINITE set of even numbers (greater than 0), there is an > even number in the set and greater than the cardinality of the set. The infinite set is nothing but the supremum of the set of finite sets. The supremum cannot be very diferent, with respect to magnitude, from every finite set. If n is in the set, then n+1 is in the set. Do you see how this proceeds? N = lim {n --> oo} {1,2,3,..,n} If infinity was not the continuation of the finite, then lim {n --> oo} 1/n = 0 needed not be true. That would imply that mathematics needed not be true. Let's avoid such nonsense. === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > But a stronger result is that > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, > Yes. Note that every sequence of this form has a last element. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}| > No. Note that this sequence does not have a last element. > Note, however, that it has only natural numbers as elements - and only > that is important. No, you can't pretend that the set does not have a greatest member. I do not pretend that the set has a greatest member. But important is > that the set has only natural numbers as members. And for each of them > my statement holds. No, it holds only if the set has a greatest member. > You cannot assume that something that is true for sequence > with a last element must be true for sequences without > a last element. > But you can assume that every element of N is a natural number and > nothing else. For any set consisting only of even natural numbers I > proved that there is a nunber contained which is greater than the > cardinal number of the set. No, for any FINITE set of even numbers (greater than 0), there is an > even number in the set and greater than the cardinality of the set. The infinite set is nothing but the supremum of the set of finite > sets. The supremum cannot be very diferent, with respect to magnitude, > from every finite set. First, you don't need this supremum business. We can talk about sets of ordinals, or of cardinals, relations on them, upper bounds, least upper bounds on them, etc. And we could define whatever for other sets (such as sets of even numbers) and ordering and supremum thereof, but as it stands, you've not defined for such sets as sets of even numbers (as opposed to the CARDINALITY of such sets). Second, even granting that we could devise (for no good reason other than added rigmarole) a supremum in this way, if you use a principle that the sup must be finite, then that is your own principle that just directly contradicts set theory. And thus, so what? Anyone can adopt a principle to contradict any theory whatsoever. > If n is in the set, then n+1 is in the set. Do you see how this > proceeds? If n is a member of N then n+1 is a member of N Yes. > N = lim {n --> oo} {1,2,3,..,n} You'd have to define 'lim' notation in such a case. Notice that in analysis and other settings, each variationo 'lim' notation is DEFINED. However the following is correct: Somewhat pedantically (to work around the fact that in Z set theory there is no set of all ordinals), in any set of ordinals S such that N and all natural numbers are members of S, then, per the membership ordering (the ordinary 'less than' for ordinals) on S, we have that N is the least upper bound of N. I.e., for any n in N, we have that N is greater than n and N is the least ordinal that has the property of being greater than any n in N. More simply, N is the least ordinal that is greater than every member of N. > If infinity was not the continuation of the finite, then lim {n -- oo} 1/n = 0 needed not be true. Various definitions of 'lim' are specific to certain sets, fields (or whatever may be the case) and orderings or metrics, etc. It is correct that in the reals, lim(n=1 to oo) 1/n = 0. That is a quite separate matter from the fact that N is the sup of N in the class of ordinals. That in the reals, lim(n=1 to oo) 1/n = 0 does not contradict that N is the least ordinal (and the least cardinal) greater than every member of N. Two different statements: (1) In the reals, lim(n=1 to oo) 1/n = 0. (2) N is the least ordinal (and cardinal) greater than every member of N. And (2) does not, by any recognizable logic in mathematics, entail the negation of (1). > That would imply that mathematics > needed not be true. Let's avoid such nonsense. You need to see that your arguments are nonsense. MoeBlee === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > But a stronger result is that > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, > Yes. Note that every sequence of this form has a last element. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}| > No. Note that this sequence does not have a last element. > Note, however, that it has only natural numbers as elements - and only > that is important. > No, you can't pretend that the set does not have a greatest member. I do not pretend that the set has a greatest member. But important is > that the set has only natural numbers as members. And for each of them > my statement holds. No, it holds only if the set has a greatest member. |{2, 4, 6, ..., 2n}| / 2n < 1 holds for every natural number. > You cannot assume that something that is true for sequence > with a last element must be true for sequences without > a last element. > But you can assume that every element of N is a natural number and > nothing else. For any set consisting only of even natural numbers I > proved that there is a nunber contained which is greater than the > cardinal number of the set. > No, for any FINITE set of even numbers (greater than 0), there is an > even number in the set and greater than the cardinality of the set. The infinite set is nothing but the supremum of the set of finite > sets. The supremum cannot be very diferent, with respect to magnitude, > from every finite set. First, you don't need this supremum business. It is required. Otherwise it is immediately clear that there is no order preserving trijection between initial segments the first column, initial segments of the diagonal, and sequences of 1's in the rows of the matrix 1 11 111 ... > We can talk about sets > of ordinals, or of cardinals, relations on them, upper bounds, least > upper bounds on them, etc. And we could define whatever for other sets > (such as sets of even numbers) and ordering and supremum thereof, but > as it stands, you've not defined for such sets as sets of even numbers > (as opposed to the CARDINALITY of such sets). Second, even granting that we could devise (for no good reason other > than added rigmarole) a supremum in this way, if you use a principle > that the sup must be finite, I do not use this principle. I only use the fact that |{2, 4, 6, ..., 2n}| / 2n < 1 holds for every natural number and that it remains if we increase n without end. And exactly that is the definition of infinity. It is the way how Cantor derived his theory: Without further definitions and ado! If n is in the set, then n+1 is in the set. Do you see how this > proceeds? If n is a member of N then n+1 is a member of N Yes. N = lim {n --> oo} {1,2,3,..,n} You'd have to define 'lim' notation in such a case. No. It simply means n can grow without end. > Notice that in > analysis and other settings, each variationo 'lim' notation is > DEFINED. However the following is correct: Somewhat pedantically (to work around the fact that in Z set theory > there is no set of all ordinals), in any set of ordinals S such that N > and all natural numbers are members of S, then, per the membership > ordering (the ordinary 'less than' for ordinals) on S, we have that N > is the least upper bound of N. I.e., for any n in N, we have that N is > greater than n and N is the least ordinal that has the property of > being greater than any n in N. More simply, N is the least ordinal that is greater than every member > of N. If infinity was not the continuation of the finite, then lim {n -- oo} 1/n = 0 needed not be true. Various definitions of 'lim' are specific to certain sets, fields (or > whatever may be the case) and orderings or metrics, etc. It is correct that in the reals, lim(n=1 to oo) 1/n = 0. That is a quite separate matter from the fact that N is the sup of N > in the class of ordinals. That in the reals, lim(n=1 to oo) 1/n = 0 does not contradict that N > is the least ordinal (and the least cardinal) greater than every > member of N. I did not say that this is contradicted. I used lim(n=1 to oo) 1/n = 0 to support N = lim {n --> oo} {1,2,3,..,n}. Two different statements: (1) In the reals, lim(n=1 to oo) 1/n = 0. (2) N is the least ordinal (and cardinal) greater than every member of > N. And (2) does not, by any recognizable logic in mathematics, entail the > negation of (1). How do you come to this conclusion? The following statements do not contradict each other. In the reals, lim(n=1 to oo) 1/n = 0. N = lim {n --> oo} {1,2,3,..,n} That would imply that mathematics > needed not be true. Let's avoid such nonsense. You need to see that your arguments are nonsense. I am sorry, but I can only see that you have not understood my text. === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > But a stronger result is that > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, Yes. Note that every sequence of this form has a last element. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}| No. Note that this sequence does not have a last element. Note, however, that it has only natural numbers as elements - and only > that is important. No, two things are important. Whether the sequence contains only natural numbers. Whether the sequence has a last element. You cannot assume that something that is true for sequence > with a last element must be true for sequences without > a last element. But you can assume that every element of N is a natural number and > nothing else. For any set consisting only of even natural numbers No, Your proof only holds for sets of even natural numbers which have a largest element. -William Hughes === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > But a stronger result is that > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, > Yes. Note that every sequence of this form has a last element. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}| > No. Note that this sequence does not have a last element. Note, however, that it has only natural numbers as elements - and only > that is important. No, two things are important. Whether the sequence contains only > natural numbers. Whether the sequence has a last element. And not whether the moon is round? You mistake size of numbers and number of numbers. Both are independen from one another. An infinite set of 1's, side by side, does not reach the height 2, because no single element reaches the height 2. Nevertheless it is infinite. Similarly, an infinite set of natural numbers, side by side, does not reach an infinite height, because all numbers are finite. > You cannot assume that something that is true for sequence > with a last element must be true for sequences without > a last element. But you can assume that every element of N is a natural number and > nothing else. For any set consisting only of even natural numbers No, Your proof only holds for sets of even natural numbers which > have a largest element. A proof of this kind, which holds for every finite segment, holds for the infinite segment too. Remember: 111... is nothing but the union of all finite segments. As every finite segment is larger than all of its predecessors, we can conclude that the infinite segment is larger than all finite segments. This would not be provable, if we could not apply the facts which are valid in the finite. === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi No, Your proof only holds for sets of even natural numbers which > have a largest element. A proof of this kind, which holds for every finite segment, holds for > the infinite segment too. Then that is some axiom or inference rule that you're adding. It's tantamount to contradicting set theory. So what? Anyone can adopt an axiom or rule to contradict any theory whatsoever. > Remember: 111... is nothing but the union of > all finite segments. And you've shown no logic whatsoever that that fact entails your principle that that which holds for every finite segment, holds for the infinite segment too. If S = Ui fi, and each fi has some property does not entail that S has that property. Where in any mathematics do you find any such principle that if each fi has some property then Ui fi has that property? > As every finite segment is larger than all of its predecessors, we can > conclude that the infinite segment is larger than all finite segments. That's NOT how we prove that the infinite sequence has greater length than any of the finite sequences. > This would not be provable, if we could not apply the facts which are > valid in the finite. We DON'T use your prinicple of extrapolating from the finite to the infinite to prove that the infinite sequence has greater length than any of the finite sequences. MoeBlee === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi Remember: 111... is nothing but the union of > all finite segments. And you've shown no logic whatsoever that that fact entails your > principle that that which holds for every finite segment, holds for > the infinite segment too. If S = Ui fi, and each fi has some property > does not entail that S has that property. Where in any mathematics do > you find any such principle that if each fi has some property then Ui > fi has that property? Not some property, but a property of size. As every finite segment is larger than all of its predecessors, we can > conclude that the infinite segment is larger than all finite segments. That's NOT how we prove that the infinite sequence has greater length > than any of the finite sequences. It does not interest us how you prove or disprove! We are interested in valid mathematics. This would not be provable, if we could not apply the facts which are > valid in the finite. We DON'T use your prinicple of extrapolating from the finite to the > infinite to prove that the infinite sequence has greater length than > any of the finite sequences. It does not interest us how you prove or disprove. We are interested do mathematics which can be used in reality and can be proven by experiment. In that mathematics (which is the only useful mathematics) we find that without further ado and any further definitions and axioms lim {n --> oo} 1/n = 0 Because: 1) every term is non-negative, and 2) every term is less than its predecessor. And we find that there is no order preserving bijection between omega (the set of finite sequences of 1's in the rows) and omega (the set of 1's of the first column) of the following matrix: 1 11 111 ... This implies that omega does not exist as a set bijectable with itself, and, hence, not as a set at all. Discovery consists of seeing what everybody has seen and thinking what nobody has thought. (Szent Gy?rgi) === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > Remember: 111... is nothing but the union of > all finite segments. And you've shown no logic whatsoever that that fact entails your > principle that that which holds for every finite segment, holds for > the infinite segment too. If S = Ui fi, and each fi has some property > does not entail that S has that property. Where in any mathematics do > you find any such principle that if each fi has some property then Ui > fi has that property? Not some property, but a property of size. And even when the property is size, you've not proven any result such as: if each fi is finite then Ui fi is finite. If you adopt that as a principle, then that is a principle of your mathematics that contradicts, say, Z set theory. Fine, now we know exactly what principle of yours contradicts Z set theory (though, even that is superfluous since we already know that you have a principle that contradicts the axiom of infinity). But your principle (each fi fintie then Ui fi finite) has not been shown to be derived from set theory, nor have you shown that the negation of your principle contradicts any ordinary mathematics such as lim(n=1 to oo) 1/n = 0. > As every finite segment is larger than all of its predecessors, we can > conclude that the infinite segment is larger than all finite segments. That's NOT how we prove that the infinite sequence has greater length > than any of the finite sequences. It does not interest us how you prove or disprove! We are interested > in valid mathematics. Your valid mathematics is whatever you pull out of your brain. Other people meanwhile, submit themselves to the discipline of stating axioms (of which is it is in principle, and ordinarily even in practice, mechancially checkable that a formula is or is not an axiom) and rules of inference (of which it is in principle, and ordinarily even in practice, mechancially checkable that a purported application of the rule is or is not a correct application of that rule), and the work of others that is not so formal can still be formalized in principle, and ordinarily even in practice. > This would not be provable, if we could not apply the facts which are > valid in the finite. We DON'T use your prinicple of extrapolating from the finite to the > infinite to prove that the infinite sequence has greater length than > any of the finite sequences. It does not interest us how you prove or disprove. We are interested > do mathematics which can be used in reality and can be proven by > experiment. In that mathematics (which is the only useful mathematics) > we find that without further ado and any further definitions and > axioms lim {n --> oo} 1/n = 0 Whether you care about proof or not, the above is something that we, even if you do not, PROVE. > Because: > 1) every term is non-negative, and > 2) every term is less than its predecessor. Meanwhile, we give a rigorous proof. > And we find that there is no order preserving bijection between omega > (the set of finite sequences of 1's in the rows) and omega (the set of > 1's of the first column) of the following matrix: 1 > 11 > 111 > ... WHAT order is not preserved by an obvious bijection between N and the set of strings of 1 of finite length? > This implies that omega does not exist as a set bijectable with > itself, and, hence, not as a set at all. implies implies some form of logic. You've shown no logic that justfies your conclusion. > Discovery consists of seeing what everybody has seen and thinking what > nobody has thought. (Szent Gy?rgi) Narccissim is marked by thinking that any unuusal thought one may have is a discovery. MoeBlee === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi There is no order preserving bijection between omega > (the set of finite sequences of 1's in the rows) and omega (the set of > 1's of the first column) of the following matrix: 1 > 11 > 111 > ... WHAT order is not preserved by an obvious bijection between N and the > set of strings of 1 of finite length? The normal order of natural numbers. Please reread the thread A problem of set geometry. There you have put a lot of questions which I answered there. There is an infinite column of omega 1's. There is a diagonal of omega 1's which can be projected on the first column but there is no finite sequence of 1's in any line on which the diagonal could be projected. === Subject: Re: Two results of set geometry <874pi0885q.fsf@huxley.huxley.fi > But a stronger result is that > lim{n --> oo} |{2, 4, 6, ..., 2n}| / 2n < 1. > So there is always, i.e. for any n, a greater natural number than > |{2, 4, 6, ..., 2n}|, > Yes. Note that every sequence of this form has a last element. > and hence, there is a greater natural number > than > |{2, 4, 6, ...}| > No. Note that this sequence does not have a last element. > Note, however, that it has only natural numbers as elements - and only > that is important. No, two things are important. Whether the sequence contains only > natural numbers. Whether the sequence has a last element. And not whether the moon is round? You mistake size of numbers and > number of numbers. Both are independen from one another. An infinite > set of 1's, side by side, does not reach the height 2, because no > single element reaches the height 2. Nevertheless it is infinite. > Similarly, an infinite set of natural numbers, side by side, does not > reach an infinite height, because all numbers are finite. > You cannot assume that something that is true for sequence > with a last element must be true for sequences without > a last element. > But you can assume that every element of N is a natural number and > nothing else. For any set consisting only of even natural numbers No, Your proof only holds for sets of even natural numbers which > have a largest element. A proof of this kind, which holds for every finite segment, holds for > the infinite segment too. No. If something is true for all finite segments it may or may not be true for the infinite segment. > Remember: 111... is nothing but the union of > all finite segments. Rembember, as you yourself have pointed out, if A is the union of sets B, and X is true for all sets B, X may or may not be true for A. - William Hughes === Subject: Proposal to deal with the solution manual postings There have been over 4000 postings since the first of July in the four groups that I have posted this to asking to buy or sell solutions manuals so they can cheat on homework. I've been quietly letting this go buy but enough is enough. I propose we all drop the posting identities of everyone who has asked or will ask to buy or sell a solutions manual into a kill file for the next year. You know if they can't buy the manual to cheat on their homework they are going to be back next week asking for someone to do their homework for them. If enough of us do this then some fraction of them can just become un-persons on the net. I have vast lists of these names in my killfile already. You can easily find the names for yourself, or if there is enough interest I can put together a file with the names from the postings that you can drop in your killfile. Email address is valid, been that for more than 15 years. === Subject: Re: Proposal to deal with the solution manual postings >There have been over 4000 postings since the first of July >in the four groups that I have posted this to asking to buy >or sell solutions manuals so they can cheat on homework. I've been quietly letting this go buy but enough is enough. I propose we all drop the posting identities of everyone who >has asked or will ask to buy or sell a solutions manual into >a kill file for the next year. You know if they can't buy >the manual to cheat on their homework they are going to be >back next week asking for someone to do their homework for them. If enough of us do this then some fraction of them can just >become un-persons on the net. I have vast lists of these names in my killfile already. >You can easily find the names for yourself, or if there is >enough interest I can put together a file with the names >from the postings that you can drop in your killfile. Email address is valid, been that for more than 15 years. The same suggestion, almost word for word, was posted a few days ago. In my opinion, the suggestion is worthless (as to why, read my reply in the prior thread). quasi === Subject: Re: Proposal to deal with the solution manual postings > There have been over 4000 postings since the first of July > in the four groups that I have posted this to asking to buy > or sell solutions manuals so they can cheat on homework. Can you stop posting this in sci.electronics.design please ? It has no relevance there. Graham === Subject: Re: does the empty set contain the least member? , porky_pig_jr@my-deja.com What is true is that EVERY element of the empty set is a least element > for the empty order. What is FALSE is that the empty set contains a least element for the > empty order. See the difference? > Arturo Magidin > magidin-at-member-ams-org > > I guess I do. I was reading some staff about the empty sets, and came > across at something like this: any property P holds for EVERY (or ALL) > elements of the empty sets (this is where the vacuous truth comes to > play), and, on the other hand, there is no element in empty set (by > the virtue of its emptiness) for which the property P holds (or > something like that). I got the second statement correctly > (containment part) but didn't know how to formulate the first one, > the 'vacuously true' part. > > What I didn't mention is the context. We are working with the Natural > Numbers, at this point, and the empty set in question is the empty > subset of the set of all natural numbers. Still (I guess) it doesn't > change the whole picture. > > Regarding the exact formulation of the question --- well, there was > *no* exact formulation. The exercise was stated roughly like this: > Prove that the least element of the non-empty subset of the natural > numbers is unique. What about the empty subset? The first part was > trivial, and the second one was pretty much open-ended. Can we talk > about that at all? The subject line does not capture the question in the above paragraph. Are you discussing the formulation of the natural numbers as { {}, {{}}, {{}, {{}} }, ... }? This collection has a least element. The least element _is_ the empty set. Capice? -- Michael Press === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! Wondering if anyone has the solutions manual to Thermal Physics, 2nd Edition, by Charles Kittel === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! Looking for the complete solutions manual to: Calculus: Multivariable, 5th Edition, by James Stewart Precalculus: Mathematics for Calculus, 5th Edition, by James Stewart, Lothar Redlin & Saleem Watson Please email benkildea@gmail.com === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! > Looking for the complete solutions manual to: Calculus: Multivariable, 5th Edition, by James Stewart > Precalculus: Mathematics for Calculus, 5th Edition, by James Stewart, > Lothar Redlin & Saleem Watson Please email benkil...@gmail.com > Actually I need the solutions to: * Calculus, 5th Edition, by James Stewart Not the Multivariate edition. Sorry... === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! I need the solutions Manual for Sears and Zemansky's University Physics 11th Edition by Young & Freedman? It needs to include chapters 21-44 and all the problems from that chapter, not just the odd ones. Can someone have it or tell me when I can get this? Brooke === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! I was wondering if you had any of these solutions manual: 1. Mechanical vibrations, 4th ed, rao 2. Modern control systems, 11th ed, dorf 3. heat transfer, 9th ed, holman. Older editions are okae as long as most of the problems match up. Please e-mail with prices for each manual if you do: thesofaking420@gmail.com === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! how do i go about getting the Engineering Mechanics: Dynamics 11th Ed. > I have the comprehensive solutions manual in electronic format for > these following textbooks. They include complete solutions to all the > problems in the text, except where noted below. Payment is through > Paypal for the amount of $8.00 to $15.00 US dollars per item. > A Course in Game Theory by Osborne, Rubinstein > A Course in Algebraic Number Theory by Cohen > Adaptive Filter Theory, 4th Edition, by Haykin > Adaptive Control, 2nd. Ed., by Astrom, Wittenmark > Advanced Engineering Mathematics, 8th Editoin, by Erwin Kreyszig (even > solutions) > Advanced Engineering Mathematics, 9th Edition, by Erwin Kreyszig (even > solutions) > Advanced Macroeconomics, David Romer > Advanced Mathematical Concepts Precalculus With Applications by > Holliday [ISBN: 0028341759] > Advanced Modern Engineering Mathematics, 3rd Ed., by G. James > A First Course In Differential Equations, 7th Edition, by Zill, Cullen > Analog Integrated Circuit Design, 1st Ed., by Johns, Martin (text > ebook and solution manual) > Analysis and Design of Analog Integrated Circuits, 4th Ed., by Gray, > Hurst, Lewis, Meyer > Analytical Mechanics, 7th Edition, by Fowels, Cassiday > An Interactive Introduction to Mathematical Analysis, by Jonathan > Lewin > An Introduction to the Mathematics of Financial Derivatives, 2nd Ed., > by Neftci [ISBN: 0125153929] > Antenna Theory, 2nd Ed., by Balanis > Antennas for all Applications, 3rd Edition, Kraus, Marhefka > Applied Linear Statistical Models, 5th Ed., by Neter (Selected Sol.) > Applied Numerical Analysis, 6th Edition, by Gerald, Wheatley > Applied Numerical Methods with MATLAB for Engineers and Scientists, > 1st Ed,. by Chapra > Applied Statistics and Probability for Engineers, 3rd Ed., by > Montgomery, Runger (Selected Solutions) > Applied Strength of Materials, 4th Edition, by Mott > A Transition to Advanced Mathematics, 5th Edition, by Smith, Eggen, > Andre > Automatic Control Systems, 8th Edition, by Kuo, Golnaraghi > Basic Business Statistics: Concepts and Applications, 10th Ed., by > Berenson, Krehbiel, Levine (chap1-18) > Basic Engineering Circuit Analysis, 7th Ed., by J. David Irwin > Basic Engineering Circuit Analysis, 8th Ed., by J. David Irwin, Nelms > (Missing a chapter or 2) > Bioprocess Engineering Principles by Doran > Calculus Early Transcendental, 5th Ed., by James Stewart > Calculus - Early Transcendentals, 7th Ed., by Anton, Bivens, Davis > Calculus: Graphical, Numerical, Algebraic, 3rd Ed., Waits, Finney, > Demana, Kennedy > Calculus: Multivariable, 5th Edition, by James Stewart > Calculus: Single Variable, Early Transcendental, 5th Edition, by James > Stewart > Calculus, Single and Multivariable, 3rd Ed., by Hughes-Hallett, > McCallum > Calculus: Study and Solutions Guide, Vol. 1, 7th Ed., by Larson, > Hostetler, Edwards > Chemical and Engineering Thermodynamics, 3rd Ed., Stanley I. Sandler > Chemical Engineering Volume 1, Sixth Edition, by Richardson, Coulson, > Backhurst, Harker > Thornton > College Physics, Volume 1: 7th Edition, by Serway, Faugh > College Physics, Volume 2: 7th Edition, by Serway, Faughn > Communications Systems, 4th Ed., by Haykin > Communications Systems Engineering, 2nd Edition, by Proakis > Computational Techniques for Fluid Dynamics by Srinivas, Fletcher > Computer Networks, 4th Ed., by Andrew S. Tanenbaum > Computer Networks: A Systems Approach, 3rd Edition, by Davie > Control Systems Engineering, 4th Ed., by Norman Nise > Corporate Finance, 6th Edition, by Ross > C++ How to Program: Intro Object-Oriented Design with the UML, 3rd > Ed., by Deitel, Nieto > Data and Computer Communications, 8th Edition by Stallings > Database Management Systems, 3rd Ed., by Ramakrishnan, Gehrke (Sol. > for Chapters 2-21, odd only) > Design of Analog CMOS Integrated Circuits, 1st Edition, by Razavi > Design of Analysis of Experiments, 6th Edition, Montgomery (missing > chapter 6-8) > Design of Machinery, 3rd Ed by Robert L. Norton > Design With Operational Amplifiers and Analog Integrated Circuits, 2nd > Ed., by Sergio Franco > Design With Operational Amplifiers and Analog Integrated Circuits, 3rd > Ed., by Sergio Franco > Device Electronics for Integrated Circuits 3rd Edition by Muller > Differential Equations with Boundary Value Problems, 2nd Ed., by > Polking, Arnold > Digital And Analog Communication Systems 7th Ed., Leon W. Couch > Digital Communications, 4th Edition, by Proakis > Digital Communications: Fundamentals and Applications, 2nd Ed, Skylar > Digital Design, 4th Edition, by Mano, Ciletti > Digital Image Processing, 2nd Edition, by Gonzalez, Woods > Digital Integrated Circuits, 2nd Ed., by Rabaey (Solutions ONLY for > Chapters 3, 5, 6, 10) > Digital Signal Processing: A Computer Based Approach, 1st Ed., by > Mitra > Digital Signal Processing: A Computer Based Approach, 2nd Ed., by S. > Mitra > Digital Signal Processing: A Computer Based Approach, 3rd Ed., by S. > Mitra > Digital Signal Processing: Priciples, Algorithms and Applications, 3rd > Edition, by Proakis > Discrete Time Signal Processing, 2nd Edition, Oppenheim > Dynamics of Mechanical Systems by C.T.F. Ross > Econometric Analysis, 5th Edition, by Greene > Wooldridge > Econometrics of Financial Markets, by Adamek, Cambell, Lo, MacKinlay, > Viceira > Electrical Properties of Materials, 7th Ed., by D. Walsh, L. Solymar > Electric Circuits 6th Ed. by Nilsson > Electric Circuits 7th Ed. by Nilsson > Electric Machinery, 6th Ed., Fitzgerald, Kingsley, Umans > Electric Machinery Fundamentals, 4th Ed by Chapman > Electromagnetic Fields and Waves by Iskander (...) > Electronic Circuit Analysis, 2nd Ed., by Donald Neamen > Electronics, 2nd Ed., by Allan R. Hambley > Elementary Differential Equations, 8th Edition, by Boyce, DiPrima > (some odd/even) > Elementary Principles of Chemical Processes, 3rd Ed., by Felder, > Rousseau > Elements of Chemical Reaction Engineering, 3rd Ed., by H. Scott Fogler > Engineering and Chemical Thermodynamics, by Koretsky [ISBN: 0471385867] > (No sol. for chapt 6) > Engineering Circuit Analysis, 6th Edition, Hyat > Engineering Electromagnetics, 6th Ed W. Hayt, J. Buck > Engineering Fluids Mechanics 7th Edition by Crowe > Engineering Fluids Mechanics 8th Edition by Crowe > Engineering Mathematics, 4th Ed., by John Bird > Engineer Mechanics: Dynamics, 4th Ed., by Bedford > Engineering Mechanics: Dynamics, 10th Ed., by Russell C. Hibbeler > Engineering Mechanics: Dynamics 11th Ed. by Hibbeler > Engineering Mechanics: Dynamics 5th Ed. by Meriam, Kraige > Engineering Mechanics: Statics, 4th Edition - A. Bedford, Wallace > Fowler > Engineering Mechanics: Statics, 5th Ed., Meriam > Engineering Mechanics: Statics, 6th Ed., Meriam > Engineering Mechanics: Statics, 10th Ed., by Russell C. Hibbeler > Engineering Mechanics: Statics 11th Ed. by Hibbeler > Experiments with Economic Principles by Bergstrom, Miller > Feedback Control of Dynamic Systems, 4th Edition, by Powell, Emami- > Naeini > Financial Accounting, 4th Ed., by Libby, Short (Chap1-14) > Financial Accounting: An International Introduction, 2nd Ed., by > Alexander, Nobes > Finite Element Techniques in Structural Mechanics by Ross > Fluid Mechanics - 5th Edition by Frank M. White > Fluid Mechanics and Thermodynamics of Turbomachinery, 5th Ed., by S. > L. Dixon [ISBN: 0750678704] > Essentials of Fluid Mechanics: Fundamentals and Applications, 1st Ed., > by Cengel & Cimbala > Fluid Mechanics with Engineering Applications, 10th Edition, by > Finnemore > Fundamentals of Aerodynamics, 3rd Edition, by J. D. Anderson, Jr. > Fundamentals of Aerodynamics, 4th Edition, by Anderson > Fundamentals of Applied Electromagnetics, 2001 Media Edition, by Ulaby > Fundamentals of Applied Electromagnetics, 5th Ed., 2008 Media Edition, > by Ulaby > Fundamentals of Digital Logic with Verilog Design, 1st Edition, by > Brown, Vranesic > Fundamentals of Digital Logic with VHDL Design by Stephen Brown, > Zvonko Vranesic > Fundamentals of Electric Circuits, 2nd Edition, by Alexander > Fundamentals of Electromagnetics with Engineering Appls by Wentworth > Fundamentals of Fluid Mechanics, 5th Ed. by Munson, Young.. > Fundamentals of Heat and Mass Transfer, 4th Ed by Incropera... > Fundamentals of Heat and Mass Transfer, 5th Ed by Incropera... > Fundamentals of Heat and Mass Transfer, 6th Ed by Incropera... > Fundamentals of Logic Design, 5th Ed., by Roth Jr. > Fundamentals of Machine Component Design, 3rd Ed., by Juvinall > Fundamentals of Machine Component Design, 4th Ed., by Juvinall > Fundamentals of Machine Elements, 2nd Ed., Hamrock, Jacobson, Schmid > Fundamentals of Physics by Halliday, 7th Ed., Walker, Resnick > Fundamentals of Semiconductor Devices, 1st Edition by Anderson > Fundamentals of Structural Analysis, 2nd Ed., Chia-Ming Uang, Kenneth > Leet > Fundamentals of Thermal-Fluid Sciences, 2nd Ed. by Cengel > Fundamentals of Thermal-fluid Sciences, Int'l 2nd Ed. by Cengel > Fundamentals of Engineering Thermodynamics, 5th Ed. by Shapiro > Fundamentals of Thermodynamics, 5th Ed., by Sonntag, Borgnakke... > Fundamentals of Thermodynamics, 6th Ed., by Sonntag > Geometry, 04 Edition, by McGraw-Hill [ISBN: 0078296374] > Guide to Energy Management, 5th Edition, by Pawlik > Heat Transfer: A Practical Approach - 2nd Edition by Cengel > Hydraulics in Civil and Environmental Engineering, 4th Ed., by Andrew > Chadwick > Introduction to Algorithms, 2nd Ed by Cormen, Leiserson (Selected > Sol.) > Introduction To Chemical Engineering Thermodynamics, 7th Ed., by Van > Ness, Smith, Abbott > Introduction to Electric Circuits, 6th Ed., by Dorf, Svoboda > Introduction to Electric Circuits, 7th Ed., by Dorf, Svoboda > Introduction to Electrodynamics, 3rd Ed. by David Griffiths > Introduction to Fluid Mechanics - 5th Ed. by Fox.. > Introduction to Fluid Mechanics - 6th Ed by Fox, McDonald... > Introduction to Linear Algebra, 3rd Ed., by Gilbert Strang > Introduction to Linear Algebra, 5th Ed., Arnold, Johnson, Riess > Introduction to Quantum Mechanics, 2nd Ed. by Griffiths > Introdution to Solid State Physics, 8th Edition by Kittel > Intro to Thermal Systems Engineering: Thermodynamics, Fluid Mechanics, > and Heat Transfer by Moran, Shapiro, Munson, DeWitt > Introduction to Thermal Systems Engineering, by Moran, Shapiro > Linear Algebra, by J. Hefferon > Linear Algebra And Its Applications, 3rd Ed., by David C. Lay > Linear Algebra with Applications, 2nd Edition - by Otto Bretscher > Linear Algebra with Applications, 3rd Edition - by Otto Bretscher > Linear Circuit Analysis: Time Domain, Phasor and Laplace.., 2nd Ed, > Lin > Machine Design: An Integrated Approach, 2nd Ed., by Robert L. Norton > (same problems as third edition except for last 2-4 problems per > chapter that were added to the third edition) > Machine Design: An Integrated Approach, 3rd Ed., by Robert L. Norton > Managerial Accounting, 11th Ed., by Noreen, Brewer, Garrison > Materials Science and Engineering: An Introduction, 6th Ed. by > Callister > Materials Science and Engineering: An Introduction, 7th Ed., by > Callister (file currently unrecoverable) > Matrix Analysis and Applied Linear Algebra by Carl Meyer > MC68HC11: An Introduction: Software/Hardware Interf, 2nd Ed, by Huang > Mechanical Engineering Design, 7th Ed. by Mischke, Shigley > Mechanical Vibrations, 3rd Edition, by S. S. Rao (99% same as 4th > Edition, No Solutions for Chapters 6, 9, and 12) > Mechanics of Fluids, 8th Ed., by Bernard Massey > Mechanics of Fluids, 4th Ed., Irving H. Shames > Mechanics of Fluids, 8th Ed., by Bernard Massey > Mechanics of Materials - 3rd Ed. by Beer, Johnston, Dewolf > Mechanics of Materials - 6th Ed. by Hibbeler > Mechanics of Materials, 6th Edition by James M. Gere (missing small > portion, section 8.5) > Mechanics of Materials, 6th Ed., by Sturges, Morris, Riley (part of > Chapt 2 is missing but only #1 thru #60) > Mechanics of Solids by C.T.F. Ross > Microeconomic Analysis, 3rd Ed., by H. Varian (Ans. to Exercises: Ch.1- > Ch.25) > Microeconomic Theory, by Mas-Colell, Whinston, Green > Microelctronic Circuits, 5th Ed. by Sedra and Smith > Microelectronic Circuit Design, 2nd Edition by Jaeger, Blalock > Microelectronics: Digital and Analog Circuits and Systems by Millman > Microwave and Rf Design of Wireless Systems, 1st Edition, by Pozar > Miller & Freund's Probability and Statistics for Engineers, 7th > Edition, Johnson, Miller > Modern Compressible Flow, 3rd Edition, by Anderson > Modern Control Engineering, 3rd Edition, by Ogata > Modern Control Engineering, 4th Edition, by Ogata > Modern Digital and Analog Communication Systems, 3rd Ed., by Lathi > Modern Control Systems, 9th Ed., by Richard C. Dorf, Robert H Bishop > Modern Operating Systems,2nd Ed., by Andrew Tanenbaum > Modern Physics 4th Edition by Tipler > Monetary Theory and Policy, 2nd Edition, by Walsh > Multivariable Calculus, 5th Edition, by James Stewart > Numerical Methods, 3rd Ed., by J. Douglas Faires, Richard L. Burden > (Selected Solutions) > Operating Systems: Internals and Design Principles, 4th Edition, by > Stallings > Operating System Concepts, 7th Ed., Silberschatz, Galvin, Gagne > Options, Futures and Other Derivatives, 5th Ed., by John Hull > (Chapters 1 thru 18 ONLY) > Orbital Mechanics: For Engineering Students by Howard Curtis (includes > matlab scripts) > Organic Chemistry, 4th Ed., by Carey, Atkins (Student Study Guide and > Sol. Man.) > Partial Differential Equations with Fourier Series and Boundary Value > Problems, 2nd Ed., by Asmar (Student Solutions Manual) > Physical Chemistry - 7th Edition - by Julio de Paula, Peter Atkins > Physics, 6th Edition, by John Cutnell > Physics, 5th Edition, Vol 2 by Halliday, Resnick, Krane (Chap 25-52) > Physics for Scientist and Engineers by Knight (No Chapt 36-42) > Physics for Scientist and Engineers, 6th Ed., by Serway > Physics for Scientists and Engineers-Vol 1, 5th Edition, Serway, > Beichner (Chap. 1 - 22) > Physics for Scientists and Engineers-Vol 2, 5th Edition, Serway, > Beichner (Chap. 23 - 46) > Physics for Scientists and Engineers, 3rd Ed., by Douglas C. Giancoli > Physics for Scientist and Engineers, 5th Edition, by Tipler, Mosca > Physics: Principles with Applications, 6th Ed. by Giancoli > Power System Analysis and Design, 3rd Ed., by Glover, Sarma > Principles and Applications of Electrical Engineering 4th (Revised) Ed > by Rizzoni > Principles of Communication: Systems, Modulation Noise, 5th Ed., > Ziemer > Principles of Physics, 3rd Edition, by Serway > Principles of Statics, 10th Ed., by Russell C. Hibbeler [ISBN: > 0131866745] > Probability and Statistics for Engineers and Scientists, 3rd Edition, > Hayter > Probability and Statistics for Engineering and the Sciences, 6th Ed., > by Jay L. Devore > Probability Random Variables, and Stochastic Processes, 4th Ed., by > Papoulis, Pillai (..) > Quantum Mechanics: An Accessible Introduction, 1st Ed., by Robert > Scherrer > Recursive Macroeconomic Theory, 1st Ed., by Ljungqvist, Sargent > RF Circuit Design: Theory & Applications, by Bretchko, Ludwig > Sears and Zemansky's University Physics 11th Edition by Young.. > Semiconductor Device Fundamentals by Pierret > Semiconductor Devices: Physics and Technology, 2nd Ed., S.M. Sze > Semiconductor Physics And Devices -3rd Ed. by D. Neamen > Separation Process Principles, 2nd Ed., Seader, Henley > Signal Processing and Linear Systems by Lathi > Signals and Systems, 2nd Edition, by Haykin, Van Veen > Signals and Systems, 2nd Edition, Oppenheim, Willsky, Hamid, Nawab > Signals and Systems: Analysis Using Transform Methods and MATLAB, 1st > Ed., by M. J. Roberts > Signals, Systems, and Transforms: Charles L. Phillips, Eve A. Riskin, > John M. Parr > (No Solutions for Chap 9-12) > Shigley's Mechanical Engineering Design, 8th Ed. by Budynas, Nisbett > (No Sol. for Chapt 18 & 19) > Simply C#: An Application-Driven Tutorial Approach, by Deitel, Hoey... > (Chapters 1-32) > Soil Mechanics: Concepts and Applications, 2nd Ed., by Powrie > Solid State Electronic Devices - 5th Ed by Streetman > Solid State Electronic Devices - 6th Ed by Streetman > Statics and Mechanics of Materials: An Integrated Approach, 2nd Ed., > by Riley, Sturges, Morris > Structural Analysis, 5th Edition, by Hibbeler > System Dynamics, 3rd Edition, by Ogata > Theory and Design for Mechanical Measurements, 4th Ed., Beasley, > Figliola > Thermal Physics, 2nd Edition, by Charles Kittel > Thermal Physics, by Ralph Baierlein > Thermodynamics: An Engineering Approach by Cengel, Boles (Missing > solutions #118-149 of Chapter 7) > The Science and Engineering of Materials, 4th Ed., by Donald R. > Askeland, Pradeep P. Phule > Thomas' Calculus, Early Trans., Part 1, 10th Ed. by Thomas, Weir, > Hass, Giordano > Wireless Communications: Principles and Practice, 2nd Ed, by Rappaport i need solution mannual for wireless communication: principles & > practice,2nd ed I need a solution mannual for College Physics, Volume 1: 7th Edition, by Serway, Faugh === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! I am looking for solutions manual for the following books ASAP: * Financial Accounting - Libby Libby Short - 5th Eddition > * Complere Business Statistics - Amir D. Aczel, Jayavel Sounderpandian > - Sixth Eddition > * Principles of Microeconomics -Mankiw Kneebone Mckenzie Rowe - Third > Canadian Edition Please e-mail me if anyone can know where to get these solution > manuals and any other aids associated with these books. Sof> Complete electronic solution manual in pdf ! Get it in hours! > I have the complete electronic SOLUTION MANUAL in PDF format > containing ALL (odd and even - except where noted) solutions for the > books listed below. These are not paper books, they are ebooks and > most are only $12.00 each - paypal accepted. > Books for which I have electronic solution manual: > A Course in Game Theory by Osborne, Rubinstein > A Course in Algebraic Number Theory by Cohen > Adaptive Filter Theory, 4th Edition, by Haykin > Adaptive Control, 2nd. Ed., by Astrom, Wittenmark > Advanced Engineering Mathematics, 8th Editoin, by Erwin Kreyszig (even > solutions) > Advanced Engineering Mathematics, 9th Edition, by Erwin Kreyszig (even > solutions) > Advanced Macroeconomics, David Romer > Advanced Mathematical Concepts Precalculus With Applications by > Holliday [ISBN: 0028341759] > Advanced Modern Engineering Mathematics, 3rd Ed., by G. James > A First Course In Differential Equations, 7th Edition, by Zill, Cullen > Analysis and Design of Analog Integrated Circuits, 4th Ed., by Gray, > Hurst, Lewis, Meyer > Analytical Mechanics, 7th Edition, by Fowels, Cassiday > An Interactive Introduction to Mathematical Analysis, by Jonathan > Lewin > An Introduction to the Mathematics of Financial Derivatives, 2nd Ed., > by Neftci [ISBN: 0125153929] > Antenna Theory, 2nd Ed., by Balanis > Antennas for all Applications, 3rd Edition, Kraus, Marhefka > Applied Linear Statistical Models,5thEd., by Neter (Selected Sol.) > Applied Numerical Analysis, 6th Edition, by Gerald, Wheatley > Applied Numerical Methods with MATLAB for Engineers and Scientists, > 1st Ed,. by Chapra > Applied Statistics and Probability for Engineers, 3rd Ed., by > Montgomery, Runger (Selected Solutions) > Applied Strength of Materials, 4th Edition, by Mott > A Transition to Advanced Mathematics,5thEdition, by Smith, Eggen, > Andre > Automatic Control Systems, 8th Edition, by Kuo, Golnaraghi > Basic Business Statistics: Concepts and Applications, 10th Ed., by > Berenson, Krehbiel, Levine (chap1-18) > Basic Engineering Circuit Analysis, 7th Ed., by J. David Irwin > Basic Engineering Circuit Analysis, 8th Ed., by J. David Irwin, Nelms > (Missing a chapter or 2) > Bioprocess Engineering Principles by Doran > Calculus Early Transcendental,5thEd., by James Stewart > Calculus - Early Transcendentals, 7th Ed., by Anton, Bivens, Davis > Calculus: Graphical, Numerical, Algebraic, 3rd Ed., Waits, Finney, > Demana, Kennedy > Calculus: Multivariable,5thEdition, by James Stewart > Calculus: Single Variable, Early Transcendental,5thEdition, by James > Stewart > Calculus, Single and Multivariable, 3rd Ed., by Hughes-Hallett, > McCallum > Calculus: Study and Solutions Guide, Vol. 1, 7th Ed., by Larson, > Hostetler, Edwards > Chemical and Engineering Thermodynamics, 3rd Ed., Stanley I. Sandler > Chemical Engineering Volume 1, Sixth Edition, by Richardson, Coulson, > Backhurst, Harker > Thornton > College Physics, Volume 1: 7th Edition, by Serway, Faugh > College Physics, Volume 2: 7th Edition, by Serway, Faughn > Communications Systems, 4th Ed., by Haykin > Communications Systems Engineering, 2nd Edition, by Proakis > Computational Techniques for Fluid Dynamics by Srinivas, Fletcher > Computer Networks, 4th Ed., by Andrew S. Tanenbaum > Computer Networks: A Systems Approach, 3rd Edition, by Davie > Control Systems Engineering, 4th Ed., by Norman Nise > Corporate Finance, 6th Edition, by Ross > C++ How to Program: Intro Object-Oriented Design with the UML, 3rd > Ed., by Deitel, Nieto > Data and Computer Communications, 8th Edition by Stallings > Database Management Systems, 3rd Ed., by Ramakrishnan, Gehrke (Sol. > for Chapters 2-21, odd only) > Design of Analog CMOS Integrated Circuits, 1st Edition, by Razavi > Design of Analysis of Experiments, 6th Edition, Montgomery (missing > chapter 6-8) > Design of Machinery, 3rd Ed by Robert L. Norton > Design With Operational Amplifiers and Analog Integrated Circuits, 2nd > Ed., by Sergio Franco > Design With Operational Amplifiers and Analog Integrated Circuits, 3rd > Ed., by Sergio Franco > Device Electronics for Integrated Circuits 3rd Edition by Muller > Differential Equations with Boundary Value Problems, 2nd Ed., by > Polking, Arnold > Digital And Analog Communication Systems 7th Ed., Leon W. Couch > Digital Communications, 4th Edition, by Proakis > Digital Communications: Fundamentals and Applications, 2nd Ed, Skylar > Digital Design, 4th Edition, by Mano, Ciletti > Digital Image Processing, 2nd Edition, by Gonzalez, Woods > Digital Integrated Circuits, 2nd Ed., by Rabaey (Solutions ONLY for > Chapters 3, 5, 6, 10) > Digital Signal Processing: A Computer Based Approach, 1st Ed., by > Mitra > Digital Signal Processing: A Computer Based Approach, 2nd Ed., by S. > Mitra > Digital Signal Processing: A Computer Based Approach, 3rd Ed., by S. > Mitra > Digital Signal Processing: Priciples, Algorithms and Applications, 3rd > Edition, by Proakis > Discrete Time Signal Processing, 2nd Edition, Oppenheim > Dynamics of Mechanical Systems by C.T.F. Ross > Econometric Analysis,5thEdition, by Greene > Wooldridge > Econometrics of Financial Markets, by Adamek, Cambell, Lo, MacKinlay, > Viceira > Electrical Properties of Materials, 7th Ed., by D. Walsh, L. Solymar > Electric Circuits 6th Ed. by Nilsson > Electric Circuits 7th Ed. by Nilsson > Electric Machinery, 6th Ed., Fitzgerald, Kingsley, Umans > Electric Machinery Fundamentals, 4th Ed by Chapman > Electromagnetic Fields and Waves by Iskander (...) > Electronic Circuit Analysis, 2nd Ed., by Donald Neamen > Electronics, 2nd Ed., by Allan R. Hambley > Elementary Differential Equations, 8th Edition, by Boyce, DiPrima > (some odd/even) > Elements of Chemical Reaction Engineering, 3rd Ed., by H. Scott Fogler > Engineering and Chemical Thermodynamics, by Koretsky [ISBN: > 0471385867] > (No sol. for chapt 6) > Engineering Circuit Analysis, 6th Edition, Hyat > Engineering Electromagnetics, 6th Ed W. Hayt, J. Buck > Engineering Fluids Mechanics 7th Edition by Crowe > Engineering Fluids Mechanics 8th Edition by Crowe > Engineering Mathematics, 4th Ed., by John Bird > Engineer Mechanics: Dynamics, 4th Ed., by Bedford > Engineering Mechanics: Dynamics, 10th Ed., by Russell C. Hibbeler > Engineering Mechanics: Dynamics 11th Ed. by Hibbeler > Engineering Mechanics: Dynamics5thEd. by Meriam, Kraige > Engineering Mechanics: Statics, 4th Edition - A. Bedford, Wallace > Fowler > Engineering Mechanics: Statics,5thEd., Meriam > Engineering Mechanics: Statics, 6th Ed., Meriam > Engineering Mechanics: Statics, 10th Ed., by Russell C. Hibbeler > Engineering Mechanics: Statics 11th Ed. by Hibbeler > Experiments with Economic Principles by Bergstrom, Miller > Feedback Control of Dynamic Systems, 4th Edition, by Powell, Emami- > Naeini > Financial Accounting, 4th Ed., byLibby,Short(Chap1-14) > Financial Accounting: An International Introduction, 2nd Ed., by > Alexander, Nobes > Finite Element Techniques in Structural Mechanics by Ross > Fluid Mechanics -5thEdition by Frank M. White > Fluid Mechanics and Thermodynamics of Turbomachinery,5thEd., by S. > L. Dixon [ISBN: 0750678704] > Essentials of Fluid Mechanics: Fundamentals and Applications, 1st Ed., > by Cengel & Cimbala > Fluid Mechanics with Engineering Applications, 10th Edition, by > Finnemore > Fundamentals of Aerodynamics, 3rd Edition, by J. D. Anderson, Jr. > Fundamentals of Aerodynamics, 4th Edition, by Anderson > Fundamentals of Applied Electromagnetics, 2001 Media Edition, by Ulaby > Fundamentals of Applied Electromagnetics,5thEd., 2008 Media Edition, > by Ulaby > Fundamentals of Digital Logic with Verilog Design, 1st Edition, by > Brown, Vranesic > Fundamentals of Digital Logic with VHDL Design by Stephen Brown, > Zvonko Vranesic > Fundamentals of Electric Circuits, 2nd Edition, by Alexander > Fundamentals of Electromagnetics with Engineering Appls by Wentworth > Fundamentals of Fluid Mechanics,5thEd. by Munson, Young.. > Fundamentals of Heat and Mass Transfer, 4th Ed by Incropera... > Fundamentals of Heat and Mass Transfer,5thEd by Incropera... > Fundamentals of Heat and Mass Transfer, 6th Ed by Incropera... > Fundamentals of Logic Design,5thEd., by Roth Jr. > Fundamentals of Machine Component Design, 3rd Ed., by Juvinall > Fundamentals of Machine Component Design, 4th Ed., by Juvinall > Fundamentals of Machine Elements, 2nd Ed., Hamrock, Jacobson, Schmid > Fundamentals of Physics by Halliday, 7th Ed., Walker, Resnick > Fundamentals of Semiconductor Devices, 1st Edition by Anderson > Fundamentals of Structural Analysis, 2nd Ed., Chia-Ming Uang, Kenneth > Leet > Fundamentals of Thermal-Fluid Sciences, 2nd Ed. by Cengel > Fundamentals of Thermal-fluid Sciences, Int'l 2nd Ed. by Cengel > Fundamentals of Engineering Thermodynamics,5thEd. by Shapiro > Fundamentals of Thermodynamics,5thEd., by Sonntag, Borgnakke... > Fundamentals of Thermodynamics, 6th Ed., by Sonntag > Geometry, 04 Edition, by McGraw-Hill [ISBN: 0078296374] > Guide to Energy Management,5thEdition, by Pawlik > Heat Transfer: A Practical Approach - 2nd Edition by Cengel > Hydraulics in Civil and Environmental Engineering, 4th Ed., by Andrew > Chadwick > Introduction to Algorithms, 2nd Ed by Cormen, Leiserson (Selected > Sol.) > Introduction To Chemical Engineering Thermodynamics, 7th Ed., by Van > Ness, Smith, Abbott > Introduction To Chemical Engineering Thermodynamics, 7th Ed., by Van > Ness, Smith, Abbott > Introduction to Electric Circuits, 6th Ed., by Dorf, Svoboda > Introduction to Electric Circuits, 7th Ed., by Dorf, Svoboda > Introduction to Electrodynamics, 3rd Ed. by David Griffiths > Introduction to Fluid Mechanics -5thEd. by Fox.. > Introduction to Fluid Mechanics - 6th Ed by Fox, McDonald... > Introduction to Linear Algebra, 3rd Ed., by Gilbert Strang > Introduction to Linear Algebra,5thEd., Arnold, Johnson, Riess > Introduction to Quantum Mechanics, 2nd Ed. by Griffiths > Introdution to Solid State Physics, 8th Edition by Kittel > Intro to Thermal Systems Engineering: Thermodynamics,... > read more E If the calc 5e by stewart has chapter 7 as being inverse functions i > need it. are u still looking for that calc 5e book ? cause i have it if u are! i was wondering also if u have the solution book for accounting 5th edition === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! Cc: hector.s.ga...@gmail.com > how do i go about getting the Engineering Mechanics: Dynamics 11th Ed. > I have the comprehensive solutions manual in electronic format for > these following textbooks. They include complete solutions to all the > problems in the text, except where noted below. Payment is through > Paypal for the amount of $8.00 to $15.00 US dollars per item. > A Course in Game Theory by Osborne, Rubinstein > A Course in Algebraic Number Theory by Cohen > Adaptive Filter Theory, 4th Edition, by Haykin > Adaptive Control, 2nd. Ed., by Astrom, Wittenmark > Advanced Engineering Mathematics, 8th Editoin, by Erwin Kreyszig (even > solutions) > Advanced Engineering Mathematics, 9th Edition, by Erwin Kreyszig (even > solutions) > Advanced Macroeconomics, David Romer > Advanced Mathematical Concepts Precalculus With Applications by > Holliday [ISBN: 0028341759] > Advanced Modern Engineering Mathematics, 3rd Ed., by G. James > A First Course In Differential Equations, 7th Edition, by Zill, Cullen > Analog Integrated Circuit Design, 1st Ed., by Johns, Martin (text > ebook and solution manual) > Analysis and Design of Analog Integrated Circuits, 4th Ed., by Gray, > Hurst, Lewis, Meyer > Analytical Mechanics, 7th Edition, by Fowels, Cassiday > An Interactive Introduction to Mathematical Analysis, by Jonathan > Lewin > An Introduction to the Mathematics of Financial Derivatives, 2nd Ed., > by Neftci [ISBN: 0125153929] > Antenna Theory, 2nd Ed., by Balanis > Antennas for all Applications, 3rd Edition, Kraus, Marhefka > Applied Linear Statistical Models, 5th Ed., by Neter (Selected Sol.) > Applied Numerical Analysis, 6th Edition, by Gerald, Wheatley > Applied Numerical Methods with MATLAB for Engineers and Scientists, > 1st Ed,. by Chapra > Applied Statistics and Probability for Engineers, 3rd Ed., by > Montgomery, Runger (Selected Solutions) > Applied Strength of Materials, 4th Edition, by Mott > A Transition to Advanced Mathematics, 5th Edition, by Smith, Eggen, > Andre > Automatic Control Systems, 8th Edition, by Kuo, Golnaraghi > Basic Business Statistics: Concepts and Applications, 10th Ed., by > Berenson, Krehbiel, Levine (chap1-18) > Basic Engineering Circuit Analysis, 7th Ed., by J. David Irwin > Basic Engineering Circuit Analysis, 8th Ed., by J. David Irwin, Nelms > (Missing a chapter or 2) > Bioprocess Engineering Principles by Doran > Calculus Early Transcendental, 5th Ed., by James Stewart > Calculus - Early Transcendentals, 7th Ed., by Anton, Bivens, Davis > Calculus: Graphical, Numerical, Algebraic, 3rd Ed., Waits, Finney, > Demana, Kennedy > Calculus: Multivariable, 5th Edition, by James Stewart > Calculus: Single Variable, Early Transcendental, 5th Edition, by James > Stewart > Calculus, Single and Multivariable, 3rd Ed., by Hughes-Hallett, > McCallum > Calculus: Study and Solutions Guide, Vol. 1, 7th Ed., by Larson, > Hostetler, Edwards > Chemical and Engineering Thermodynamics, 3rd Ed., Stanley I. Sandler > Chemical Engineering Volume 1, Sixth Edition, by Richardson, Coulson, > Backhurst, Harker > Thornton > College Physics, Volume 1: 7th Edition, by Serway, Faugh > College Physics, Volume 2: 7th Edition, by Serway, Faughn > Communications Systems, 4th Ed., by Haykin > Communications Systems Engineering, 2nd Edition, by Proakis > Computational Techniques for Fluid Dynamics by Srinivas, Fletcher > Computer Networks, 4th Ed., by Andrew S. Tanenbaum > Computer Networks: A Systems Approach, 3rd Edition, by Davie > Control Systems Engineering, 4th Ed., by Norman Nise > Corporate Finance, 6th Edition, by Ross > C++ How to Program: Intro Object-Oriented Design with the UML, 3rd > Ed., by Deitel, Nieto > Data and Computer Communications, 8th Edition by Stallings > Database Management Systems, 3rd Ed., by Ramakrishnan, Gehrke (Sol. > for Chapters 2-21, odd only) > Design of Analog CMOS Integrated Circuits, 1st Edition, by Razavi > Design of Analysis of Experiments, 6th Edition, Montgomery (missing > chapter 6-8) > Design of Machinery, 3rd Ed by Robert L. Norton > Design With Operational Amplifiers and Analog Integrated Circuits, 2nd > Ed., by Sergio Franco > Design With Operational Amplifiers and Analog Integrated Circuits, 3rd > Ed., by Sergio Franco > Device Electronics for Integrated Circuits 3rd Edition by Muller > Differential Equations with Boundary Value Problems, 2nd Ed., by > Polking, Arnold > Digital And Analog Communication Systems 7th Ed., Leon W. Couch > Digital Communications, 4th Edition, by Proakis > Digital Communications: Fundamentals and Applications, 2nd Ed, Skylar > Digital Design, 4th Edition, by Mano, Ciletti > Digital Image Processing, 2nd Edition, by Gonzalez, Woods > Digital Integrated Circuits, 2nd Ed., by Rabaey (Solutions ONLY for > Chapters 3, 5, 6, 10) > Digital Signal Processing: A Computer Based Approach, 1st Ed., by > Mitra > Digital Signal Processing: A Computer Based Approach, 2nd Ed., by S. > Mitra > Digital Signal Processing: A Computer Based Approach, 3rd Ed., by S. > Mitra > Digital Signal Processing: Priciples, Algorithms and Applications, 3rd > Edition, by Proakis > Discrete Time Signal Processing, 2nd Edition, Oppenheim > Dynamics of Mechanical Systems by C.T.F. Ross > Econometric Analysis, 5th Edition, by Greene > Wooldridge > Econometrics of Financial Markets, by Adamek, Cambell, Lo, MacKinlay, > Viceira > Electrical Properties of Materials, 7th Ed., by D. Walsh, L. Solymar > Electric Circuits 6th Ed. by Nilsson > Electric Circuits 7th Ed. by Nilsson > Electric Machinery, 6th Ed., Fitzgerald, Kingsley, Umans > Electric Machinery Fundamentals, 4th Ed by Chapman > Electromagnetic Fields and Waves by Iskander (...) > Electronic Circuit Analysis, 2nd Ed., by Donald Neamen > Electronics, 2nd Ed., by Allan R. Hambley > Elementary Differential Equations, 8th Edition, by Boyce, DiPrima > (some odd/even) > Elementary Principles of Chemical Processes, 3rd Ed., by Felder, > Rousseau > Elements of Chemical Reaction Engineering, 3rd Ed., by H. Scott Fogler > Engineering and Chemical Thermodynamics, by Koretsky [ISBN: 0471385867] > (No sol. for chapt 6) > Engineering Circuit Analysis, 6th Edition, Hyat > Engineering Electromagnetics, 6th Ed W. Hayt, J. Buck > Engineering Fluids Mechanics 7th Edition by Crowe > Engineering Fluids Mechanics 8th Edition by Crowe > Engineering Mathematics, 4th Ed., by John Bird > Engineer Mechanics: Dynamics, 4th Ed., by Bedford > Engineering Mechanics: Dynamics, 10th Ed., by Russell C. Hibbeler > Engineering Mechanics: Dynamics 11th Ed. by Hibbeler > Engineering Mechanics: Dynamics 5th Ed. by Meriam, Kraige > Engineering Mechanics: Statics, 4th Edition - A. Bedford, Wallace > Fowler > Engineering Mechanics: Statics, 5th Ed., Meriam > Engineering Mechanics: Statics, 6th Ed., Meriam > Engineering Mechanics: Statics, 10th Ed., by Russell C. Hibbeler > Engineering Mechanics: Statics 11th Ed. by Hibbeler > Experiments with Economic Principles by Bergstrom, Miller > Feedback Control of Dynamic Systems, 4th Edition, by Powell, Emami- > Naeini > Financial Accounting, 4th Ed., by Libby, Short (Chap1-14) > Financial Accounting: An International Introduction, 2nd Ed., by > Alexander, Nobes > Finite Element Techniques in Structural Mechanics by Ross > Fluid Mechanics - 5th Edition by Frank M. White > Fluid Mechanics and Thermodynamics of Turbomachinery, 5th Ed., by S. > L. Dixon [ISBN: 0750678704] > Essentials of Fluid Mechanics: Fundamentals and Applications, 1st Ed., > by Cengel & Cimbala > Fluid Mechanics with Engineering Applications, 10th Edition, by > Finnemore > Fundamentals of Aerodynamics, 3rd Edition, by J. D. Anderson, Jr. > Fundamentals of Aerodynamics, 4th Edition, by Anderson > Fundamentals of Applied Electromagnetics, 2001 Media Edition, by Ulaby > Fundamentals of Applied Electromagnetics, 5th Ed., 2008 Media Edition, > by Ulaby > Fundamentals of Digital Logic with Verilog Design, 1st Edition, by > Brown, Vranesic > Fundamentals of Digital Logic with VHDL Design by Stephen Brown, > Zvonko Vranesic > Fundamentals of Electric Circuits, 2nd Edition, by Alexander > Fundamentals of Electromagnetics with Engineering Appls by Wentworth > Fundamentals of Fluid Mechanics, 5th Ed. by Munson, Young.. > Fundamentals of Heat and Mass Transfer, 4th Ed by Incropera... > Fundamentals of Heat and Mass Transfer, 5th Ed by Incropera... > Fundamentals of Heat and Mass Transfer, 6th Ed by Incropera... > Fundamentals of Logic Design, 5th Ed., by Roth Jr. > Fundamentals of Machine Component Design, 3rd Ed., by Juvinall > Fundamentals of Machine Component Design,- Hide quoted text - - Show quoted text -... read more ? hi can u send me the solution manual of introduction of chemical Faizan85@gmail.com === Subject: Re: Complete electronic solution manual in pdf ! Get it in hours! I need a solution for Data and Computer Communications, 8th Edition by Stallings i need it urgent if its possible for today === Subject: Re: JSH: SF Algorithm [JSH ] [... repetition ...] So if that theory is correct then you could factor an RSA sized number within 32 trials. [Tim Peters] >> Guess what? I won't spoil the surprise. Math people are not very bright. They are PRETEND bright. Pretend smart, but not real smart. >> So /that's/ the basis on which you want to be considered a major >> mathematician -- I knew there had to be something. [JSH] > Yuck. You're slimey. How so? You're forever pretending to knowledge you don't have, so it seems reasonable to me -- you are very much what you accuse others of being, in this and so many other respects. > I'm sure you're smiling to yourself as you read this. Ah, but like all your loyal critics, my only shot at recognition is to steal a sliver of your reflected glory. How could I not smile when the mere fact that you deigned to reply to me moved me to the top of all major search engines, if not directly to God Himself? > Maybe I will never crack the factoring problem. For example, after marcus_b gave you some really good insight into why other congruence of squares methods /were/ seen to be promising while yours was not, you ignored all the substance and lauched into a lunatic rant about racism. Acting insane doesn't disqualify you from discovering something of value, but it doesn't boost your odds of success either. > Maybe you people will win with lies, but win what? What lies? While few people bother with you anymore, at least marcus_b and Enrico told you more about this method du jour after a glance than you've been able to figure out despite your repeated claims to have finished a thorough analysis. They weren't lying. Although you did (yes, I realize you call your deliberate lies bravado -- doesn't change that they're lies). > Satisfaction at convincing people too stupid to care about the truth? I'm not trying to convince you of anything. > So many others have done that and done it better. But all you win is the death of the human species, and if that's what > you're after, then fine. Believe it or not, the human species isn't harmed in the slightest by rejecting another unremarkable (except for needless convolution) factoring method. > I'm beaten. Can't you be honest about anything, ever? You don't believe you're beaten at all. And more's the pity for you, BTW -- since you never truly believe you fail, you never take steps necessary to stop failing. > Let them die. I can do other things than continue to care. You're too far gone to have much real choice about anything anymore. > Somehow I lost when I didn't think that possible which is why it > happened. Maybe it's just a learning experience for me. Training for > bigger battles down the line when it matters. See? You have no choice but to follow your fantasy of inevitable glorious triumph. > The worst can now begin. And the real pain for the planet can now > start. You're going to increase your posting rate again? Cool. > I have stalled it as long as I can and now I'm just tired. But not too tired to rant like a loon about it -- visit, e.g., a nursing home to learn what tired means in reality. === Subject: Re: JSH: SF Algorithm Maybe you people will win with lies, but win what? course! > I'm beaten. Let them die. I can do other things than continue to > care. Come now, James. You know what to do. Increase an exponent! Introduce another meaningless independent variable! Start another blog! > Somehow I lost when I didn't think that possible which is why it > happened. Maybe it's just a learning experience for me. Training for > bigger battles down the line when it matters. Yes. Bigger battles than the death of the human species. Does this involve those space aliens with whom you communicate? > The worst can now begin. And the real pain for the planet can now > start. You're going to stop posting? That would be painful. -- So this post is a suggestion to those Bud Light people and if you've already done it or I forgot, as I'm drinking a few beers now as I post this, then sorry. -- James Harris === Subject: Re: JSH: SF Algorithm I'm sure you're smiling to yourself as you read this. > Maybe you people will win with lies, but win what? course! I'm beaten. Let them die. I can do other things than continue to > care. Come now, James. You know what to do. Increase an exponent! > Introduce another meaningless independent variable! Start > another blog! Somehow I lost when I didn't think that possible which is why it > happened. Maybe it's just a learning experience for me. Training for > bigger battles down the line when it matters. Yes. Bigger battles than the death of the human species. > Does this involve those space aliens with whom you communicate? The worst can now begin. And the real pain for the planet can now > start. You're going to stop posting? That would be painful. > -- > So this post is a suggestion to those Bud Light people and if you've > already done it or I forgot, as I'm drinking a few beers now as I > post this, then sorry. -- James Harris A real mathematican would probably solve the RSA challenge and unlike JSH, confront the stupid cocksuckers with the factors, and then throw forward a ing stupid idea to keep the cocksuckers occupied for years... I would sell out the algorithm to any agency as long as they would promise the stupid cocksuckers couldn't get hold on it muahahahha. Go eat a banana monkey ers... === Subject: Re: JSH: SF Algorithm > A real mathematican would probably solve the RSA challenge and unlike > JSH, confront the stupid cocksuckers with the factors, and then throw > forward a ing stupid idea to keep the cocksuckers occupied for > years... I would sell out the algorithm to any agency as long as they would > promise the stupid cocksuckers couldn't get hold on it muahahahha. Go eat a banana monkey ers... I understand that nowadays, decaf coffee is almost indistinguishable from the real thing. Just a thought. -- Destiny is a funny thing. Once I thought I was destined to become Emperor of Greenland, sole monarch over its 52,000 inhabitants. Then I thought I was destined to build a Polynesian longship in my garage. I was wrong then, but I've got it now. -- The Tick === Subject: Re: JSH: SF Algorithm > So it all works, except for one crucial little > fact: if you compute y^2 as noted above, y is not > necessarily an integer. That is where your method > breaks down. > And see, it is easier to just bypass everything > you have done and proceed as follows:... > Real easy. Much easier than all that messing > around with k and n and factoring of some surrogate. > So what's wrong with it? > Nothing, except it is nothing more than Fermat's > method! Crap! You mean I took all of my money out of the stock market for > nothing? But seriously though, surrogate factoring does have one big advantage > that you're overlooking : it got James some sitting time with God. (I > guess the SF does have a sense of humor after all.) I would wager that > that is probably *not* something you get with Fermat's method. M You people are so damn dumb. Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that x^2>y^2 so x=floor(sqrt(T)) is too small, so add one. Here's the corrected algorithm: Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: Note, still with x^2 = y^2 mod T, and k = 2x mod T. So then, with x = floor(sqrt(T))+1, k = 2x, n_min = floor((4(x+k-1) - 2k^2)/T) and n_max = floor(((x+k)^2 - 2k^2)/T) , you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, OK, let's try this. Let T = 403 = 13 * 31. x = floor(sqrt(403)) + 1 = 21. k = 2x = 42 nmin = -8 nmax = 1 y^2 = (x + k)^2 - 2*k^2 - nT So, letting n run from nmin to nmax, you get the following table: n y^2 y ----- ----- ----- -8 3665 60.54 -7 3262 57.11 -6 2859 53.47 -5 2456 45.31 -4 2053 45.31 -3 1650 40.62 -2 1247 35.31 -1 844 29.05 0 441 21.00 1 38 6.16 The only perfect square among the y^2 is when n = 0 and y = x = 21. This of course does not give a useful factorization. So adding 1 to your previous value for x does not help. Nothing in the range from nmin to nmax works. You're just floundering around. Clearly you have no theory that makes any sense. For this particular problem, Fermat's method also starts with x = floor(sqrt(T)) + 1 = 21. This gives x^2 mod T = 38. So increment x by 1; x = 22. This gives x^2 mod T = 81. So let y = 9. Note that x + y = 31 and x - y = 13, the two factors of T. Two steps, and Fermat has it. Plus, note that with your current method, you have totally abandoned surrogate factoring. You specify x, k, and n, and then compute y^2 = (x + k)^2 - 2*k^2 - nT and no factoring of any surrogate is involved at all. What you are doing is nothing but a wasteful inefficient approach to Fermat's method. > looking for a > perfect square and the theory says there must be at least one, Clearly the theory is wrong again. And please. Don't come back and say Oops! Let x = floor(sqrt(T)) + 2. Other counterexamples are lined up waiting. > if for > a prime factor p_1 of T with -(k+2xT)/T < m_1 < -(k + 2xp_1)/T the absolute value of m_1 is 1 or greater. Guess I should add that now x+y must have one prime factor of T, and > x- y must have another, if you find an integer y, for at LEAST ONE of > the n's so you take a gcd with T. So the theory says it will give you a solution to a difference of > squares, AND that for at least one of the n's that solution must non- > trivially factor T. Rightly so. It is easy to see it doesn't work from the example above. > That tests my theory here and intriguingly enough in case some > peoplethink there are a lot of n's, there are 32. 32 n's. So if that theory is correct then you could factor an RSA sized number > within 32 trials. > The theory is wrong. See above. > Math people are not very bright. They are PRETEND bright. > You spelled out a method. Clearly, trivially, it doesn't work. So who is not very bright ? > Pretend smart, but not real smart. > Who? Marcus. > James Harris === Subject: Re: JSH: SF Algorithm > So it all works, except for one crucial little > fact: if you compute y^2 as noted above, y is not > necessarily an integer. That is where your method > breaks down. > And see, it is easier to just bypass everything > you have done and proceed as follows:... > Real easy. Much easier than all that messing > around with k and n and factoring of some surrogate. > So what's wrong with it? > Nothing, except it is nothing more than Fermat's > method! > Crap! You mean I took all of my money out of the stock market for > nothing? > But seriously though, surrogate factoring does have one big advantage > that you're overlooking : it got James some sitting time with God. (I > guess the SF does have a sense of humor after all.) I would wager that > that is probably *not* something you get with Fermat's method. > M You people are so damn dumb. Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that x^2>y^2 so x=floor(sqrt(T)) is too small, so add one. Here's the corrected algorithm: Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: Note, still with x^2 = y^2 mod T, and k = 2x mod T. So then, with x = floor(sqrt(T))+1, k = 2x, n_min = floor((4(x+k-1) - 2k^2)/T) and n_max = floor(((x+k)^2 - 2k^2)/T) , you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, OK, let's try this. Let T = 403 = 13 * 31. x = floor(sqrt(403)) + 1 = 21. k = 2x = 42 nmin = -8 nmax = 1 y^2 = (x + k)^2 - 2*k^2 - nT So, letting n run from nmin to nmax, you get the following table: n y^2 y > ----- ----- ----- > -8 3665 60.54 > -7 3262 57.11 > -6 2859 53.47 > -5 2456 45.31 > -4 2053 45.31 > -3 1650 40.62 > -2 1247 35.31 > -1 844 29.05 > 0 441 21.00 > 1 38 6.16 The only perfect square among the y^2 is when n = 0 and > y = x = 21. This of course does not give a useful factorization. So adding 1 to your previous value for x does not help. > Nothing in the range from nmin to nmax works. You're just floundering around. Clearly you have no > theory that makes any sense. > Yeah your suggestion it turns out was stupid. It doesn't work. Hey, for your benefit, for a while I thought it might fly. Can you figure out why your idea does not work? Hint: Not every x will work with x^2 = y^2 mod T, if abs(x)>abs(y) is why. James Harris === Subject: Re: JSH: SF Algorithm > So it all works, except for one crucial little > fact: if you compute y^2 as noted above, y is not > necessarily an integer. That is where your method > breaks down. > And see, it is easier to just bypass everything > you have done and proceed as follows:... > Real easy. Much easier than all that messing > around with k and n and factoring of some surrogate. > So what's wrong with it? > Nothing, except it is nothing more than Fermat's > method! > Crap! You mean I took all of my money out of the stock market for > nothing? > But seriously though, surrogate factoring does have one big advantage > that you're overlooking : it got James some sitting time with God. (I > guess the SF does have a sense of humor after all.) I would wager that > that is probably *not* something you get with Fermat's method. > M > You people are so damn dumb. > Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that > x^2>y^2 > so x=floor(sqrt(T)) is too small, so add one. > Here's the corrected algorithm: > Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: > Note, still with x^2 = y^2 mod T, and k = 2x mod T. > So then, with x = floor(sqrt(T))+1, k = 2x, > n_min = floor((4(x+k-1) - 2k^2)/T) > and > n_max = floor(((x+k)^2 - 2k^2)/T) , > you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, OK, let's try this. Let T = 403 = 13 * 31. x = floor(sqrt(403)) + 1 = 21. k = 2x = 42 nmin = -8 nmax = 1 y^2 = (x + k)^2 - 2*k^2 - nT So, letting n run from nmin to nmax, you get the following table: n y^2 y > ----- ----- ----- > -8 3665 60.54 > -7 3262 57.11 > -6 2859 53.47 > -5 2456 45.31 > -4 2053 45.31 > -3 1650 40.62 > -2 1247 35.31 > -1 844 29.05 > 0 441 21.00 > 1 38 6.16 The only perfect square among the y^2 is when n = 0 and > y = x = 21. This of course does not give a useful factorization. So adding 1 to your previous value for x does not help. > Nothing in the range from nmin to nmax works. You're just floundering around. Clearly you have no > theory that makes any sense. Yeah your suggestion it turns out was stupid. It doesn't work. > I'm crushed. But you must have been equally stupid to think it would work. > Hey, for your benefit, for a while I thought it might fly. > Now I'm elated again. > Can you figure out why your idea does not work? > Of course. > Hint: Not every x will work with x^2 = y^2 mod T, if abs(x)>abs(y) is > why. > Obvious from what I posted. Yes, there are examples where your method may factor even if x^2 = y^2 mod T is not true. You considered T = 606007. You factored it after 26 or so probes. Let's compare that with Fermat's method. Fermat starts with x = 779. 779^2 mod T is not a square. Try x = 780. Similar. Try x = 781. Similar. Try x = 782. Similar. Try x = 783. Same result. Try x = 784. 784^2 mod T = 93^2. So let y = 93. Note that x + y = 784 + 93 = 877 and x - y = 691. And T = 877 * 691. So Fermat's method works after just 6 probes. This is substantially (and typically) faster than your method. I would also point out that you redefined nmin. Previously you said it should be: nmin = int((4*(x + k - 1) - 2*k*k)/T) That formula gives nmin = -7 (when x = 778, k = 2x). You computed it to be -15. What formula for that are you using now? Marcus. > James Harris === Subject: Re: JSH: SF Algorithm > So it all works, except for one crucial little > fact: if you compute y^2 as noted above, y is not > necessarily an integer. That is where your method > breaks down. > And see, it is easier to just bypass everything > you have done and proceed as follows:... > Real easy. Much easier than all that messing > around with k and n and factoring of some surrogate. > So what's wrong with it? > Nothing, except it is nothing more than Fermat's > method! > Crap! You mean I took all of my money out of the stock market for > nothing? > But seriously though, surrogate factoring does have one big advantage > that you're overlooking : it got James some sitting time with God. (I > guess the SF does have a sense of humor after all.) I would wager that > that is probably *not* something you get with Fermat's method. > M > You people are so damn dumb. > Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that > x^2>y^2 > so x=floor(sqrt(T)) is too small, so add one. > Here's the corrected algorithm: > Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: > Note, still with x^2 = y^2 mod T, and k = 2x mod T. > So then, with x = floor(sqrt(T))+1, k = 2x, > n_min = floor((4(x+k-1) - 2k^2)/T) > and > n_max = floor(((x+k)^2 - 2k^2)/T) , > you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, > OK, let's try this. Let T = 403 = 13 * 31. > x = floor(sqrt(403)) + 1 = 21. > k = 2x = 42 > nmin = -8 > nmax = 1 > y^2 = (x + k)^2 - 2*k^2 - nT > So, letting n run from nmin to nmax, you get the following table: > n y^2 y > ----- ----- ----- > -8 3665 60.54 > -7 3262 57.11 > -6 2859 53.47 > -5 2456 45.31 > -4 2053 45.31 > -3 1650 40.62 > -2 1247 35.31 > -1 844 29.05 > 0 441 21.00 > 1 38 6.16 > The only perfect square among the y^2 is when n = 0 and > y = x = 21. This of course does not give a useful factorization. > So adding 1 to your previous value for x does not help. > Nothing in the range from nmin to nmax works. > You're just floundering around. Clearly you have no > theory that makes any sense. Yeah your suggestion it turns out was stupid. It doesn't work. I'm crushed. But you must have been equally stupid to think > it would work. Hey, for your benefit, for a while I thought it might fly. Now I'm elated again. > Brainstorming. That means to me let it fly and see later why it's a stupid idea. Your idea turned out to be stupid, but no reason not to let it go forth, you know? That's modern problem solving technique. > Can you figure out why your idea does not work? Of course. Hint: Not every x will work with x^2 = y^2 mod T, if abs(x)>abs(y) is > why. Obvious from what I posted. Yes, there are examples where your method may factor even > if x^2 = y^2 mod T is not true. You considered T = 606007. You factored it after 26 or so probes. Let's compare that with Fermat's method. Fermat starts with x = 779. 779^2 mod T is not a square. Try x = 780. Similar. Try x = 781. Similar. Try x = 782. Similar. Try x = 783. Same result. Try x = 784. 784^2 mod T = 93^2. So let y = 93. > Note that x + y = 784 + 93 = 877 and x - y = 691. And T = 877 * 691. So Fermat's method works after just 6 probes. This is substantially (and typically) faster than your method. > Except the guarantee is that with my surrogate factoring you'll get a factorization within 16 surrogates--no matter how big the number. So that average drops precipitously as you get bigger and bigger numbers until its not even worth comparing. The danger in my surrogate factoring is not with dinky numbers but with RSA sized numbers. With those numbers there is no comparison between surrogate factoring and any other technique. Even the number field sieve will go through millions of possibles while surrogate factoring trots through far fewer needing only 16 surrogates. Some may wonder why I say 16 versus 32, which I said earlier, when 32 comes in with x = floor(sqrt(2T)) when it turns out you only need x = floor(sqrt(T)), and 16 surrogates. So talk about probes all you want with tiny numbers but the surrogate factoring theory says that by factoring 16 surrogates you can factor ANY target composite T out to infinity. To infinity. The problem is factoring the 16. James Harris === Subject: Re: JSH: SF Algorithm > So it all works, except for one crucial little > fact: if you compute y^2 as noted above, y is not > necessarily an integer. That is where your method > breaks down. > And see, it is easier to just bypass everything > you have done and proceed as follows:... > Real easy. Much easier than all that messing > around with k and n and factoring of some surrogate. > So what's wrong with it? > Nothing, except it is nothing more than Fermat's > method! > Crap! You mean I took all of my money out of the stock market for > nothing? > But seriously though, surrogate factoring does have one big advantage > that you're overlooking : it got James some sitting time with God. (I > guess the SF does have a sense of humor after all.) I would wager that > that is probably *not* something you get with Fermat's method. > M > You people are so damn dumb. > Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that > x^2>y^2 > so x=floor(sqrt(T)) is too small, so add one. > Here's the corrected algorithm: > Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: > Note, still with x^2 = y^2 mod T, and k = 2x mod T. > So then, with x = floor(sqrt(T))+1, k = 2x, > n_min = floor((4(x+k-1) - 2k^2)/T) > and > n_max = floor(((x+k)^2 - 2k^2)/T) , > you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, > OK, let's try this. Let T = 403 = 13 * 31. > x = floor(sqrt(403)) + 1 = 21. > k = 2x = 42 > nmin = -8 > nmax = 1 > y^2 = (x + k)^2 - 2*k^2 - nT > So, letting n run from nmin to nmax, you get the following table: > n y^2 y > ----- ----- ----- > -8 3665 60.54 > -7 3262 57.11 > -6 2859 53.47 > -5 2456 45.31 > -4 2053 45.31 > -3 1650 40.62 > -2 1247 35.31 > -1 844 29.05 > 0 441 21.00 > 1 38 6.16 > The only perfect square among the y^2 is when n = 0 and > y = x = 21. This of course does not give a useful factorization. > So adding 1 to your previous value for x does not help. > Nothing in the range from nmin to nmax works. > You're just floundering around. Clearly you have no > theory that makes any sense. > Yeah your suggestion it turns out was stupid. It doesn't work. I'm crushed. But you must have been equally stupid to think > it would work. > Hey, for your benefit, for a while I thought it might fly. Now I'm elated again. Brainstorming. That means to me let it fly and see later why it's a stupid idea. Your idea turned out to be stupid, but no reason not to let it go > forth, you know? That's modern problem solving technique. > So how come Fermat's method, which is regarded as fairly crude, was almost 5 times faster than your method? > Can you figure out why your idea does not work? Of course. > Hint: Not every x will work with x^2 = y^2 mod T, if abs(x)>abs(y) is > why. Obvious from what I posted. Yes, there are examples where your method may factor even > if x^2 = y^2 mod T is not true. You considered T = 606007. You factored it after 26 or so probes. Let's compare that with Fermat's method. Fermat starts with x = 779. 779^2 mod T is not a square. Try x = 780. Similar. Try x = 781. Similar. Try x = 782. Similar. Try x = 783. Same result. Try x = 784. 784^2 mod T = 93^2. So let y = 93. > Note that x + y = 784 + 93 = 877 and x - y = 691. And T = 877 * 691. So Fermat's method works after just 6 probes. This is substantially (and typically) faster than your method. Except the guarantee is that with my surrogate factoring you'll get a > factorization within 16 surrogates--no matter how big the number. > Let's see your proof of that. > So that average drops precipitously as you get bigger and bigger > numbers until its not even worth comparing. > That too. I think you are bluffing. > The danger in my surrogate factoring is not with dinky numbers but > with RSA sized numbers. > How do you know that? Why does your proof - assuming you have one - work for big numbers but not for small ones? > With those numbers there is no comparison between surrogate factoring > and any other technique. > Big talk. Prove it. > Even the number field sieve will go through millions of possibles > while surrogate factoring trots through far fewer needing only 16 > surrogates. > Where's your proof? > Some may wonder why I say 16 versus 32, which I said earlier, when 32 > comes in with x = floor(sqrt(2T)) when it turns out you only need x = floor(sqrt(T)), and 16 surrogates. So talk about probes all you want with tiny numbers but the > surrogate factoring theory says that by factoring 16 surrogates you > can factor ANY target composite T out to infinity. > Where's the proof? > To infinity. The problem is factoring the 16. > I bet not. Let's see the details of your computations for T = 10658390671. Marcus. > James Harris- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - === Subject: Re: JSH: SF Algorithm > So it all works, except for one crucial little > fact: if you compute y^2 as noted above, y is not > necessarily an integer. That is where your method > breaks down. > And see, it is easier to just bypass everything > you have done and proceed as follows:... > Real easy. Much easier than all that messing > around with k and n and factoring of some surrogate. > So what's wrong with it? > Nothing, except it is nothing more than Fermat's > method! > Crap! You mean I took all of my money out of the stock market for > nothing? > But seriously though, surrogate factoring does have one big advantage > that you're overlooking : it got James some sitting time with God. (I > guess the SF does have a sense of humor after all.) I would wager that > that is probably *not* something you get with Fermat's method. > M > You people are so damn dumb. > Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that > x^2>y^2 > so x=floor(sqrt(T)) is too small, so add one. > Here's the corrected algorithm: > Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: > Note, still with x^2 = y^2 mod T, and k = 2x mod T. > So then, with x = floor(sqrt(T))+1, k = 2x, > n_min = floor((4(x+k-1) - 2k^2)/T) > and > n_max = floor(((x+k)^2 - 2k^2)/T) , > you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, > OK, let's try this. Let T = 403 = 13 * 31. > x = floor(sqrt(403)) + 1 = 21. > k = 2x = 42 > nmin = -8 > nmax = 1 > y^2 = (x + k)^2 - 2*k^2 - nT > So, letting n run from nmin to nmax, you get the following table: > n y^2 y > ----- ----- ----- > -8 3665 60.54 > -7 3262 57.11 > -6 2859 53.47 > -5 2456 45.31 > -4 2053 45.31 > -3 1650 40.62 > -2 1247 35.31 > -1 844 29.05 > 0 441 21.00 > 1 38 6.16 > The only perfect square among the y^2 is when n = 0 and > y = x = 21. This of course does not give a useful factorization. > So adding 1 to your previous value for x does not help. > Nothing in the range from nmin to nmax works. > You're just floundering around. Clearly you have no > theory that makes any sense. > Yeah your suggestion it turns out was stupid. It doesn't work. > I'm crushed. But you must have been equally stupid to think > it would work. > Hey, for your benefit, for a while I thought it might fly. > Now I'm elated again. Brainstorming. That means to me let it fly and see later why it's a stupid idea. Your idea turned out to be stupid, but no reason not to let it go > forth, you know? That's modern problem solving technique. So how come Fermat's method, which is regarded as > fairly crude, was almost 5 times faster than > your method? > Can you figure out why your idea does not work? > Of course. > Hint: Not every x will work with x^2 = y^2 mod T, if abs(x)>abs(y) is > why. > Obvious from what I posted. > Yes, there are examples where your method may factor even > if x^2 = y^2 mod T is not true. > You considered T = 606007. > You factored it after 26 or so probes. > Let's compare that with Fermat's method. > Fermat starts with x = 779. 779^2 mod T is not a square. > Try x = 780. Similar. > Try x = 781. Similar. > Try x = 782. Similar. > Try x = 783. Same result. > Try x = 784. 784^2 mod T = 93^2. So let y = 93. > Note that > x + y = 784 + 93 = 877 and x - y = 691. > And T = 877 * 691. > So Fermat's method works after just 6 probes. > This is substantially (and typically) faster than your method. Except the guarantee is that with my surrogate factoring you'll get a > factorization within 16 surrogates--no matter how big the number. Let's see your proof of that. So that average drops precipitously as you get bigger and bigger > numbers until its not even worth comparing. That too. I think you are bluffing. The danger in my surrogate factoring is not with dinky numbers but > with RSA sized numbers. How do you know that? Why does your proof - assuming > you have one - work for big numbers but not for > small ones? With those numbers there is no comparison between surrogate factoring > and any other technique. Big talk. Prove it. Even the number field sieve will go through millions of possibles > while surrogate factoring trots through far fewer needing only 16 > surrogates. Where's your proof? Some may wonder why I say 16 versus 32, which I said earlier, when 32 > comes in with x = floor(sqrt(2T)) when it turns out you only need x = floor(sqrt(T)), and 16 surrogates. So talk about probes all you want with tiny numbers but the > surrogate factoring theory says that by factoring 16 surrogates you > can factor ANY target composite T out to infinity. Where's the proof? To infinity. The problem is factoring the 16. I bet not. Let's see the details of your > computations for T = 10658390671. Marcus. James Harris- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - T 10658390671 x = floor(sqrt(T)) +1 103240 k = 2x 206480 n_max = floor(((x+k)^2 - 2k^2)/T) 1 n_min = floor((4(x+k-1) - 2k^2)/T) -8 n = 1 (x+k)^2 95926478400 S =2k^2 + nT 95926371471 F2 = Factor of S 309393 F1 = Factor of S 310047 Gcd((F2 - k, T) 102913 Gcd((F1 - k, T) 103567 X=(F1+F2-2K)/2 103240 Y=(F2-F1)/2 -327 mod(X^2,T) 106929 mod(Y^2,T) 106929 Enrico === Subject: Re: JSH: SF Algorithm > So it all works, except for one crucial little > fact: if you compute y^2 as noted above, y is not > necessarily an integer. That is where your method > breaks down. > And see, it is easier to just bypass everything > you have done and proceed as follows:... > Real easy. Much easier than all that messing > around with k and n and factoring of some surrogate. > So what's wrong with it? > Nothing, except it is nothing more than Fermat's > method! > Crap! You mean I took all of my money out of the stock market for > nothing? > But seriously though, surrogate factoring does have one big advantage > that you're overlooking : it got James some sitting time with God. (I > guess the SF does have a sense of humor after all.) I would wager that > that is probably *not* something you get with Fermat's method. > M > You people are so damn dumb. > Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that > x^2>y^2 > so x=floor(sqrt(T)) is too small, so add one. > Here's the corrected algorithm: > Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: > Note, still with x^2 = y^2 mod T, and k = 2x mod T. > So then, with x = floor(sqrt(T))+1, k = 2x, > n_min = floor((4(x+k-1) - 2k^2)/T) > and > n_max = floor(((x+k)^2 - 2k^2)/T) , > you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, > OK, let's try this. Let T = 403 = 13 * 31. > x = floor(sqrt(403)) + 1 = 21. > k = 2x = 42 > nmin = -8 > nmax = 1 > y^2 = (x + k)^2 - 2*k^2 - nT > So, letting n run from nmin to nmax, you get the following table: > n y^2 y > ----- ----- ----- > -8 3665 60.54 > -7 3262 57.11 > -6 2859 53.47 > -5 2456 45.31 > -4 2053 45.31 > -3 1650 40.62 > -2 1247 35.31 > -1 844 29.05 > 0 441 21.00 > 1 38 6.16 > The only perfect square among the y^2 is when n = 0 and > y = x = 21. This of course does not give a useful factorization. > So adding 1 to your previous value for x does not help. > Nothing in the range from nmin to nmax works. > You're just floundering around. Clearly you have no > theory that makes any sense. > Yeah your suggestion it turns out was stupid. It doesn't work. > I'm crushed. But you must have been equally stupid to think > it would work. > Hey, for your benefit, for a while I thought it might fly. > Now I'm elated again. > Brainstorming. > That means to me let it fly and see later why it's a stupid idea. > Your idea turned out to be stupid, but no reason not to let it go > forth, you know? > That's modern problem solving technique. So how come Fermat's method, which is regarded as > fairly crude, was almost 5 times faster than > your method? > Can you figure out why your idea does not work? > Of course. > Hint: Not every x will work with x^2 = y^2 mod T, if abs(x)>abs(y) is > why. > Obvious from what I posted. > Yes, there are examples where your method may factor even > if x^2 = y^2 mod T is not true. > You considered T = 606007. > You factored it after 26 or so probes. > Let's compare that with Fermat's method. > Fermat starts with x = 779. 779^2 mod T is not a square. > Try x = 780. Similar. > Try x = 781. Similar. > Try x = 782. Similar. > Try x = 783. Same result. > Try x = 784. 784^2 mod T = 93^2. So let y = 93. > Note that > x + y = 784 + 93 = 877 and x - y = 691. > And T = 877 * 691. > So Fermat's method works after just 6 probes. > This is substantially (and typically) faster than your method. > Except the guarantee is that with my surrogate factoring you'll get a > factorization within 16 surrogates--no matter how big the number. Let's see your proof of that. > So that average drops precipitously as you get bigger and bigger > numbers until its not even worth comparing. That too. I think you are bluffing. > The danger in my surrogate factoring is not with dinky numbers but > with RSA sized numbers. How do you know that? Why does your proof - assuming > you have one - work for big numbers but not for > small ones? > With those numbers there is no comparison between surrogate factoring > and any other technique. Big talk. Prove it. > Even the number field sieve will go through millions of possibles > while surrogate factoring trots through far fewer needing only 16 > surrogates. Where's your proof? > Some may wonder why I say 16 versus 32, which I said earlier, when 32 > comes in with > x = floor(sqrt(2T)) > when it turns out you only need x = floor(sqrt(T)), and 16 surrogates. > So talk about probes all you want with tiny numbers but the > surrogate factoring theory says that by factoring 16 surrogates you > can factor ANY target composite T out to infinity. Where's the proof? > To infinity. > The problem is factoring the 16. I bet not. Let's see the details of your > computations for T = 10658390671. Marcus. > James Harris- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - T 10658390671 > x = floor(sqrt(T)) +1 103240 > k = 2x 206480 > n_max = floor(((x+k)^2 - 2k^2)/T) 1 > n_min = floor((4(x+k-1) - 2k^2)/T) -8 > n = 1 > (x+k)^2 95926478400 > S =2k^2 + nT 95926371471 > F2 = Factor of S 309393 > F1 = Factor of S 310047 > Gcd((F2 - k, T) 102913 > Gcd((F1 - k, T) 103567 > X=(F1+F2-2K)/2 103240 > Y=(F2-F1)/2 -327 > mod(X^2,T) 106929 > mod(Y^2,T) 106929 Enrico I happened by chance to have picked an example which can be factored by Fermat's method in one step. I wonder if you might try applying the Harris algorithm to: T = 9524208139 Marcus. === Subject: Re: JSH: SF Algorithm > So it all works, except for one crucial little > fact: if you compute y^2 as noted above, y is not > necessarily an integer. That is where your method > breaks down. > And see, it is easier to just bypass everything > you have done and proceed as follows:... > Real easy. Much easier than all that messing > around with k and n and factoring of some surrogate. > So what's wrong with it? > Nothing, except it is nothing more than Fermat's > method! > Crap! You mean I took all of my money out of the stock market for > nothing? > But seriously though, surrogate factoring does have one big advantage > that you're overlooking : it got James some sitting time with God. (I > guess the SF does have a sense of humor after all.) I would wager that > that is probably *not* something you get with Fermat's method. > M > You people are so damn dumb. > Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that > x^2>y^2 > so x=floor(sqrt(T)) is too small, so add one. > Here's the corrected algorithm: > Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: > Note, still with x^2 = y^2 mod T, and k = 2x mod T. > So then, with x = floor(sqrt(T))+1, k = 2x, > n_min = floor((4(x+k-1) - 2k^2)/T) > and > n_max = floor(((x+k)^2 - 2k^2)/T) , > you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, > OK, let's try this. Let T = 403 = 13 * 31. > x = floor(sqrt(403)) + 1 = 21. > k = 2x = 42 > nmin = -8 > nmax = 1 > y^2 = (x + k)^2 - 2*k^2 - nT > So, letting n run from nmin to nmax, you get the following table: > n y^2 y > ----- ----- ----- > -8 3665 60.54 > -7 3262 57.11 > -6 2859 53.47 > -5 2456 45.31 > -4 2053 45.31 > -3 1650 40.62 > -2 1247 35.31 > -1 844 29.05 > 0 441 21.00 > 1 38 6.16 > The only perfect square among the y^2 is when n = 0 and > y = x = 21. This of course does not give a useful factorization. > So adding 1 to your previous value for x does not help. > Nothing in the range from nmin to nmax works. > You're just floundering around. Clearly you have no > theory that makes any sense. > Yeah your suggestion it turns out was stupid. It doesn't work. > I'm crushed. But you must have been equally stupid to think > it would work. > Hey, for your benefit, for a while I thought it might fly. > Now I'm elated again. > Brainstorming. > That means to me let it fly and see later why it's a stupid idea. > Your idea turned out to be stupid, but no reason not to let it go > forth, you know? > That's modern problem solving technique. So how come Fermat's method, which is regarded as > fairly crude, was almost 5 times faster than > your method? > Can you figure out why your idea does not work? > Of course. > Hint: Not every x will work with x^2 = y^2 mod T, if abs(x)>abs(y) is > why. > Obvious from what I posted. > Yes, there are examples where your method may factor even > if x^2 = y^2 mod T is not true. > You considered T = 606007. > You factored it after 26 or so probes. > Let's compare that with Fermat's method. > Fermat starts with x = 779. 779^2 mod T is not a square. > Try x = 780. Similar. > Try x = 781. Similar. > Try x = 782. Similar. > Try x = 783. Same result. > Try x = 784. 784^2 mod T = 93^2. So let y = 93. > Note that > x + y = 784 + 93 = 877 and x - y = 691. > And T = 877 * 691. > So Fermat's method works after just 6 probes. > This is substantially (and typically) faster than your method. > Except the guarantee is that with my surrogate factoring you'll get a > factorization within 16 surrogates--no matter how big the number. Let's see your proof of that. > So that average drops precipitously as you get bigger and bigger > numbers until its not even worth comparing. That too. I think you are bluffing. > The danger in my surrogate factoring is not with dinky numbers but > with RSA sized numbers. How do you know that? Why does your proof - assuming > you have one - work for big numbers but not for > small ones? > With those numbers there is no comparison between surrogate factoring > and any other technique. Big talk. Prove it. > Even the number field sieve will go through millions of possibles > while surrogate factoring trots through far fewer needing only 16 > surrogates. Where's your proof? > Some may wonder why I say 16 versus 32, which I said earlier, when 32 > comes in with > x = floor(sqrt(2T)) > when it turns out you only need x = floor(sqrt(T)), and 16 surrogates. > So talk about probes all you want with tiny numbers but the > surrogate factoring theory says that by factoring 16 surrogates you > can factor ANY target composite T out to infinity. Where's the proof? > To infinity. > The problem is factoring the 16. I bet not. Let's see the details of your > computations for T = 10658390671. Marcus. > James Harris- Hide quoted text - > - Show quoted text -- Hide quoted text - > - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - T 10658390671 > x = floor(sqrt(T)) +1 103240 > k = 2x 206480 > n_max = floor(((x+k)^2 - 2k^2)/T) 1 > n_min = floor((4(x+k-1) - 2k^2)/T) -8 > n = 1 > (x+k)^2 95926478400 > S =2k^2 + nT 95926371471 > F2 = Factor of S 309393 > F1 = Factor of S 310047 > Gcd((F2 - k, T) 102913 > Gcd((F1 - k, T) 103567 > X=(F1+F2-2K)/2 103240 > Y=(F2-F1)/2 -327 > mod(X^2,T) 106929 > mod(Y^2,T) 106929 Enrico- Hide quoted text - - Show quoted text - Interesting. However Harris has reverted to x = floor(sqrt(T)) instead of x = floor(sqrt(T)) + 1. Your calculation shows that this is an instance of a Fermat factorization anyway which is attained in one step. Marcus. === Subject: Re: JSH: SF Algorithm > Except the guarantee is that with my surrogate factoring you'll get a > factorization within 16 surrogates--no matter how big the number. BUT ... (see later) > So that average drops precipitously as you get bigger and bigger > numbers until its not even worth comparing. > > The danger in my surrogate factoring is not with dinky numbers but > with RSA sized numbers. > > With those numbers there is no comparison between surrogate factoring > and any other technique. No comparison in poor performance you mean? See above for more details. > Even the number field sieve will go through millions of possibles > while surrogate factoring trots through far fewer needing only 16 > surrogates. See above. Also, I don't quite see how only 16 surrogates are needed. > Some may wonder why I say 16 versus 32, which I said earlier, when 32 > comes in with > > x = floor(sqrt(2T)) > > when it turns out you only need x = floor(sqrt(T)), and 16 surrogates. > > So talk about probes all you want with tiny numbers but the > surrogate factoring theory says that by factoring 16 surrogates you > can factor ANY target composite T out to infinity. See above. > To infinity. > > The problem is factoring the 16. IIUC, the surrogates are, on average, *larger* than the current number. So to factor those larger numbers, you need to run this program recursively (after all, you claim, it is the only one capable of handling large numbers easily). Those 16 numbers need to factor 16 surrogates each, for a total of 256. But wait! Those 256 need factoring, so you really have to factor 4096 more. There's more! those need to be factored, yielding 65536 more. And it continues until a surrogate suddenly gets a magic number and can terminate its recursive branch. Hmm... You are basing your *entire* claim of surrogate factoring's prowess on the ability to factor the surrogates instantly through a magic oracle. I think I'll steal your oracle and just use that to factor T, thank you very much. > > > James Harris > > > > === Subject: Re: JSH: SF Algorithm >That tests my theory here and intriguingly enough in case some people >think there are a lot of n's, there are 32. So if that theory is correct then you could factor an RSA sized number >with 32 trials. So why babble about anything else? You can shoot down my latest theory with a single counterexample. Not wishing to babble, 606007 is such a counterexample, it needs to go through 50 n's before it finds a factor. Below is the output from my instrumented version of your program. This was the x = ceil(sqrt(T)) version. The previous version with x = floor(sqrt(T)) did even worse, needing 58 different n's before finding a factor. You appear to have been right to make the change, at least in this case. rossum *** Begin Program Output *** target = 606007 k = 1558, x = 779 trying n = -7, surrogate = 612679 trying n = -6, surrogate = 1218686 trying n = -5, surrogate = 1824693 trying n = -4, surrogate = 2430700 trying n = -3, surrogate = 3036707 trying n = -2, surrogate = 3642714 trying n = -1, surrogate = 4248721 trying n = 0, surrogate = 4854728 trying n = 1, surrogate = 5460735 k = 1560, x = 780 trying n = -8, surrogate = 19144 trying n = -7, surrogate = 625151 trying n = -6, surrogate = 1231158 trying n = -5, surrogate = 1837165 trying n = -4, surrogate = 2443172 trying n = -3, surrogate = 3049179 trying n = -2, surrogate = 3655186 trying n = -1, surrogate = 4261193 trying n = 0, surrogate = 4867200 trying n = 1, surrogate = 5473207 k = 1562, x = 781 trying n = -8, surrogate = 31632 trying n = -7, surrogate = 637639 trying n = -6, surrogate = 1243646 trying n = -5, surrogate = 1849653 trying n = -4, surrogate = 2455660 trying n = -3, surrogate = 3061667 trying n = -2, surrogate = 3667674 trying n = -1, surrogate = 4273681 trying n = 0, surrogate = 4879688 trying n = 1, surrogate = 5485695 k = 1564, x = 782 trying n = -8, surrogate = 44136 trying n = -7, surrogate = 650143 trying n = -6, surrogate = 1256150 trying n = -5, surrogate = 1862157 trying n = -4, surrogate = 2468164 trying n = -3, surrogate = 3074171 trying n = -2, surrogate = 3680178 trying n = -1, surrogate = 4286185 trying n = 0, surrogate = 4892192 trying n = 1, surrogate = 5498199 k = 1566, x = 783 trying n = -8, surrogate = 56656 trying n = -7, surrogate = 662663 trying n = -6, surrogate = 1268670 trying n = -5, surrogate = 1874677 trying n = -4, surrogate = 2480684 trying n = -3, surrogate = 3086691 trying n = -2, surrogate = 3692698 trying n = -1, surrogate = 4298705 trying n = 0, surrogate = 4904712 trying n = 1, surrogate = 5510719 k = 1568, x = 784 trying n = -8, surrogate = 69192 BINGO! 186 succeeded factor = 691 606007 = 691 * 877 2113 probes total. 50 values of n tried. *** End Program Output *** === Subject: Re: JSH: SF Algorithm think there are a lot of n's, there are 32. So if that theory is correct then you could factor an RSA sized number >with 32 trials. So why babble about anything else? You can shoot down my latest theory with a single counterexample. Not wishing to babble, 606007 is such a counterexample, it needs to go > through 50 n's before it finds a factor. Below is the output from my > instrumented version of your program. This was the x = ceil(sqrt(T)) version. The previous version with x = > floor(sqrt(T)) did even worse, needing 58 different n's before finding > a factor. You appear to have been right to make the change, at least > in this case. rossum *** Begin Program Output *** target = 606007 k = 1558, x = 779 > trying n = -7, surrogate = 612679 > trying n = -6, surrogate = 1218686 > trying n = -5, surrogate = 1824693 > trying n = -4, surrogate = 2430700 > trying n = -3, surrogate = 3036707 > trying n = -2, surrogate = 3642714 > trying n = -1, surrogate = 4248721 > trying n = 0, surrogate = 4854728 > trying n = 1, surrogate = 5460735 k = 1560, x = 780 > trying n = -8, surrogate = 19144 > trying n = -7, surrogate = 625151 > trying n = -6, surrogate = 1231158 > trying n = -5, surrogate = 1837165 > trying n = -4, surrogate = 2443172 > trying n = -3, surrogate = 3049179 > trying n = -2, surrogate = 3655186 > trying n = -1, surrogate = 4261193 > trying n = 0, surrogate = 4867200 > trying n = 1, surrogate = 5473207 k = 1562, x = 781 > trying n = -8, surrogate = 31632 > trying n = -7, surrogate = 637639 > trying n = -6, surrogate = 1243646 > trying n = -5, surrogate = 1849653 > trying n = -4, surrogate = 2455660 > trying n = -3, surrogate = 3061667 > trying n = -2, surrogate = 3667674 > trying n = -1, surrogate = 4273681 > trying n = 0, surrogate = 4879688 > trying n = 1, surrogate = 5485695 k = 1564, x = 782 > trying n = -8, surrogate = 44136 > trying n = -7, surrogate = 650143 > trying n = -6, surrogate = 1256150 > trying n = -5, surrogate = 1862157 > trying n = -4, surrogate = 2468164 > trying n = -3, surrogate = 3074171 > trying n = -2, surrogate = 3680178 > trying n = -1, surrogate = 4286185 > trying n = 0, surrogate = 4892192 > trying n = 1, surrogate = 5498199 k = 1566, x = 783 > trying n = -8, surrogate = 56656 > trying n = -7, surrogate = 662663 > trying n = -6, surrogate = 1268670 > trying n = -5, surrogate = 1874677 > trying n = -4, surrogate = 2480684 > trying n = -3, surrogate = 3086691 > trying n = -2, surrogate = 3692698 > trying n = -1, surrogate = 4298705 > trying n = 0, surrogate = 4904712 > trying n = 1, surrogate = 5510719 k = 1568, x = 784 > trying n = -8, surrogate = 69192 > BINGO! 186 succeeded factor = 691 > 606007 = 691 * 877 > 2113 probes total. > 50 values of n tried. *** End Program Output *** Counter example using surrogate Factoring T 606007 x = floor(sqrt(T))+1 779 k = 2x 1558 n_max = floor(((x+k)^2 - 2k^2)/T) 1 n_min = floor((4(x+k-1) - 2k^2)/T) -8 Forced n -17 *** Its less than 32 tries for n *** S =2k^2 + nT -5447391 Factor of S 16069 S / Factor of S -339 Gcd((Factor of S)-k, T) 691 Surrogate Factoring. Its alive! Its aliiive!!! Enrico === Subject: Re: JSH: SF Algorithm That tests my theory here and intriguingly enough in case some people >think there are a lot of n's, there are 32. >So if that theory is correct then you could factor an RSA sized number >with 32 trials. >So why babble about anything else? >You can shoot down my latest theory with a single counterexample. Not wishing to babble, 606007 is such a counterexample, it needs to go > through 50 n's before it finds a factor. Below is the output from my > instrumented version of your program. This was the x = ceil(sqrt(T)) version. The previous version with x = > floor(sqrt(T)) did even worse, needing 58 different n's before finding > a factor. You appear to have been right to make the change, at least > in this case. rossum *** Begin Program Output *** target = 606007 k = 1558, x = 779 > trying n = -7, surrogate = 612679 > trying n = -6, surrogate = 1218686 > trying n = -5, surrogate = 1824693 > trying n = -4, surrogate = 2430700 > trying n = -3, surrogate = 3036707 > trying n = -2, surrogate = 3642714 > trying n = -1, surrogate = 4248721 > trying n = 0, surrogate = 4854728 > trying n = 1, surrogate = 5460735 k = 1560, x = 780 > trying n = -8, surrogate = 19144 > trying n = -7, surrogate = 625151 > trying n = -6, surrogate = 1231158 > trying n = -5, surrogate = 1837165 > trying n = -4, surrogate = 2443172 > trying n = -3, surrogate = 3049179 > trying n = -2, surrogate = 3655186 > trying n = -1, surrogate = 4261193 > trying n = 0, surrogate = 4867200 > trying n = 1, surrogate = 5473207 k = 1562, x = 781 > trying n = -8, surrogate = 31632 > trying n = -7, surrogate = 637639 > trying n = -6, surrogate = 1243646 > trying n = -5, surrogate = 1849653 > trying n = -4, surrogate = 2455660 > trying n = -3, surrogate = 3061667 > trying n = -2, surrogate = 3667674 > trying n = -1, surrogate = 4273681 > trying n = 0, surrogate = 4879688 > trying n = 1, surrogate = 5485695 k = 1564, x = 782 > trying n = -8, surrogate = 44136 > trying n = -7, surrogate = 650143 > trying n = -6, surrogate = 1256150 > trying n = -5, surrogate = 1862157 > trying n = -4, surrogate = 2468164 > trying n = -3, surrogate = 3074171 > trying n = -2, surrogate = 3680178 > trying n = -1, surrogate = 4286185 > trying n = 0, surrogate = 4892192 > trying n = 1, surrogate = 5498199 k = 1566, x = 783 > trying n = -8, surrogate = 56656 > trying n = -7, surrogate = 662663 > trying n = -6, surrogate = 1268670 > trying n = -5, surrogate = 1874677 > trying n = -4, surrogate = 2480684 > trying n = -3, surrogate = 3086691 > trying n = -2, surrogate = 3692698 > trying n = -1, surrogate = 4298705 > trying n = 0, surrogate = 4904712 > trying n = 1, surrogate = 5510719 k = 1568, x = 784 > trying n = -8, surrogate = 69192 > BINGO! 186 succeeded factor = 691 > 606007 = 691 * 877 > 2113 probes total. > 50 values of n tried. *** End Program Output *** Counter example using surrogate Factoring T 606007 > x = floor(sqrt(T))+1 779 > k = 2x 1558 > n_max = floor(((x+k)^2 - 2k^2)/T) 1 > n_min = floor((4(x+k-1) - 2k^2)/T) -8 > Forced n -17 *** Its less than 32 tries for n *** S =2k^2 + nT -5447391 > Factor of S 16069 > S / Factor of S -339 > Gcd((Factor of S)-k, T) 691 Surrogate Factoring. Its alive! Its aliiive!!! Enrico Well at least you people are putting up big enough examples for me to refute you easily with my own programs. Looks like you got multiple things wrong including n_min and m_max: n_min=-15 n_max=0 n_max-n_min=15 Surrogate factorization: factored? Yes. ( -1 )( 29 )( 146477 ) Product: -4247833 Surrogate factorization: factored? Yes. ( -1 )( 2 )( 3 )( 606971 ) Product: -3641826 Surrogate factorization: factored? Yes. ( -1 )( 3035819 ) Product: -3035819 Surrogate factorization: factored? Yes. ( -1 )( 2^2 )( 7^4 )( 11 )( 23 ) Product: -2429812 Number factored. k=1556 n=-12 n_max=0 Total all combinations: 26 Time: 0 Time/combination: 0.0 Surrogate: ( -1 )( 2^2 )( 7^4 )( 11 )( 23 ) Product: -2429812 Surrogate combinations checked: 26 Initial Factorization: f_1=691 f_2=877 Now checking its factors... Success! Factors: ( 691 )( 877 ) Product: 606007 In coming is 606007 Surrogate factorization data for target: Surrogates factored : 4 Surrogates not factored : 0 Factored fuel percentage: 100% Data about all surrogates including those from recursions: Factored fuel : 4 Fuel not factored: 0 Factored fuel percentage: 100% Processing time: 93 Number of digits: 6 bitLength=20 ------------------------------------------------------ I'm not surprised at that result as it's consistent with what I see with dinky numbers. James Harris === Subject: Re: JSH: SF Algorithm >Well at least you people are putting up big enough examples for me to >refute you easily with my own programs. Looks like you got multiple things wrong including n_min and m_max: n_min=-15 >n_max=0 >n_max-n_min=15 No James, it is you who are getting things wrong. Look at your OP for this thread, it says: n_min = floor((4(x+k-1) - 2k^2)/T) That was the formula I used for n_min - your own formula. Let's put in the numbers for T = 606007: T = 606007 x = ceil(sqrt(T)) = 779 initially x | k (=2x) | n_min = floor((4(x+k-1) - 2k^2)/T) --------------------------------------------------- 779 | 1558 | -7 = floor(-4845384/606007) 780 | 1560 | -8 = floor(-4857844/606007) 781 | 1562 | -8 = floor(-4870320/606007) 782 | 1564 | -8 = floor(-4882812/606007) 783 | 1566 | -8 = floor(-4895320/606007) 784 | 1568 | -8 = floor(-4907844/606007) 785 | 1570 | -8 = floor(-4920384/606007) Those are the numbers that the method you posted generate. At no point in this range does n_min reach as low as -15. You are obviously using a different formula for generating n_min. I can only test the method that you post. If you tell me what your latest change to your method is then I can test that. My tests were correct for the method you posted initially, I make no guarantees that they will still be correct for changed versions of your method. What formula do you use to calculate n_min = -15? rossum === Subject: Re: JSH: SF Algorithm >> [JSH ] >> [... repetition ...] >> So if that theory is correct then you could factor an RSA sized number >> within 32 trials. >> Guess what? I won't spoil the surprise. >> Math people are not very bright. They are PRETEND bright. >> Pretend smart, but not real smart. >> So /that's/ the basis on which you want to be considered a major >> mathematician -- I knew there had to be something. Yuck. You're slimey. I'm sure you're smiling to yourself as you read >this. Maybe I will never crack the factoring problem. Maybe you people will win with lies, but win what? You really don't realize the contradiction here, eh? You say that maybe you won't crack the problem. Seems like a good guess. Now _if_ you fail to solve the problem then the people who've been explaining that you haven't solved it have nonetheless been _lying_ about that? >Satisfaction at convincing people too stupid to care about the truth? So many others have done that and done it better. But all you win is the death of the human species, and if that's what >you're after, then fine. I'm beaten. Let them die. I can do other things than continue to >care. Somehow I lost when I didn't think that possible which is why it >happened. Maybe it's just a learning experience for me. Training for >bigger battles down the line when it matters. The worst can now begin. And the real pain for the planet can now >start. I have stalled it as long as I can and now I'm just tired. Say it isn't so! The planet's been depending on you. ___JSH ************************ David C. Ullrich === Subject: Re: JSH: SF Algorithm > Marcus pointed out a problem with my earlier algorithm as it turns out > I have an assumption that x^2>y^2 so x=floor(sqrt(T)) is too small, so add one. Here's the corrected algorithm: Kind of odd really, but fascinating to contemplate as by the theory > the following algorithm should be applicable against an RSA sized > number factoring it in a maximum of 32 trials: Note, still with x^2 = y^2 mod T, and k = 2x mod T. So then, with x = floor(sqrt(T))+1, k = 2x, n_min = floor((4(x+k-1) - 2k^2)/T) and n_max = floor(((x+k)^2 - 2k^2)/T) , you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min > to n_max, looking for a > perfect square and the theory says there must be at least one, if for > a prime factor p_1 of T with -(k+2xT)/T < m_1 < -(k + 2xp_1)/T the absolute value of m_1 is 1 or greater. Guess I should add that now x+y must have one prime factor of T, and > x- y must have another, if you find an integer y, for at LEAST ONE of > the n's so you take a gcd with T. So the theory says it will give you a solution to a difference of > squares, AND that for at least one of the n's that solution must non- > trivially factor T. > How can *this* be the theory rejected by the Bulletin? Marcus made rejected *this* theory. They rejected a similar theory that you now acknowledge was wrong. No? -- Jesse F. Hughes Contrariwise, continued Tweedledee, if it was so, it might be, and if it were so, it would be; but as it isn't, it ain't. That's logic! -- Lewis Carroll === Subject: Re: JSH: SF Algorithm Oh well, enough bravado on my part as I'm not certain this will work > did reject. Why would they change their minds between now and then? The expert opinion is noted. Here is what my research says, which > presumably then will not work, but I do not know why it would not. Given a target composite T, from theory using x^2 = y^2 mod T and k = > 2x mod T, it can be proven that (x+k)^2 = y^2 + 2k^2 mod T must be true for any solution of a difference of squares. Explicitly to solve you need solutions for (x+k)^2 = y^2 + 2k^2 + nT. The algorithm picks x directly, choosing x = floor(sqrt(T)), so k = > 2x, and then ranges for the n's from You started off right, requiring, or desiring, that x^2 = y^2 mod > T. > But now you start with x = floor(sqrt(T)) and k = 2x, and you choose n > in a certain range. Here is how it works for, say, T = 77: T = 77 x = floor(sqrt(77)) = 8 k = 2x = 16. n_max = floor((24^2 - 2*16^2)/77) = > = floor(576 - 512)/77 = floor(64/77) = 0 n_min = floor((4*23 - 256)/77) = floor(-420/77) = -5. So choose n = -1. Then from what you say above, (x + k)^2 = y^2 + 2k^2 + nT, or 24^2 = y^2 + 512 - 77 = y^2 + 435, so y^2 = 576 - 435 = 141 which is not the perfect square of an integer; > y = 11.874342... Incidentally, 576 - 435 = 141 = 36^2 = 41^2 mod 77, which shows that n should not be chosen in advance. 36+8, 36-8, 41+8, 41-8 all have a nontrivial gcd with 77 and JSH's method fails to find them in due time (if at all). [Note here, incidentally, that the computation of y can > be carried out without any factoring of the surrogate.] You wanted INTEGERS x and y such that x^2 = y^2 mod T; > if your algorithm had a high probability of achieving this, > it would very likely be competitive with others. But > you have provided no rationale for having a high probability > that (x + k)^2 - 2k^2 - nT will be a perfect square of an integer, and clearly in general > it is not. n_max = floor(((x+k)^2 - 2k^2)/T) and n_min = floor((4(x+k-1) - 2k^2)/T) which with my program has meant roughly 32 surrogates to factor. By the theory, if you can fully factor all 32 surrogates for any > target T, then you will non-trivially factor T. If you cannot factor all 32 with the given x, you can increment it by > 1 and try again, indefinitely. Note that you can also use x = floor(sqrt(2T)) to have about 64 > surrogates and much greater odds but I'm not clear how that works > exactly and besides if you can factor 32 with the first one then you > have the target in hand. It is so weirdly simple and I think the theory is correct, but I guess > I could be wrong. See above. You have included no justification for the key > ingredient: that your algorithm has a high probability of > finding an integer value for y. I have tried to implement with my own programs but as I pointed out in > a previous post, I use recursion and with big numbers fewer and fewer > of the surrogates get factored, so it craps out. That should be a manageable technical problem. The more important > problem occurs as described above. I am not confident that I can work that problem out so what I said > earlier was bravado on my part. Really! Well, that has happened a few times before ... Marcus. James Harris === Subject: Re: JSH: SF Algorithm did reject. Why would they change their minds between now and then? The expert opinion is noted. Here is what my research says, which >presumably then will not work, but I do not know why it would not. Given a target composite T, from theory using x^2 = y^2 mod T and k = >2x mod T, it can be proven that (x+k)^2 = y^2 + 2k^2 mod T must be true for any solution of a difference of squares. Explicitly to solve you need solutions for (x+k)^2 = y^2 + 2k^2 + nT. The algorithm picks x directly, choosing x = floor(sqrt(T)), so k = >2x, and then ranges for the n's from n_max = floor(((x+k)^2 - 2k^2)/T) and n_min = floor((4(x+k-1) - 2k^2)/T) which with my program has meant roughly 32 surrogates to factor. By the theory, if you can fully factor all 32 surrogates for any >target T, then you will non-trivially factor T. If you cannot factor all 32 with the given x, you can increment it by >1 and try again, indefinitely. Note that you can also use x = floor(sqrt(2T)) to have about 64 >surrogates and much greater odds but I'm not clear how that works >exactly and besides if you can factor 32 with the first one then you >have the target in hand. It is so weirdly simple and I think the theory is correct, but I guess >I could be wrong. I have tried to implement with my own programs but as I pointed out in >a previous post, I use recursion and with big numbers fewer and fewer >of the surrogates get factored, so it craps out. I am not confident that I can work that problem out so what I said >earlier was bravado on my part. James Harris As I have done previously, I tested James' latest version of his > surrogate method on 500 random composite odd numbers that are > multiples of two different primes, each in the range 500 to 1000. The > results are compared to Fermat's method, trial factorisation (both > forward and reverse) and random picking. Fermat average = 7.01 probes. > JSH average = 892.29 probes. > Probe ratio = 1 : 127.361 > Trial average = 120.62 probes. > Reverse average = 11.72 probes. > Random average = 754.83 probes. > 500 trials, 0 misfactors found. Average k's tried per factorisation: 4.276 > Average n's tried per factorisation: 26.344 > Average n's tried per k: 7.736 As seems usual with James' methods it will factorise the target > number, but more slowly than existing methods. As a side point, I made two errors in the initial version of my code: > I swapped the expressions for nMin and nMax, and I miscoded the > expression for nMax (which I was assigning to nMin). In effect I had: nMin = ((x+k) - (2*k*k))/T; // (x+k) should be (x+k)*(x+k) > nMax = (4*(x+k+1) - (2*k*k))/T; With these two errors in place my results were: Fermat average = 7.09 probes. > JSH average = 300.20 probes. > Probe ratio = 1 : 42.318 > Trial average = 120.78 probes. > Reverse average = 11.63 probes. > Random average = 710.27 probes. > 500 trials, 0 misfactors found. Average k's tried per factorisation: 7.742 > Average n's tried per factorisation: 8.110 > Average n's tried per k: 1.055 This version runs faster than both random picking and James' current > version. It uses more k's than James' version, but within each k it > needs fewer n's. On balance this version needs fewer probes to find a > factor. It might be interesting to analyse why my mistakes had such a > beneficial effect - they might point out a way to improve the method > further. rossum My conjecture is that *any* change away from JSH's method is an improvement :) === Subject: Re: JSH: SF Algorithm > My conjecture is that *any* change away from JSH's method is an > improvement :) Actually, that's historically true of many JSH factoring methods. With so many pointless free variables and searches, there are many ways to rearrange the details, but time after time he somehow manages to suggest what turns out to be the (or very close to the) least efficient way to proceed with the mess. I figured that's one of the reasons he became increasingly unwilling to suggest an algorithm to flesh out his piles of equations and excruciatingly muddy technical prose; i.e., plain embarrassment. Much easier for him to maintain the (self-)illusion of genius if he can rant at others for being too stupid to make turn his brilliant insights into a kick-ass algorithm. === Subject: Re: JSH: SF Algorithm >Marcus pointed out a problem with my earlier algorithm as it turns out >I have an assumption that x^2>y^2 so x=floor(sqrt(T)) is too small, so add one. Here's the corrected algorithm: Kind of odd really, but fascinating to contemplate as by the theory >the following algorithm should be applicable against an RSA sized >number factoring it in a maximum of 32 trials: Note, still with x^2 = y^2 mod T, and k = 2x mod T. So then, with x = floor(sqrt(T))+1, k = 2x, Minor point, why not use the ceiling function, x = ceil(sqrt(T)) ? n_min = floor((4(x+k-1) - 2k^2)/T) and n_max = floor(((x+k)^2 - 2k^2)/T) , you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min >to n_max, looking for a >perfect square and the theory says there must be at least one, if for >a prime factor p_1 of T with -(k+2xT)/T < m_1 < -(k + 2xp_1)/T the absolute value of m_1 is 1 or greater. Guess I should add that now x+y must have one prime factor of T, and >x- y must have another, if you find an integer y, for at LEAST ONE of >the n's so you take a gcd with T. So the theory says it will give you a solution to a difference of >squares, AND that for at least one of the n's that solution must non- >trivially factor T. > No it is not. The theory they rejected had x = floor(sqrt(T)), your new version has x = ceil(sqrt(T)). That is a different theory. They rejected your old version, not the current version. Since you yourself have now also rejected the old version, perhaps you might want to stop criticising them on this because they now agree with you - the original version was flawed. Any further criticism of the yourself. rossum That tests my theory here and intriguingly enough in case some >peoplethink there are a lot of n's, there are 32. 32 n's. So if that theory is correct then you could factor an RSA sized number >within 32 trials. Math people are not very bright. They are PRETEND bright. Pretend smart, but not real smart. >James Harris === Subject: Re: JSH: SF Algorithm I have an assumption that x^2>y^2 so x=floor(sqrt(T)) is too small, so add one. Here's the corrected algorithm: Kind of odd really, but fascinating to contemplate as by the theory >the following algorithm should be applicable against an RSA sized >number factoring it in a maximum of 32 trials: Note, still with x^2 = y^2 mod T, and k = 2x mod T. So then, with x = floor(sqrt(T))+1, k = 2x, Minor point, why not use the ceiling function, x = ceil(sqrt(T)) ? n_min = floor((4(x+k-1) - 2k^2)/T) and n_max = floor(((x+k)^2 - 2k^2)/T) , you loop through y = sqrt((x+k)^2 - 2k^2 - nT), with n's from n_min >to n_max, looking for a >perfect square and the theory says there must be at least one, if for >a prime factor p_1 of T with -(k+2xT)/T < m_1 < -(k + 2xp_1)/T the absolute value of m_1 is 1 or greater. Guess I should add that now x+y must have one prime factor of T, and >x- y must have another, if you find an integer y, for at LEAST ONE of >the n's so you take a gcd with T. So the theory says it will give you a solution to a difference of >squares, AND that for at least one of the n's that solution must non- >trivially factor T. > No it is not. The theory they rejected had x = floor(sqrt(T)), your > new version has x = ceil(sqrt(T)). That is a different theory. They > rejected your old version, not the current version. Since you > yourself have now also rejected the old version, perhaps you might > want to stop criticising them on this because they now agree with you > - the original version was flawed. Any further criticism of the > yourself. rossum > I will criticize myself as necessary. Some corrections: 16 surrogates and not 32 are needed. You get 32 with x = floor(sqrt(2T)). Also the size issue is a granularity one that goes away as T increases in size. What baffles me now as the picture gets clearer is how it is all playing out like this on newsgroups? As for the Bulletin, they're stupid. It's that simple. No one sensible would want to go down in history for rejecting a new factoring method as powerful as this one clearly is. But Susan Friedlander is point and center as the editor whose name I have on the rejection. She may now have a dubious place in histroy. Oh and what is it with you putting up flawed supposed counterexamples like the one I just shot down in a previous post? Did you forget I have my own program to check you on assertions? James Harris === Subject: Re: JSH: SF Algorithm But Susan Friedlander is point and center as the editor whose name I > have on the rejection. She may now have a dubious place in histroy. > I know who she is, and she is anything but careless or stupid. Her PhD was done under the supervision of Louis Howard, a mathematician I got to know when I was at FSU. He is about as careful and thoughtful mathematician as you'll find on the planet. OTOH however if peer-review is not your thing (and after all, in your opinion they are not your peers), why don't you publish in ArXiv? After all, it was good enough for Perelman. M === Subject: Re: JSH: SF Algorithm I have stalled it as long as I can and now I'm just tired. ___JSH You doctor could help you with that. === Subject: PING Art Deco from the FNVW (Re: UB NOMINATION: Re: The German Archives) > It is hard to believe that [hanson] has never won a kook award even > though he has been ranting thusly for many moons. > > Unabomber Surprise > > This award is intended to honor those who have the gift for saying > the least with the most. Those who, for some reason only they may > fathom, more than all others crave the art of saying infinite amounts > of Nothing... and are PROUD OF IT. > > Also note the kookdance with TommY Crackpotter. > > I therefore nominate [hanson] for the Unabomber Surprise. > > Seconds? Art, you're going to have to withdraw this nomination in order for me to count your nomination of Archimedes Plutonium for the Unabomber Surprise. However, neither Tom Potter nor [hanson] have won Busted Urinal, an award both of whom richly deserve for their anti-Semitism. -- Pinku-Sensei Co-FNVW of AUK Acting Pollmaster of AFA-B Official Overseer of Kooks & Trolls in rec.arts.marching.drumcorps http://www.caballista.org/auk/index.html === Subject: Re: PING Art Deco from the FNVW (Re: UB NOMINATION: Re: The German Archives) Mother Pinky of silly auk coffeeboys > Home of the auk fake award really Big Whoppers > >>It is hard to believe that [hanson] has never won a kook award even >>though he has been ranting thusly for many moons. >>Unabomber Surprise >>This award is intended to honor those who have the gift for saying >>the least with the most. Those who, for some reason only they may >>fathom, more than all others crave the art of saying infinite amounts >>of Nothing... and are PROUD OF IT. >>Also note the kookdance with TommY Crackpotter. >>I therefore nominate [hanson] for the Unabomber Surprise. >>Seconds? Denied >Silly auk Mother Pinky > > Art, you're going to have to withdraw this nomination in order for me to > count your nomination of Archimedes Plutonium for the Unabomber Surprise. > However, neither Tom Potter nor [hanson] have won Busted Urinal, an award > both of whom richly deserve for their anti-Semitism. Auk's hopeless, clueless, and silly to boot, Science Officer hanson is one of the finest Peer science researchers and respected posters on the net today, so there. Oh the humanity! Officer Honest John these two auk mutts are running amuck again therefore please escort them back into the Captain's stowfile please. They are harassing my profound AA Officers especially esteemed Science Officer hanson. when will they ever learn, the nightbat === Subject: Re: PING Art Deco from the FNVW (Re: UB NOMINATION: Re: The German Archives) > Mother Pinky of silly auk coffeeboys Home of the auk fake award really Big Whoppers >It is hard to believe that [hanson] has never won a kook award even >>though he has been ranting thusly for many moons. >>Unabomber Surprise >>This award is intended to honor those who have the gift for saying >>the least with the most. Those who, for some reason only they may >>fathom, more than all others crave the art of saying infinite amounts >>of Nothing... and are PROUD OF IT. >>Also note the kookdance with TommY Crackpotter. >>I therefore nominate [hanson] for the Unabomber Surprise. >>Seconds? Denied Silly auk Mother Pinky Art, you're going to have to withdraw this nomination in order for me to > count your nomination of Archimedes Plutonium for the Unabomber Surprise. > However, neither Tom Potter nor [hanson] have won Busted Urinal, an award > both of whom richly deserve for their anti-Semitism. Auk's hopeless, clueless, and silly to boot, Science Officer hanson is > one of the finest Peer science researchers and respected posters on the > net today, so there. Oh the humanity! Officer Honest John these two auk > mutts are running amuck again therefore please escort them back into the > Captain's stowfile please. They are harassing my profound AA Officers > especially esteemed Science Officer hanson. when will they ever learn, > the nightbat I will get right on that task! C.H.J. === ... ... of times)) number of times)-plex = G(g) ... plex;... ... ... ... ... G((g^g) raised to the (g^g) a (g^g) number of times; === Summary: UDP === This server is temporarily under UDP because of recent massive hipcrime attacks, until the problem is solved by their administrators. We are sorry for the inconvenience. See news.admin.net-abuse.policy and news.admin.net-abuse.usenet for more details. === Subject: A Futile Attempt at Infinity A Study of Large Numbers or A Futile Attempt at Infinity ... ... ... plex; ... ... ... plex */ power. */ ... ... ... of times */ times */ 1 number of times) power */ /* Let us call this ... */ === Subject: Re: A Futile Attempt at Infinity > A Study of Large Numbers > or > A Futile Attempt at Infinity -- And I wish some of you would grow past thinking that you've discovered some extraordinary thing [...] as if you found the Holy Grail or something, when I acknowledge a mistake. After all, I've had to do it quite a few times. It's not like it's news. --James S. Harris === Subject: Re: A Futile Attempt at Infinity > A Study of Large Numbers > or > A Futile Attempt at Infinity ... > ... > ... > plex; > ... > ... > ... plex */ > power. */ ... > ... > ... of times */ > times */ > 1 number of times) power */ > /* Let us call this ... */ You still have a finite number, you know. No infinity. === Summary: UDP === This server is temporarily under UDP because of recent massive hipcrime attacks, until the problem is solved by their administrators. We are sorry for the inconvenience. See news.admin.net-abuse.policy and news.admin.net-abuse.usenet for more details. === Subject: The Extreme Function The extreme function f(0) = 10 = 1 with 1 zero f(1) = f(0) ^ f(0) = 10^10 = 10,000,000,000 = 10B = 1 with 10 zeros = 1 with (1 * (1 with 1 zero)) zeros f(2) = f(1) ^ f(1) = 10,000,000,000 ^ 10,000,000,000 = 1 with 100,000,000,000 zeros = 1 with (10 * (1 with 10 zeros)) zeros f(3) = f(2) ^ f(2) = (1 with 100B zeros) ^ (1 with 100B zeros) = 1 with (100B * (1 with 100B zeros)) zeros f(4) = f(3) ^ f(3) = (1 with (100B * (1 with 100B zeros)) zeros) ^ (1 with (100B * (1 with 100B zeros)) zeros) = 1 with ( (100B * (1 with 100B zeros)) * 1 with (100B * (1 with 100B zeros)) zeros ) zeros f(n) = f(n-1) ^ f(n-1) = (1 with x zeros) ^ ( 1 with x zeros) = 1 with (x * (1 with x zeros)) zeros === Subject: Re: The Extreme Function > The extreme function Not that extreme. Look for, say, Ackerman's function A(x, y); even something as simple as A(6, 6) will dwarf whatever you get from your extreme function. > > f(0) = 10 = 1 with 1 zero > f(1) = f(0) ^ f(0) = 10^10 = 10,000,000,000 = 10B = 1 with 10 zeros = 1 > with (1 * (1 with 1 zero)) zeros > f(2) = f(1) ^ f(1) = 10,000,000,000 ^ 10,000,000,000 = 1 with > 100,000,000,000 zeros = 1 with (10 * (1 with 10 zeros)) zeros f(3) = > f(2) ^ f(2) = (1 with 100B zeros) ^ (1 with 100B zeros) = 1 with (100B * > (1 with 100B zeros)) zeros > f(4) = f(3) ^ f(3) = (1 with (100B * (1 with 100B zeros)) zeros) ^ (1 > with (100B * (1 with 100B zeros)) zeros) = 1 with ( > > (100B * (1 with 100B zeros)) * 1 with (100B * (1 with 100B zeros)) zeros > ) zeros > > f(n) = f(n-1) ^ f(n-1) = (1 with x zeros) ^ ( 1 with x zeros) = 1 with > (x * (1 with x zeros)) zeros === Summary: UDP === This server is temporarily under UDP because of recent massive hipcrime attacks, until the problem is solved by their administrators. We are sorry for the inconvenience. See news.admin.net-abuse.policy and news.admin.net-abuse.usenet for more details. === Subject: Re: MULTIPLE NOMINATIONS: Re: China freer than USA? :-) >>I urge all folks to take a look at the following web site >>to see how racist bigots like Richard Herring and Art Deco operate. > > *ding* > > Good to see you are still trained, TommY. >>Draw your own conclusions. > > Will do. > > Bolo Bullis Foam Duck > > The spoonerism for dumb , for those who've lost more marbles > than a Chinese Checker factory will ever make. > > Yep, that sounds like Crackpotter, so nominated. > > Seconds? You also nominated Warhol for this award. Which one do you want as yours? > Busted Urinal Award > > The reward for the Lamest of the Lame on Usenet. One who clogs and > stinks up the joint like a busted urinal, you could say. > > Oh yes, that is Crackpotter, too. > > Seconds? Forget my message about [hanson] not winning this. Crackpotter is seconded. Now all that's needed is an opponent. > Tony Sidaway Drama Queen Award > > For those logically consistent persons who leave Usenet for ever, > rather more often than the Cabal's rules require. It isn't necessary > to leave Usenet for ever three times in a single week to win this > award, but it helps. Operatic flouncing of other kinds, including > unprovoked meltdowns and ridiculous ranting may also lead to a > nomination for this award. Named after Drama Empress Tony Sidaway. > > Note the ridiculous ranting part -- no one rants like Crackpotter. > > Seconds? Someone needs to find an opponent. I'm not going to let Tom run for this one alone. > Joseph Bartlo Pathetic Anal Pineapple Award > > The Jojo Pineapple is awarded every month to the Usenet poster who > best establishes a pattern of stupidity, idiocy, moronism, silliness, > cretinism and general kookery in a way exemplified by the eponymous > founder of this award, Joseph Foamboi Bartlo. > > Anyone who points out TommY's nonsense is automatically a lying, > bigoted, racist, brainwashed, needs a restraining order, etc. > > Seconds? As FNVW, I can nominate and/or second multiple candidates for the same award, and I'm going to do that here. Seconded. Now, Art, it's going to be two of your favorites, Tom Potter and Nightbat, running against each other for the PAP (or its successor). How do you like that choice? > [alt.fan.howard-stern, alt.law-enforcement, talk.politics.guns, > alt.politics.bush, alt.politics.clinton, > alt.sports.basketball.nba.sa-spurs snecked] Why Potter thinks that lits of groups is where he's on-topic is beyond me. > I see that alt.astronomy has gotten better and better since I've been > on sabbatical. It does help that I keep suggesting that kooks go there or that Nightbat recruit them. After all, last spring I wagered several tens of quatloos that Tom Potter would be one of Nightbat's l33t Spaced-Out Science Officers by Labor Day. Looks like I was correct, too. -- Pinku-Sensei Co-FNVW of AUK Acting Pollmaster of AFA-B Official Overseer of Kooks & Trolls in rec.arts.marching.drumcorps http://www.caballista.org/auk/index.html === Subject: energy vs power spectral density Say x(t) is a signal (function from R -> C) and its fourier transform (which exists) is X(w). X(w) can be wirtten as X(w) = |X(w)| e^(j*phi(w)). my book calls |X(w)| the magnitude spectrum, and phi(w) the phase spectrum. question: 1) Are the terms magnitude spectrum, amplitude spectrum, energy density, and energy spectral density all equivalent? 2) What exactly is it's relationship to the power spectrum (which my book defines only as the fourier transform of the autocorrelation of x)? Obviously, if you have the magnitude and phase spectrums, you can reconstruct the original signal. Can I reconstruct the signal if I have the power spectrum and the phase spectrum? Tom === Subject: Re: (easy?) trapezoid question , > > Hi all! This time the situation is: > > 'Consider trapezioid ABCD as a right trapezoid (DCB = CBA = 90). > is the bisector of angle BAD. If DC = 6 cm and AB = 10 cm, what is the > area of the trapezoid?' > > This question came with a drawing, and I redid it in the computer so > that you can have a better idea of the picture: > > http://img208.imageshack.us/my.php?image=trapezoidur6.jpg > > > 32*sqrt(5) I get 32.sqrt{15}. Drop a perpendicular from B to AD, meeting AD at F, and therefore MD/EF = 10/4 EF = 2/5 MD = 1/5.CD BE = 4/5.CD By the theorem on angle bisectors in triangles BE/EF = AB/4 and since BE/EF = 4 we have AB = 16 Now apply the Pythogorean theorem. BF^2 = 16^2 - 4^2 = 4^2.15 BF = 4.sqrt{15}. Area{ABCD} = 6.BF + 1/2.AF.BF = 6.BF + 1/2.4.BF = 8.BF = 32.sqrt{15}. -- Michael Press === Subject: the cyberstalker s michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor w === Subject: EXPOSE MICHAEL LALONDE michael lalonde DF mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === === === === Subject: Integration of functions similar to x^x? Hi. Suppose one defines a function rho(x) as follows: rho(x) = int_{0...x} t^t dt. With this in hand, it is obvious that, say, int x^x dx = rho(x) + C int 2x^x dx = 2 rho(x) + C int (2x)^(2x) dx = 1/2 rho(2x) + C. But, is it possible to integrate something like x^(2x) using our rho function, or not? If so, how would one do it? Or something, like, say, x^(x^2 - 1)? If so, how would one do it, anyway? Is this rho function pretty much a useless thing, then? === Subject: continuous mappings Does there exist a continuous mapping f:R^n->R^n, D(f)=R^n, such that for all y in f(R^n) the equation f(x)=y has exactly two solutions? I want to know this at least for n=1, n=2. === Subject: Re: continuous mappings > Does there exist a continuous mapping f:R^n->R^n, > D(f)=R^n, such that for all y in f(R^n) the equation > f(x)=y has exactly two solutions? > > I want to know this at least for n=1, n=2. It's pretty obvious if you draw a graph that this can't happen for n=1. If f(a) = f(b), with a < b, take any c in (a,b). If f(c) < f(a) then it is easy to see that f(x) > f(a) if x < a or x > b. So f(x) must actually be bounded below. Similarly if f(c) > f(a) then f(x) is bounded above. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: email hi i m amarnath,send interesting msges from your emails and so ons === Subject: mike lalonde sudbury michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: mike michael mike lalonde michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: michael lalonde stalker sudbury we michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: michael lalonde michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: Re: michael lalonde > Vocals: Gilles Mallette > Lead Guitar: Mike Lalonde > Rhythm Guitar: Craig Jefferson > Bass Guitar: James Roy > Drums: Mark Tessarolo > > http://www.garageband.com/artist/66Bloor Apart from the fact that 'sci.math' is not the proper forum for posting this stuff, they still have _a lot_ to improve when compared with e.g. Beethoven's 9th. Han de Bruijn === Subject: mike lalonde kjh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: Re: Using the definition to prove convexity On Sep 11, 7:29 pm, The World Wide Wade It is well-known that > f(x) := x^2, x in R, > is a convex function. However, I would like to prove that through the > definition of convex function. I have tried it, but with no success. > Perhaps, not being a professional mathematician, I am missing some > well-known inequality that would solve the problem. Some help would be > very appreciated. We want (tx + (1-t)y)^2 <= tx^2 + (1-t)y^2. The LHS is t^2x^2 + > 2t(1-t)xy + (1-t)^2y^2. So we want 2t(1-t)xy <= (t-t^2)x^2 + > ((1-t)-(1-t)^2)y^2 = t(1-t)x^2 + t(1-t)y^2. Cancelling t(1-t), we're > left with 2xy <= x^ + y^2, which should look friendly. Of course I meant x^2 + y^2 in the last line. I would like to thank you all for your solutions. Paul === Subject: Re: Continuous maps between boolean/stone spaces > >> Great! I think this should help Jose to find out whether his conjecture >> is true or not, since we are always dealing with duals of Boolean >> algebras which should be complete I suspect. I am really not a >> specialist in this area, but certainly nice to know things about it. >> Jose, what can you do with that? Can you prove it now? Let's see ... >> -- >> Best wishes, >> J. >> Ok.. so I know for boolean algebras surjections and epimorphisms are > the same.. but I seem to have a counterexample to the conjecture Attempted Counterproof for the conjecture: Let f : Y --> X be a continuous surjective map between an extremally > disconnected Stone space and a non-extremally-disconnected Stone > space (Exercise: Such a map exists Hint: Consider a non-Baer regular ring > embedded in its Baer Hull and then take its spectral map.. discussion > open =^) ) Let U be an open set in X such that the closure is not clopen (such a > U exist by assumption of X). Then cl(f^-1(U)) is clopen and f(cl(f^-1(U))) = cl(U) (because both X and Y are Hausdorff and compact) But If f were open then cl(U) would be clopen, thus f cannot be open. A slightly simpler example: Let X be the converging sequence: X = {0}cup{1/n:n in N} define f:X->X by f(0)=0, f(1)=0 and f(1/(n-1))=f(1/n). This map is continuous and onto but the isolated point 1 gets mapped to the non-isolated point 0, so f is not open. KP -- E-MAIL: K.P.Hart@TUDelft.NL PAPER: Faculteit EWI PHONE: +31-15-2784572 TU Delft FAX: +31-15-2787245 Postbus 5031 URL: http://fa.its.tudelft.nl/~hart 2600 GA Delft the Netherlands . === Subject: Re: Continuous maps between boolean/stone spaces >> I have this strange impression that continuous surjective maps between >> boolean/stone spaces are also automatically open.. is this true? > > I think this is true - up to checks of the sketch of proof given below: I scanned the following which turned out to be wrong to find the spots where it breaks down. > By Stone's representation theorem (cf., e.g., > http://planetmath.org/encyclopedia/MHStonesRepresentationTheorem.html) > we can assume that the surjection of Stone spaces comes from a > homomorphism phi: A -> B of Boolean algebras. We denote the > corresponding continuous map of Stone spaces B* -> A* by phi*. > > Here A* denotes the dual of A, i.e. A* = Hom(A,E) with the trivial > Boolean algebra E with underlying set {0,1} (0 <> 1), A* is equipped > with the topology induced by {0,1}^A which is a compact and Hausdorff > space, by Tychonov. Since A* is closed in {0,1}^A, A* is compact and > Hausdorff, too. > > The assumption that phi* is surjective implies that phi is injective, > hence we can assume w.l.o.g. that A is a subalgebra of B. > > Let's assume in the first step that A and B are *free* Boolean algebras > with basis S and T, resp., with S contained in T. Then the Stone spaces > A* and B* are homeomorphic to {0,1}^S and {0,1}^T, resp., and the map B* > -> A* induced by the restriction from T to S is clearly open. These homeomorphism do not seem to hold true. > Now we try to reduce the general situation to the free case above. For > this purpose let S be a generating system of A as a Boolean algebra. > Denoting the free Boolean algebra over S by Fr(S), the homomorphism > Fr(S) -> A (induced by the inclusion S c-> A) is a surjection inducing > the map A* -> Fr(S)* which is injective and the topology of A* coincides > with the trace topology induced by Fr(S*). > > With this in mind, let S=A and T=B systems generating A and B, resp. > Then we have the surjection > > Fr(A) ->> A > > of Boolean algebras yielding the (regular) injection > > A* c-> Fr(A)* (*A) > > and likewise for B (With regular trace topology is meant). And unfortunately the claim that A* carries the trace topology fails to be true in general. > Since the > surjections B* -> A* are compatible with the mappings (*A) and (*B), the > openness of B* -> A* follows from all the facts noted before. > > This should complete the proof. > > > Jose, please let me know whether you could verify all the details and > the claim holds true. I appreciate an email in any case. > -- Best wishes, J. === Subject: Re: Continuous maps between boolean/stone spaces I have this strange impression that continuous surjective maps between boolean/stone spaces are also automatically open.. is this true? >> I think this is true - up to checks of the sketch of proof given below: > > I scanned the following which turned out to be wrong to find the spots > where it breaks down. > >> By Stone's representation theorem (cf., e.g., >> http://planetmath.org/encyclopedia/MHStonesRepresentationTheorem.html) >> we can assume that the surjection of Stone spaces comes from a >> homomorphism phi: A -> B of Boolean algebras. We denote the >> corresponding continuous map of Stone spaces B* -> A* by phi*. >> Here A* denotes the dual of A, i.e. A* = Hom(A,E) with the trivial >> Boolean algebra E with underlying set {0,1} (0 <> 1), A* is equipped >> with the topology induced by {0,1}^A which is a compact and Hausdorff >> space, by Tychonov. Since A* is closed in {0,1}^A, A* is compact and >> Hausdorff, too. >> The assumption that phi* is surjective implies that phi is injective, >> hence we can assume w.l.o.g. that A is a subalgebra of B. >> Let's assume in the first step that A and B are *free* Boolean >> algebras with basis S and T, resp., with S contained in T. Then the >> Stone spaces A* and B* are homeomorphic to {0,1}^S and {0,1}^T, resp., >> and the map B* -> A* induced by the restriction from T to S is clearly >> open. > > These homeomorphism do not seem to hold true. This is almost meaningless. :-( I meant: They are not open. >> Now we try to reduce the general situation to the free case above. For >> this purpose let S be a generating system of A as a Boolean algebra. >> Denoting the free Boolean algebra over S by Fr(S), the homomorphism >> Fr(S) -> A (induced by the inclusion S c-> A) is a surjection inducing >> the map A* -> Fr(S)* which is injective and the topology of A* >> coincides with the trace topology induced by Fr(S*). >> With this in mind, let S=A and T=B systems generating A and B, resp. >> Then we have the surjection >> Fr(A) ->> A >> of Boolean algebras yielding the (regular) injection >> A* c-> Fr(A)* (*A) >> and likewise for B (With regular trace topology is meant). > > And unfortunately the claim that A* carries the trace topology fails to > be true in general. > >> Since the surjections B* -> A* are compatible with the mappings (*A) >> and (*B), the openness of B* -> A* follows from all the facts noted >> before. >> This should complete the proof. >> Jose, please let me know whether you could verify all the details and >> the claim holds true. I appreciate an email in any case. > -- Best wishes, J. === Subject: Re: Need Intructor's Solution Manual I desperately need the instructor's solution manual to: > A First Course In Differential Equations by Dennis G. Zill > ISBN- 0-534-37388-7 Its the Classic Fifth Edition have you found it yet? I'm also looking === Subject: need help!! Hello! I'm currently taking a course that's utilizing a textbook that I cannot find a solutions manual to. It's called Physical Chemistry... 4th Ed, by Tinoco, Sauer, Wang, Publisi.The solutions manual is called Solution manual for Principles and Applications in Biological Sciences, 4th ed. Tinoco, Sauer, Wang and Puglisi, 2002. The professor isn't doing such a good job explaining the concepts, and I'd like to be able to check my work after doing the exercise problems. === Subject: Convexity and to grade or not to grade Draw a Cantor set C on the circle and consider the set A of all chords between points of C. (a) Prove that A is compact ***(b) Is A convex? (a) is very simple, but point (b) is supposed to be *very* difficult according to the three stars (but keep in mind it's an introductory book to Analysis). That's strange because I solved it in 5 minutes. Either my answer is wrong (but I don't think so) or the exercise is simpler that the author thought. Perhaps I was lucky to see an easy way to solve it. How would you attack (b)? I won't give my solution right now because I don't want to influence you in any way. Kiuhnm === Subject: Re: Convexity and to grade or not to grade > Draw a Cantor set C on the circle and consider the set A of all chords > between points of C. > (a) Prove that A is compact > ***(b) Is A convex? You consider the SET of chords (as opposed to the UNION of chords), but which topology on that set? Haussdorff-metric maybe? And for b: In which vector space do they live? (a) is very simple, but point (b) is supposed to be *very* difficult > according to the three stars (but keep in mind it's an introductory book > to Analysis). That's strange because I solved it in 5 minutes. > Either my answer is wrong (but I don't think so) or the exercise is > simpler that the author thought. > Perhaps I was lucky to see an easy way to solve it. How would you attack (b)? > I won't give my solution right now because I don't want to influence you > in any way. Kiuhnm === Subject: Re: Convexity and to grade or not to grade >> Draw a Cantor set C on the circle and consider the set A of all chords >> between points of C. >> (a) Prove that A is compact >> ***(b) Is A convex? > > You consider the SET of chords (as opposed to the UNION of chords), > but which topology on that set? Haussdorff-metric maybe? > And for b: In which vector space do they live? Ouch, now that you say it, I must confess I considered the set of the points in R^2 of all the chords. Otherwise we miss some information, as you pointed out. Then I was mistaken even before starting!? Kiuhnm === Subject: michael lalonde the stalker don't believe the hype believe the rock music!!!!!!! michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: many solution manual this is my new solution list , please email to getsolution(at) hotmai.com,change the at as @, Adaptive Control 2nd. Edt. by Karl.J.Astrom - solution manuel Engineering Circuit Analysis 6Ed - Hayt Solutions Manual.pdf Introduction to electrodynamics 3edt. (Solutions Manual) Solutions Manual for Introduction to Linear Algebra, 3/e Strang, (McGraw- Hill)_(Instructor's_Manual)_Electric_Machinery_Fundamentals_4th_Edition_(Ste phen_J_Chapma.pdf . Goldstein, C.P. Poole, & J.L. Safko, Classical Mechanics, 3/e with Partial Solutions Manual to 2/e .C. 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Buck, Engineering Electromagnetics, 6/e, with Solutions Manual weatherwax_bertsimas_solutions_manual.pdf Wiley - 1999 - Calculus one and several variables Instructor solutions manual.pdf Wiley - Advanced Engineering Mathematics 8Ed - Erwin Kreyszig - Solutions Manual.pdf Wiley - Theory.and.Applications.of.OFDM.and.CDMA.Wideband.Wireless.Communications. (2005).solutions.manual.pdf Wiley_-_Salas_et_al._- _Instructor's_Solutions_Manual_to_Accompany_Calculus- _1_and_Several_Variables__8th_Edition_pdf.zip William D. Callister Jr, Solutions Manual:Materiala Science and Engineering An Introduction 6 ed William H. Greene, Solutions Manual to Econometric Analysis (5th Edition) William Stallings - Operating systems - Instructors Manual- Solutions.doc Wireless Communications 2Ed - Theodore Rappaport - Solutions Manual.pdf wireless_communications_2ed_-_theodore_rappaport_-_solutions_manual Yunus A. Cengel,& Yunus Cengel, Heat Transfer: A Practical Approach 2 ed === Subject: solutions guide I need the solutions to Fluid Mechanics: Fundamentals and Applications, 1st Ed., by Cengel & Cimbala, can anyone help me out? === Subject: Continuous functions between topological spaces Let's say I have a continuous map f : X --> Y between top. spaces X and Y. If for any open subset U of X, there is an open set V such that f(U) contains V (I don't suppose that then this map becomes open, does it?) .. is there a name for such a function? Jose Capco === Subject: Re: Continuous functions between topological spaces > Let's say I have a continuous map f : X --> Y between top. spaces X > and Y. > If for any open subset U of X, there is an open set V such that f(U) > contains V (I don't suppose that then this map becomes open, does > it?) .. is there a name for such a function? > Jose Capco Perhaps nonconstant almost everywhere? An example would be f: R -> R, given by x |-> x^2. It's not an open map. -- Dave Seaman Oral Arguments in Mumia Abu-Jamal Case heard May 17 U.S. Court of Appeals, Third Circuit === Subject: Google Posting Troubles, disappearing post, and poor cummunication channel- is it just me? Is anyone else having the following difficulties? When I post to an existing thread the thread does not come up on Later dates are shown and it is as if I never replied unless someone specifically finds that thread. Now I have posted a complaint such as this one that has disappeared from my own listing. I do have successful posts, but without them coming into a topic list communication is fouled. As a result my threads are likely dying prematurely. When I create a new post it does come up in the list. This is really weird behavior and as with all such things I trouble over whether someone is screwing with me though I know that the likelihood is not. Still, such behavior should then be observed by others and I hope that you will give feedback here. I've studies some of the protocol fields but I don't see anything spooky. So one possible conclusion is that Google's interface is giving me a substandard channel, but these are the people who do no evil right? So a form of internet paranoia is established and understandable but not provable. === Subject: Re: Google Posting Troubles, disappearing post, and poor cummunication channel- is it just me? On Sep 13, 7:13 am, Timothy Golden BandTechnology.com > Is anyone else having the following difficulties? When I post to an existing thread the thread does not come up on > Later dates are shown and it is as if I never replied unless someone > specifically finds that thread. > Now I have posted a complaint such as this one that has disappeared > from my own listing. > I do have successful posts, but without them coming into a topic list > communication is fouled. > As a result my threads are likely dying prematurely. > When I create a new post it does come up in the list. > This is really weird behavior and as with all such things I trouble > over whether someone is screwing with me though I know that the > likelihood is not. Still, such behavior should then be observed by > others and I hope that you will give feedback here. > I've studies some of the protocol fields but I don't see anything > spooky. So one possible conclusion is that Google's interface is > giving me a substandard channel, but these are the people who do no > evil right? So a form of internet paranoia is established and > understandable but not provable. yes posts are disappearing even from the middle of threads more and more frequently too it is strange that those i have seen disappear have so far been those i regularly try to read but that could just be selective search at work on my part others have suggested switching newsreader (in your old threads) (search, no need to fetch, ...) i like using two readers to get the best of both -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- galathaea: prankster, fablist, magician, liar === Subject: Re: Google Posting Troubles, disappearing post, and poor cummunication channel- is it just me? > On Sep 13, 7:13 am, Timothy Golden BandTechnology.com Is anyone else having the following difficulties? When I post to an existing thread the thread does not come up on > Later dates are shown and it is as if I never replied unless someone > specifically finds that thread. > Now I have posted a complaint such as this one that has disappeared > from my own listing. > I do have successful posts, but without them coming into a topic list > communication is fouled. > As a result my threads are likely dying prematurely. > When I create a new post it does come up in the list. > This is really weird behavior and as with all such things I trouble > over whether someone is screwing with me though I know that the > likelihood is not. Still, such behavior should then be observed by > others and I hope that you will give feedback here. > I've studies some of the protocol fields but I don't see anything > spooky. So one possible conclusion is that Google's interface is > giving me a substandard channel, but these are the people who do no > evil right? So a form of internet paranoia is established and > understandable but not provable. yes posts are disappearing > even from the middle of threads more and more frequently too it is strange that those i have seen disappear > have so far been those i regularly try to read > but that could just be selective search at work on my part others have suggested switching newsreader > (in your old threads) > (search, no need to fetch, ...) i like using two readers > to get the best of both -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=- > galathaea: prankster, fablist, magician, liar Who do you use and does it cost $? Also I just posted to your star -Tim === Subject: Re: Relationships between e and pi without an i in sight? ... Was: Re: AP forgot about Ramanujan. Was: Re: #28K (pi) and (e) and bifurcation of Reals Re: ATOM TOTALITY (Atom Universe) THEORY REPLACES BIG BANG THEORY IN PHYSICS > [AP complains about the equation e^(i pi) = -1 having an i in it, >> in addition to e and pi, and says there's no direct connection >> between e and pi.] >> There are several near-misses, in that e^(pi *sqrt(d)) is nearly an >> integer [for some values of d]. [...] >> >> I just got done reading _An Imaginary Tale: The Story of Sqrt(-1)_, >> which has the following equation (not approximation!) in it: >> >> e = pi / (integral of cos(x)/(1+x^2) from -infinity to >> +infinity) >> >> This is an odd result, in that the right-hand side only involves pi, a >> trig function, and the derivative of an inverse trig function; no e in >> sight! Well, if you look closely enough at cos x, you'll see >it's (1/2)(e^{ix} + e^{-ix}). That wouldn't have occurred to me, but I don't see how to use it. Rearranging: pi / e = (integral ... ) Let u = ix integration limits: -infinite_x = -infinite_u / i = infinite_u * i infinite_x = infinite_u / i = -infinite_u * i dx = (1 / i) du x^2 = -u^2 cos x = (1/2)(e^{u} + e^{-u}) = cosh u so: pi / e = (1 / 2 / i) (integral of (e^{u} + e^{-u}) / (1 - u^2) from -{infinity / i} to {infinity / i}) or pi / e = (1 / i) (integral of cosh(u) / (1 - u^2) from -{infinity / i} to {infinity / i}) Integration by parts looks messy. Where to from there? Adam -- === Subject: Re: Relationships between e and pi without an i in sight? ... Was: Re: AP forgot about Ramanujan. Was: Re: #28K (pi) and (e) and bifurcation of Reals Re: ATOM TOTALITY (Atom Universe) THEORY REPLACES BIG BANG THEORY IN PHYSICS > [AP complains about the equation e^(i pi) = -1 having an i in it, >> in addition to e and pi, and says there's no direct connection >> between e and pi.] >> There are several near-misses, in that e^(pi *sqrt(d)) is nearly an >> integer [for some values of d]. [...] >> >> I just got done reading _An Imaginary Tale: The Story of Sqrt(-1)_, >> which has the following equation (not approximation!) in it: >> >> e = pi / (integral of cos(x)/(1+x^2) from -infinity to >> +infinity) >> >> This is an odd result, in that the right-hand side only involves pi, a >> trig function, and the derivative of an inverse trig function; no e in >> sight! Well, if you look closely enough at cos x, you'll see >it's (1/2)(e^{ix} + e^{-ix}). Interesting. Let u = ix cos x = (1/2)(e^{u} + e^{-u}) = cosh u so: pi / e = (1 / 2 / i) (integral of (e^{u} + e^{-u}) / (1 - u^2) from -{infinity * i} to {infinity * i}) or pi / e = (1 / i) (integral of cosh(u) / (1 - u^2) from -{infinity * i} to {infinity * i}) Where to from there? Integration by parts looks messy because (integral of 1 / (1 - u^2)) = inverse_tanh(u) A way to simplify the product {inverse_tanh(u) * cosh(u)} isn't apparent to me. Adam -- === Subject: Adaptive Control 2nd. Edt. by Karl.J.Astrom - solution manuel Adaptive Control 2nd. Edt. by Karl.J.Astrom - solution manuel Engineering Circuit Analysis 6Ed - Hayt Solutions Manual.pdf Introduction to electrodynamics 3edt. (Solutions Manual) Solutions Manual for Introduction to Linear Algebra, 3/e Strang, (McGraw- Hill) (Instructor's Manual) Electric Machinery Fundamentals 4th Edition (Stephen J Chapma.pdf . Goldstein, C.P. Poole, & J.L. Safko, Classical Mechanics, 3/e with Partial Solutions Manual to 2/e .C. Montgomery,& G.C.Runger, Applied Statistics and Probability for Engineers 3/e Book+Solutions Manual Advanced Modern Engineering Mathematics, 3rd Edt by Glyn James - solution m advanced engineering mathematics 8ed - erwin kreyszig - solutions manual ANTENNAS FOR ALL APPLICATIONS, THIRD EDITION Applied Numerical Analysis 7Ed - Curtis F. Gerald, Patrick O. 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Cengel,& Yunus Cengel, Heat Transfer: A Practical Approach 2 ed PS: These are part of my solutions, if the solution you want isn't on the list, do not give up, just contact with me: My email is getsolution(at)hotmail.com( please replace the (at) with @ ) NOTE: if the solutions you want is on the list renewed, please mention in your email,thank you! Solution manual for the list: http://rapidshare.com/files/52408080/list.doc I will reply with your Email within one minitues! === Subject: solution manuel Adaptive Control 2nd. Edt. by Karl.J.Astrom - solution manuel Engineering Circuit Analysis 6Ed - Hayt Solutions Manual.pdf Introduction to electrodynamics 3edt. (Solutions Manual) Solutions Manual for Introduction to Linear Algebra, 3/e Strang, (McGraw- Hill) (Instructor's Manual) Electric Machinery Fundamentals 4th Edition (Stephen J Chapma.pdf . Goldstein, C.P. Poole, & J.L. Safko, Classical Mechanics, 3/e with Partial Solutions Manual to 2/e .C. 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Stokey-Lucas - Irigoyen, Claudio; Rossi-Hansberg, Esteban; Wright, Mark L. J. Solutions Manual for Recursive Methods in Economic Dynamics Student Solutions Manual to accompany Boyce Elementary Differential Equations and Boundary Value Problems Study Guide and Solutions Manual to Accompany Organic Chemistry (Atkins, Carey)(4ed) sze solutions manual for semiconductor devices physics and technology 2ed. Taylor & Francis,.Solutions Manual for the Guide to Energy Management, Fifth Edition. Theodore S. Rappaport - Wireless Communications Principles and Practice Solutions Manual (1996 eBook). Undergraduate Econometrics Solutions Manual - Hill, Judge And Griffiths Van Vylen - Solutions Manual 5th edition.pdf W. H. Hayt, & J. A. 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Cengel,& Yunus Cengel, Heat Transfer: A Practical Approach 2 ed PS: These are part of my solutions, if the solution you want isn't on the list, do not give up, just contact with me: My email is getsolution(at)hotmail.com( please replace the (at) with @ ) NOTE: if the solutions you want is on the list renewed, please mention in your email,thank you! Solution manual for the list: I will reply with your Email within one minitues! === Subject: many solution manuals Many Solutions Manuals and Ebooks in Electronic (PDF)Format! PS: These are part of my solutions, if the solution you want isn?t on the list, do not give up, just contact with me: My email is solutionpay(at)hotmail.com( please replace the (at) with @ ) NOTE: if the solutions you want is on the list renewed,please mention in your email,thank you! Solution manual for the list:? I will reply with your Email within 12 hours!! A Course in Modern Mathematical Physics By Peter Szekeres advanced engineering mathematics (8/e) by ERWIN KREYSZIG advanced engineering mathematics (8/e)( korean version) by ERWIN KREYSZIG advanced engineering mathematics?9/e? (even solutions) by ERWIN KREYSZIG advanced macroeconomics By Romer AISC Manual of Steel Construction: Load and Resistance Factor Design, Third Edition (LRFD 3rd Edition) By AISC Manual Committee Analytical Mechanics (7/e) By Grant R. Fowles, George L. Cassiday Analytical Mechanics, 5th ,By Grant R. Fowles, George L. Cassiday, APPLIED MATHEMATICS AND MODELLING FOR CHEMICAL ENGINEERS, 8th, By Erwin Kreyszig Applied Statisticsand Probability for Engineers,3rd?by Douglas C. Montgomery(select problem) Applied Strength of Materials (4th Edition) by Robert MoTT C How to Program, 3RD Edition 2000 By Harvey M. Deitel Calculus I, II, Transidentals 10e BY George B. 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Kraige(chapter1,2,3,4) Engineering Mechanics: Dynamics 5 th ed,By J. L. Meriam, L. G. Kraige Engineering Mechanics: Statics By R.C. Hibbeler Engineering Mechanics: Statics ,10th ,by R.C.Hibbeler, field and wave electromagnetics (2/e) by David Cheng Fourier and Laplace Transform ,by Antwoorden Fundamental of Heat and Mass Transfer by Frank P. Incropera and David P.DeWitt (ANOTHER EDITION) Fundamentals of engineering Thermodynamics Fundamentals of Logic Design 5Ed by CharlesRoth Fundamentals of Chemical Reaction Engineering By Mark E. E. Davis, Robert J. J. Davis Fundamentals of Engineering Thermodynamics by Moran, M.J. & Shapiro H.N. Fundamentals Of Fluid Mechanics 3Rd And 4Th Edition Fundamentals of Fluid Mechanics, 5th by By Bruce,R. Munson, Donald, Theodore H. 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Kasap Probability & Statistics for Engineers & Scientists, 8th by Sharon Myers , Keying Ye, Walpole Probability and Statistics for Engineering and the Sciences 7/e JAY L.DEVORE Probability and Statistics for Engineers and Scientists by HAYLER Probability, Random Variables and Stochastic Processes,3rd, by Athanasios Papoulis Probability, Random Variables, and Stochastic Processes,4th,By Athanasios Papoulis ,S.Unnikrishna Pillai Probability,Random Variables and Stochastic Processes,4th,by Athanasios Papoulis Quantum Field Theory (draft version) ,By Mark Srednicki Quantum Physics, Third Edition, By Stephen Gasiorowicz Recursive Methods in Economic Dynamics By Claudio Irigoyen, Esteban Rossi-Hansberg, Semiconductor Devices Second Edition By S.M.Sze Semiconductor Physics and Devices Third Edition By Donald Neamen Separation Process Principles, 2nd Ed., by Seader, Henley Signal Processing and Linear Systems (2001) by B P Lathi Signal Processing and Linear Systems, By B.P. Lathi Signals and Systems (2nd Edition) By Oppenheim, Willsky and Nawab Signals and Systems 2nd by Haykin System Dynamics 3rd Ed ,By Katsuhiko Ogata Thermodynamics: An Engineering Approach,6th Ed. by Cengel Thomas' Calculus (11th Edition) by George B Thomas Thomas' Calculus, Early Trascendentals 10th ed Instructors Solutions Manual Two-Dimensional Incompressible Navier-Stokes Equations- Maciej Matyk University Physics with Modern Physics:,11 ed By Hugh D. Young, Roger A. Freedman, Vector Mechanics for Engineers: Dynamics, 7th By === Subject: Re: solution manuel DO NOT BUY FROM THIS GUY! THE MANUALS ARE FAKE! THEY HAVE WRONG ANSWERS! RIP - OFF === Subject: A monotone function Let a fucntion G:RxR->R satisfy the following properties: 1. It is increasing in each variable. 2. If G(x,y) > G(x',y') then G(a*x+b,a*y+b) > G(a*x'+b,a*y'+b) for each a>0, b. It is required to prove (or disprove) that for each x and y there exists z such that G(x,y)=G(z,z). Mikhail === Subject: An exercise in Algebraic Geometry Elementary Geometry of Algebraic Curves: An Undergraduate Introduction by Christopher G. Gibson I am doing the Exercise 17.1.1 . I can complete the first part by choosing a distinct point on the curve E and then applying Lemma 16.6 and Bezout's Theorem. However, I get confused with the second part. I don't know how to deduce the remaining component is of degree (d-c). I have considered to construct a Linear System of dimension (d-c)... I don't know how to approach it ... or is it not necessary to do so? Chi Ho, Chan Image of the Exercise 17.1.1 http://x60.xanga.com/e3dc0054d5332147088156/b109166680.jpg Link of Lemma 16.6 === Subject: Re: Pizza Puzzle!! <46e60ae5$1@news.702com.net | What is the minimum and the maximum number of people the teacher could > | invite to his celebration if he uses just 3 cuts? 4cuts? 5cuts? N cuts? These numbers are also known as pancake numbers, elsewhere as > lazy caterer's sequence, and also as Hogben's central polyonal > numbers. The formula is: pieces = n * (n+1)/2 + 1 The series is (starting with zero cuts, then 1 cut, ..., 20 cuts): 1 2 4 7 11 16 22 29 37 46 56 67 79 92 106 121 137 154 172 191 211 ... Curious. These are just triangle numbers plus one, i.e., p = T(n) + 1, where T(n) is the nth triangle number, T(0) = 0, T(n+1) = T(n) + (n+1). => 0 1 3 6 10 15 21 28 36 45 55 ... === Subject: Re: Pizza Puzzle!! <46e60ae5$1@news.702com.net> On Sep 11, 5:15 am, Gerry Myerson > | The math teacher decided to have a pizza party for his outstanding >> | class. Knowing how hungry teenagers can be he decided to order the >> | largest pizza the world had ever seen. >> | >> | >> | The pizza was so big that it was going to cost a small fortune just to >> | cut it. The World's Largest Pizza Cutting Corporation quoted him $1000 >> | each straight cut. Nobody else was willing to cut such a large pizza. >> | Obviously it was important to slice up the pizza in as few cuts as >> | possible to save money. The pizza was plenty large enough that one >> | slice would be enough even for the hungriest teenager. So the teacher >> | decided on a policy of one slice per person. He was also hungry >> | himself so one slice would go to him as well. >> | >> | >> | What is the minimum and the maximum number of people the teacher could >> | invite to his celebration if he uses just 3 cuts? 4cuts? 5cuts? N cuts? > These numbers are also known as pancake numbers, elsewhere as >> lazy caterer's sequence, and also as Hogben's central polyonal >> numbers. > The formula is: pieces = n * (n+1)/2 + 1 > The series is (starting with zero cuts, then 1 cut, ..., 20 cuts): > 1 2 4 7 11 16 22 29 37 46 56 67 79 92 106 121 137 154 172 191 211 ... > Yes, but, the trouble with this method is that the pieces will be >> of very different sizes. A couple of questions that I once asked >> at a conference: > 1. How large can you make the smallest piece if you do things >> this way (that is, n cuts, 1 + n (n + 1) / 2 pieces)? > 2. How many pieces can you make with n cuts, if all pieces >> are required to be of equal size? It's easy to make 2 n pieces; >> is it ever possible to make more? > Those seem very difficult questions. Did you get any answers? No. So far as I know, both questions are wide open. For n=3, I can prove the maximum is 6. As the puzzle does not limit the cuts to straight line cuts, if you make a circular cut with the same center and r = R/sqrt(2) and then two perpendicular cuts by through center, you will have 8 equal pieces with n = 3 cuts. Ludovicus === Subject: Re: Pizza Puzzle!! <46e60ae5$1@news.702com.net> > On Sep 11, 5:15 am, Gerry Myerson > | The math teacher decided to have a pizza party for his outstanding >> | class. Knowing how hungry teenagers can be he decided to order the >> | largest pizza the world had ever seen. >> | >> | >> | The pizza was so big that it was going to cost a small fortune just to >> | cut it. The World's Largest Pizza Cutting Corporation quoted him $1000 >> | each straight cut. Nobody else was willing to cut such a large pizza. >> | Obviously it was important to slice up the pizza in as few cuts as >> | possible to save money. The pizza was plenty large enough that one >> | slice would be enough even for the hungriest teenager. So the teacher >> | decided on a policy of one slice per person. He was also hungry >> | himself so one slice would go to him as well. >> | >> | >> | What is the minimum and the maximum number of people the teacher could >> | invite to his celebration if he uses just 3 cuts? 4cuts? 5cuts? N cuts? >> These numbers are also known as pancake numbers, elsewhere as >> lazy caterer's sequence, and also as Hogben's central polyonal >> numbers. >> The formula is: pieces = n * (n+1)/2 + 1 >> The series is (starting with zero cuts, then 1 cut, ..., 20 cuts): >> 1 2 4 7 11 16 22 29 37 46 56 67 79 92 106 121 137 154 172 191 211 ... >> Yes, but, the trouble with this method is that the pieces will be >> of very different sizes. A couple of questions that I once asked >> at a conference: >> 1. How large can you make the smallest piece if you do things >> this way (that is, n cuts, 1 + n (n + 1) / 2 pieces)? >> 2. How many pieces can you make with n cuts, if all pieces >> are required to be of equal size? It's easy to make 2 n pieces; >> is it ever possible to make more? >> Those seem very difficult questions. Did you get any answers? >No. So far as I know, both questions are wide open. For n=3, I can prove the maximum is 6. As the puzzle does not limit the cuts to straight line cuts, > if you make a circular cut with the same center and r = R/sqrt(2) > and then two perpendicular cuts by through center, > you will have 8 equal pieces with n = 3 cuts. The original question refers to straight cuts. If you allow cuts that aren't straight lines then you can create any number of pieces with any number of cuts (even with just one cut), and make the pieces any sizes you want, so the problems become uninteresting. === Subject: michael lalonde the stalker gotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: michael lalonde klj2gotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: michael lalonde cyberstalker poop cow gotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: michael lalonde the stalker 2gotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: michael lalonde mike lalonde da stalker gotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor sss === Subject: michael lalonde kick da cow gotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: mike lalonde 2kjh gotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor sss === Subject: michael lalonde kjhkhgotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor ss === Subject: mike lalonde kjh gotcher attention eh michael lalonde mike lalonde sudbury it's true mike lalonde is the cyberstalker. you will find him stalking fans and then blowing their minds with his wicked guitar licks. Mike Lalonde is the madman of the guitar. You know Michael Lalonde as the Ozzy fanatic. The Ozzmium saw it share of visits. Well Mike Lalonde has taken his love for Ozzy and Black Sabbath to the stage. Their band 66Bloor has toured many Northern Ontario towns. They have been likened to a cross between the Headstones and Ugly Kid Joe, but with more past issues So keep your eyes and ears open, you'll be hearing more of these guys. The bands covers are awesome too. Band members are Vocals: Gilles Mallette Lead Guitar: Mike Lalonde Rhythm Guitar: Craig Jefferson Bass Guitar: James Roy Drums: Mark Tessarolo http://www.garageband.com/artist/66Bloor === Subject: Re: Mathematics: art or science? <871wdi1dom.fsf@huxley.huxley.fi> consistency, or consistency of one in another. We can prove the consistency of Peano arithmetic in the same sense as any theorem of ordinary mathematics: by producing a correct argument from principles we accept as correct and indeed employ all the time. Of course if one does not accept these principles one will not accept the proof either, but there is nothing special in consistency statements in this respect; the observation applies as well to Kruskal's tree theorem or any result of ordinary mathematics. > There is also the question of consistency of > inaccessible ordinals. In case of inaccessible cardinals the situation is different, since we know that their consistency can't be proved using the usual principles employed in ordinary mathematics as it is today. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Mathematics: art or science? > If the numerical solution converges to the analytical solution then of > course the analytical solution exists. But for the other direction you > need consistency and stability plus the Lax-Richtmeyer theorem, which > in turn needs machinery from analysis. > > You know all this perfectly well. Yes. And I find this consistency and stability extremely irritating. Consider the differential equation d^u/dx^2 + u = 0 . Its analytical solution is a sine function. Of course, this solution isn't stable. It's inherently _un_stable. In this way, _lots_ of quite interesting (ODE's and) PDE's are beyond the scope of the Lax-Richtmeyer theorem. Apart from this, there are examples of Finite Element Methods which, according to standard theory, converge. But they do it so slowly that these methods are nevertheless completely useless in practice. An example is the Least Squared Finite Element Method with common (linear) triangles: And this is only the top of the iceberg ... Han de Bruijn === Subject: LOWER MY PAYMENTS AND INTEREST.... WE CAN HELP LOWER MY PAYMENTS AND INTEREST.... WE CAN HELP www.BigLoanGuide.com === Subject: Re: #28F perhaps an easy path to a proof that (pi) and (e) are the only two independent Transcendental Numbers Re: ATOM TOTALITY (Atom Universe) THEORY REPLACES BIG BANG THEORY IN PHYSICS > ... > Do you say that temperature is formally defined > in terms of the material properties of water, > or any other substance? > > One metric of temperature is: > http://physics.nist.gov/cuu/Units/current.html > ... and history > http://physics.nist.gov/cuu/Units/kelvin.html > > Temperature itself is devoid of being tied to any single > material. Yes, I know that, and you would know that I know that if you read my participation in this thread. -- Michael Press === Subject: Specialized Search Engines along with Google Search Capability Specialized Search Engines along with Google Search Capability (2 in 1): http://specialized-search-engines.blogspot.com/ Billions of websites are available on the web and plenty of extremely good search engines are there like Google, Yahoo and Live to name few of them. Though this search engines have extremely efficient, complex and beautiful algorithms designed by gems of the industry, but still they may not deliver best results for the advanced users due to following reasons: (1) Normally a user (and specially advanced user) is not willing to look more than 20-30 results to find out exact information they are looking for and it really doesn't matter that one search engine gives fifty thousand results and other search engine gives three lack results to the end user. (2) As the search engines tend to index maximum number of sites they generally tend to include some irrelevant sites in the results as they may find matching text in completely different context. Though these and some more problems does not bother normal user and even to advance users in most of the cases, but they are cases when it really matters (specially when you are searching for the topics of your personal interests). Google has come up with Custom Search Engine when you can build your own search engine based on your taste or interests. I have built one such pool of specialized/custom search engines: http://specialized-search-engines.blogspot.com/ The home page (link give above) contains GOOGLE SEARCH capability only page of this site), which is what most of the users uses most of the time. And you will find links for the specialized search engines on this page. So you can bookmark the home page and can perform your normal searches as usual and can use specialized search capabilities whenever for their feedbacks because of whom such improvements are materialized. Currently four different search engines are available for: Software Search Developer's Search - Beta Physics Search - Beta Contemplative Search You like it or not, still would like to here your praise/complaints: passion_0155@yahoo.com This search engines have just born and will keep evolving and to do this your feedback is really important. If you own any website or blog on any of the above topics and you are interested in providing such search ability into your site then contact me at passion_0155@yahoo.com. - passion_0155@yahoo.com === Subject: Taking dual distributes over tensor product? If V and W are finite dimensional, is it true that (V otimes W)* is naturally isomorphic to V* otimes W* ? It is easy to cook up what the natural isomorphism should be: Define phi : V* otimes W* ---> (V otimes W)* on simple tensors v otimes w by phi (f otimes g) (v otimes w) = f(v) otimes g(w) However, this is defined on only a generated set of V* otimes W*, so how do we know that it extends to a linear map? I came across this when trying to connect two definitions of the tensor bundle of a manifold. I am used to defining it as simply replacing each fiber F_p of the tangent bundle with T_l^k(F_p) (i.e. the set of multilinear maps from k copies of V and l copies of V* to R). However, another way I have seen it defined is by simply taking the tensor product of some number of the tangent bundle and some number of the cotangent bundle. Showing that these bundles are isomorphic boils down to show that the above two vector spaces are naturally isomorphic. === Subject: 23.- Unprovable nature of the Goldbach Conjecture in Peano Arithmetic. Any formal proof of the Goldbach Conjecture in PA would imply the existence of a sequence (A_1,..., A_n) of well formed formulas in PA which constitute a proof of the Goldbach Conjecture in the infinite set: S = {a: (a even number) and (a>=16) and (a/2 non- prime) and (a-3 non -prime)} This would imply the existence of a proof of the Goldbach Conjecture (A_1(t),...,A_n(t)) (t=t(u), u>1, t time) valid in infinite isomorphic structures (N(u),+(u),*(u)) to the usual one (N,+,*). For u=1 we have another structure (N(1), +(1), *(1)) isomorphic to the usual one (N,+,*), but (A_1(1),...,A_n (1)) would not be a proof of the GC due to the fact that for u=1 the characterization in S of the Goldbach Conjecture via essential points in the corresponding dynamical process has been lost. All of this , has to do with the convenience of consider natural numbers associated to states of time and these ones associated to hyperbolas. So we avoid unidirectional time ( prime factorization). I have sometimes thought that the profound mystery which envelops our conceptions relative to prime numbers depends upon the limitations of our faculties in regard to time which like space may be in essence poly-dimensional and that this and other such sort of truths would become self-evident to a being whose mode of perception is according to superficially as opposed to our own limitation to linearly extended time. (J.J. Sylvester) P.S. For previous comments see: http://mathforum.org/kb/thread.jspa?threadID=1620940&messageID=5903846#59038 46 Fernando. === Subject: Roots to Power Series Compute the roots to any Power Series. http://mypeoplepc.com/members/jon8338/math/id2.html === Subject: Re: Proposal: 1 year killfile of everyone who has asked for a solutions manual dontdont@gmail.com hath wroth: >This year there have been about 2000 >postings in the last 2.5 months begging to buy the homework >answers just in the four newsgroups that I have posted to. I don't think a killfile would do much to reduce the number of cheaters. Methinks a poison pill approach would be more useful. Collect some of the solutions manuals and change the answers and methods to render them useless. Then, post the modified solutions manuals to all the usual places (file sharing and download sites). Printed copies could be sold on eBay. I predict a sudden decrease in student test scores. I'm not sure how many modified solutions books will be required to produce the desired results. It may only be necessary to edit the most popular undergraduate solutions books and do the rest as time permits. It could also be performed as a group project, where volunteers edit only specific solutions books to avoid duplication. The predicted dramatic drop in test scrores will probably be immediately attributed to global warming, Lead in kids toys, RF pollution, or genetic damage, which will surely promote a corresponding increase in research grants and published papers. This may justify the time spent on the project. -- Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558 === Subject: Re: Proposal: 1 year killfile of everyone who has asked for a solutions manual dontdont@gmail.com says... > I've been quietly watching from the sidelines the continual > stream of people begging for a solutions manual for their > textbooks this year, in sci.math and the other newsgroups > I've posted this to. This year there have been about 2000 > postings in the last 2.5 months begging to buy the homework > answers just in the four newsgroups that I have posted to. > I do realize once one kid posts that ten others see this and > think what a great idea, why don't I buy my homework > answers too? > > I've resisted doing this for a while but finally enough is > enough. > > I propose we collectively killfile the name of everyone who > has asked and will ask for solutions manuals for the next > year, in all newsgroups. You know if they are begging for > the solutions manual and they don't get it they will be back > begging for someone to do their homework for them next week. > > My killfile already has a lot of these names and if anyone > wants a list to add to their killfile I will happily spend > the time to go back through old postings and carefully > confirm the names and send you a list so that you can drop > them into your killfile too. If enough of us killfile them > they become unpersons. > > I have considered this and think I realize the limitations > and consequences. > I kinda like what they do over in comp.arch. They answer obvious homework questions with funny, plausible, and wrong answers. It's good for a laugh and if the poser is as dumb as they usually sound, will get a second laugh later. The only problem is that the C.A crowd doesn't get to be in on the last laugh. -- Keith === Subject: Re: Proposal: 1 year killfile of everyone who has asked for a solutions manual >Nice try, but it didn't work! Top posting, Usenet retard! === Subject: Re: Proposal: 1 year killfile of everyone who has asked for a solutions manual > >Nice try, but it didn't work! > > > Top posting, Usenet retard! But it wasn't a _response_. It was a re-presentation. The only new content was the _introduction_ to what followed. Introductions come *before* the thing they introduce. Soon we'll have you counting on both hands, hopefully. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Proposal: 1 year killfile of everyone who has asked for a solutions manual On 12 Sep 2007 13:31:37 +0300, Phil Carmody >> >>Nice try, but it didn't work! >> >> >> Top posting, Usenet retard! >But it wasn't a _response_. It was a re-presentation. >The only new content was the _introduction_ to what >followed. Introductions come *before* the thing they >introduce. Soon we'll have you counting on both hands, hopefully. Phil I used slide rules long before my first scientific calculator purchase, and that was in the mid seventies, dip. You likely don't even no the date we pass through the Galactic Equator, and you think you are going to give me a primer on math? Fat chance. We'll... No, You and whomever you include with yourself won't be having me doing anything, asswipe.