mm-106 You must stop smoking that stuff! Its having a terrible effect onyour mind.>While we all eagerly await the publicists results about my important>paper which rights the contradictions of the transnitude and sets>ablaze the torch of absolute truth, I will in the meantime divulge the>general idea upon which the proof is built. Of course this is not the>proof itself, being merely a laymans terms version, and the actual>proof requires much more rigour. Nevertheless I anticipate that even>this brief sample will shed some enlightenment upon you and convince>you of my sincerity and credibility. :-)>>I assert, then, that Cantors principals are based erroneously upon an>axiom which is very subtle and ghostlike, so much so in fact that even>the most learned of the mathematical community are blind to it, though>it is right under their noses. To wit, I refer to that ever-assumed>hypothesis that we can construct a set whose members are uncountable>in the rst place. But wait!, cry the critics, What of the set of>real numbers? But do not laugh at them, for these are very subtle>matters and their lack of understanding is worthy of our sympathy. >Let us, then, address their concern. For, indeed, we may say Let R>then be all such numbers which cannot be represented as the quotient>of two integers, and a unique well-dened set named R does thus>arise, there is no arguing about this. However, how do we know R is>uncountable? Because Cantor asserts that it is. But now I am the one>to cry But wait!. Cantors proof subtly assumes that, even if a>transnitude does exist (which we will for the moment allow, just to>give him the benet of the doubt), that it is then possible to>construct a set of such magnitude. So when I assert that this>constructability is unproven, and you offer R as a counterexample with>Cantors assertian that R possesses such magnitude, it is clear as day>that you are using circular reasoning. But I do not hold this against>you, because this is a matter of very great subtlety which is very>easily overlooked, so do not hit your head for being stupid. You are>not alone.>Now I see that you are greatly confused, now that the matter of this>assumed, implicit axiom about the constructability of transnite sets>(even given the existance of a transnitude) has been laid out before>your eyes. For, surely this axiom must hold true, you reason, else>many works of mathematics will be rendered invalid, many great proofs>of famous reknown rendered erroneous. Alas, I say, this is not the>case. They are unsalvageable, and not just the proofs but the>theorems themselves. But do not be saddened, for the mappings I will>soon propose will be of such profound insight into mathematics that>they will more than compensate for all the works thus lost.>To wit, I will in short time present for your examination a>well-dened mapping, accessible even to the amateur mathematician in>its simplicity, yet inspirational to even the most learned>mathematician for its ingenuity, which establishes a one-to-one>correspondance between the set of integers and the set of all real>numbers. I will present a common-sense, laymans terms, unrigorous>sample of this mapping at a later time to this worthy forum, and the>nal, rigorous version will be presented at a still later time when>the publicists have overcome their sheer astoundment upon learning the>true nature of these things. At that time a link will be given freely>to you to a webpage where you will be able to examine all of these>matters in all their profound rigour with as much skepticism as you>like, and to learn that all such skepticism is misled.>>I eagerly look forward to sharing these great subtle truths with you>:-)>>Nathan the Great>Age == It was in the early Seventies, when I was in Mainz. There wassomething in the air.A certain Je ne sais quoiPerhaps even a Je ne veux pas savoir - a foreign thing.Something transnite on the breeze. A breath of Cantor.Then in the later Seventies, as I proceeded with my work on EulersGamma function, I found the Fantasy Maths. Just like the Alephs ofCantor.I began to wax poetic - with SILICONE wax. The best.I conjugated all the verbs into their complex conjugates.I declined the nouns. I dont like those. Not even when their angle ofdeclination is dened.Am I to be overtaken by Nathan the Great?What is the deep secret of his eternal youth? He has been 11 since atleast 1994.Is it his use of UNCOUNTABLE NUMBERS? Will we ever know?I put something vague about the Fantasy Maths on my website athttp://wehner.org/eulerPerhaps I shall return one day to Mainz. It is 50 degrees North, when the weather is right. Celsius or Fahrenheit - whatever!Charles Douglas Wehner I assert, then, that Cantors principals are based erroneously upon an> axiom which is very subtle and ghostlike, >[...]> hypothesis that we can construct a set whose members are uncountableMathematicians will never accept a common sensedenition of construct as invalidation of theirown abstract, surreal constructions, no matterhow compelling that common sense denition may be.If you want to beat the mathematicians at their own game, you would have to nd a formal inconsistencyin their logic, where they have made up the rulesof logic. I think its unlikely you can succeedthat way.The key to beating the Cantorians is to focus on a notion of reality. Mathematics, before Cantor camealong, always had something to do with computation.And computation is connected to reality. Cantors theory has nothing to do with computation, and hencehas nothing to do with reality.Someday, the Cantorians are going to lose it. The key to beating the Cantorians is to focus on a notion of> reality. Mathematics, before Cantor came along, always had> something to do with computation.Oh, really? Euclid is about computation?> And computation is connected to reality. Oh, really? > Cantors theory has nothing to do with computation, and hence has> nothing to do with reality.>> Someday, the Cantorians are going to lose it.Not as a result of your big brain, apparently.Cantor was not the paradigm-shift that you imagine.-- A set having three members is a single thing wholly constituted byits members but distinct from them. After this, the theologicaldoctrine of the Trinity as three in one should be childs play. --Max Black, _Caveats and Critiques_ > The key to beating the Cantorians is to focus on a notion of> reality. Mathematics, before Cantor came along, always had> something to do with computation.>> Oh, really? Euclid is about computation?>> And computation is connected to reality.>> Oh, really?>> Cantors theory has nothing to do with computation, and hence has> nothing to do with reality.>> Someday, the Cantorians are going to lose it.>> Not as a result of your big brain, apparently.>> Cantor was not the paradigm-shift that you imagine.>> -- > A set having three members is a single thing wholly constituted by> its members but distinct from them. After this, the theological> doctrine of the Trinity as three in one should be childs play.> --Max Black, _Caveats and Critiques_Why the ames? Why make disparaging remarks about Petrys brain? Petryspoint of view goes way back, including Kronecker, Brouwer, and recently,Bishop. They have always been in a minority, because their rules about whatconstitutes valid mathematics are much stricter. In any case, you certainlycant disparage the brains of these men.Poincare had changing feelings over Cantors work. At rst he stronglyendorsed it, but had serious reservations later.Bishop was able to recreate most of basic analysis (real analysis, complexanalysis, functional analysis) in a very strict constructive setting. Seehis book, _Constructive Analysis_. Be forewarned that its rough going,though.If you want a statement that you can really get riled up about, try BishopsTopology is a vampire that sucks the lifeblood of mathematics. (quotefrom memory, not exact)The entire topic is certainly an interesting one for discussion, forthoughtful give-and-take, for each individual to try personally to come toterms philosophically with these issues. <874r1dem8d.fsf@phiwumbda.localnet> > reality. Mathematics, before Cantor came along, always had> something to do with computation.>> Oh, really? Euclid is about computation?>> And computation is connected to reality.>> Oh, really?>> Cantors theory has nothing to do with computation, and hence has>> nothing to do with reality.>> Someday, the Cantorians are going to lose it.>> Not as a result of your big brain, apparently.>> Cantor was not the paradigm-shift that you imagine.> Why the ames? Why make disparaging remarks about Petrys brain?Perhaps the ames were out of line. I apologize to David andappreciate the remark from Will.> Petrys point of view goes way back, including Kronecker, Brouwer,> and recently, Bishop. They have always been in a minority, because> their rules about what constitutes valid mathematics are much> stricter. In any case, you certainly cant disparage the brains of> these men.Its not a matter of whether or not Cantors set theory was a newmathematical and even philosophical direction. I tend to agree thatit was a new trend in mathematics, although I hesitate to be moreexplicit in how it differed from existing mathematics without a morethorough background in the history of mathematics.However, to claim that all mathematics before Cantor was aboutcomputation is simply ridiculous. Also, to claim that computationis connected with reality is either ridiculous or, more likely,meaningless.> Poincare had changing feelings over Cantors work. At rst he strongly> endorsed it, but had serious reservations later.>> Bishop was able to recreate most of basic analysis (real analysis, complex> analysis, functional analysis) in a very strict constructive setting. See> his book, _Constructive Analysis_. Be forewarned that its rough going,> though.>> If you want a statement that you can really get riled up about, try Bishops> Topology is a vampire that sucks the lifeblood of mathematics. (quote> from memory, not exact)>> The entire topic is certainly an interesting one for discussion, for> thoughtful give-and-take, for each individual to try personally to come to> terms philosophically with these issues.I am more or less familiar with constructivism and related topics inthe philosophy of mathematics. If Petry was attempting to give aconstructivist critique of modern set theory, then it was whollyunrecognizable in his words.Perhaps I was harsh in my reaction, but I have yet to see a seriousphilosopher of mathematics anywhere discuss the so-called Cantorians.Cantors result is an important example of the distinction betweenconstructive and non-constructive mathematics, perhaps, but it is a(now obvious) theorem of the predominant set theory. If one wants toreject Cantors theorem, one must reject either the power set axiom orthe axiom of innity. This is a rejection not of Cantor and hisarguments but of a particular mathematical theory. I tend to believe that anyone referring to Cantorians must miss thisbasic and obvious point. Cantors argument is clearly awless, sotheir disagreements are either simply wrong or they are not withCantors proof at all, but with the prevailing axiomatic theory forsets.-- No feeling sympathy for mathematicians who start marching with signslike Will work for food in the future... I will not show mercygoing forward. I was trained as a soldier in the United States Armyafter all... == Lamely following up to my damn self.> Perhaps I was harsh in my reaction, but I have yet to see a serious> philosopher of mathematics anywhere discuss the so-called Cantorians.> Cantors result is an important example of the distinction between> constructive and non-constructive mathematics, perhaps, but it is a> (now obvious) theorem of the predominant set theory. If one wants to> reject Cantors theorem, one must reject either the power set axiom or> the axiom of innity.Or the comprehension axiom, of course, but I havent seen anyonedispute the comprehension axioms personally. I am not sure whetherthe power set axiom is controversial in constructivist circles. Iassume it is, in fact would be shocked if not, but I dont recall anydiscussions on the topic.Clearly, as well, the axiom of innity isnt relevant to the proofthat the cardinality of any set is strictly less than the cardinalityof its power set. But, in the nite case, there is no disagreementwith this theorem at all. Hence, I include the axiom of innity inthe list of dubious axioms above.Of course, to reject the axiom of innity in order to avoid thetransnite hierarchy is a bit odd. We get rid of different sizes onnite sets in this way, but only by getting rid of innite setsaltogether!-- Sorry, wakeup to the real world. Youre on your own dependent on meas your guide. Luckily for you, Im self-correcting to a large extent,so if the proof were wrong, Id tell you. Its not wrong. --- James Harris conrms that his proof is correct. However, to claim that all mathematics before Cantor was about> computation is simply ridiculous. Also, to claim that computation> is connected with reality is either ridiculous or, more likely,> meaningless.Heres the idea. We can think of the computer as a microscope which allowsus to peer deeply into the world of computation. This worldof computation is, unlike Cantors fantasy world, somethingthat we all agree exists. Its a world in which we can performexperiments and make observations. We can think of mathematicsas a science which studies phenomena in this world. As such,mathematics can be seen as a science which studies a real,observable world.Euclid does t within this paradigm, if we think of thehuman brain as a computer with special hardware for performingcalculations in three dimensional space.> Cantors argument is clearly awless, so> their disagreements are either simply wrong or they are not with> Cantors proof at all, but with the prevailing axiomatic theory for> sets.The axioms of set theory were chosen so as to makeCantors argument work formally. So your assertion haslittle content. Starblade Darksquall> Proof:>> 1) Ax(BxC) != (AxB)xC> Since Ax(BxC) = B(A*C)-C(A*B), and (AxB)xC = -Cx(AxB) = -A(C*B)+B(C*A)> = B(A*C)-A(B*C) which is clearly NOT B(A*C)-C(A*B) since A(B*C) ! C(A*B) except for special cases, namely, when A(B*C)-C(A*B) = 0, IE,> when Bx(AxC) = 0, then clearly cross products are noncommunative.> 2) AxA = 0> This follows from the denition of the cross product. The cross product of parallel vectors is zero.> 3) (A+B)xC = AxC+BxC> Since the vector cross product IS distributive over addition. This is> easy enough to verify.> 4) Ax(BxC)+Bx(CxA)+Cx(AxB) = 0> Changing it to (B(A*C)-C(A*B))+(C(B*A)-A(B*C))+(A(C*B)-B(C*A)) makes> it obvious that this identity is true, since that the dot product is> communative.> 5) AxB=-BxA> This is as easy to verify as 2.>> So clearly, we have known about a type of lie algebra for longer than> we let on. In fact, we use lie algebra type mathematics in dealing> with even classical physics all the time, particularly in> electromagnetics.> Er, sort of. The old vector product AxB is the same as *(A^B) where * is the> Hodge dual and ^ is the (antisymmetric) exterior product. (Im not sure if> Hodge dual is standard jargon, but its easy to see what the term is> intended to mean.)> LH> Are you talking about tensors, or something else entirely?> (...Starblade Riven Darksquall...)it is a well known fact that the linear space R^3 with the crossproduct is a lie algebra. let A, B in R^3, then the hodge dual (thestar * operator) mapsA^B to A cross B, and vice versa. as such, it is a linear spaceisomorphism between R^3 and linear space of bivectors. regardingelectromagnetism, it can certainly be formulated using the exterioralgebra, or, in more compact form, the clifford algebra on R^3. M.T. == I am working on a problem from Dummit & Footes book, Abstract Algebra.The problem is tofind the remainder of 37^100 when divided by 29.I have played around with 37 = 8 (mod 29) by raising it to powers, but Ihave yet to discover the right path. I tried37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). I have also thought aboutresidue classes, but I just cant seem to make any connections. Couldanybody give me an idea?TIALurch I am working on a problem from Dummit & Footes book, Abstract Algebra.>The problem is tofind the remainder of 37^100 when divided by 29.>>I have played around with 37 = 8 (mod 29) by raising it to powers, but I>have yet to discover the right path. I tried>>37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). I have also thought about>residue classes, but I just cant seem to make any connections. Could>anybody give me an idea?As you note, 37 = 8 (mod 29). What happens to 8 as you raise it tosuccessive powers, modulo 29?What is 8^{28} (mod 29), according to Fermats Little Theorem? What is8^{29} (mod 29)? If a^b = 1 (mod c), then how much is a^{k*b} (mod c)? == I am working on a problem from Dummit & Footes book, Abstract Algebra.>The problem is tofind the remainder of 37^100 when divided by 29.>I have played around with 37 = 8 (mod 29) by raising it to powers, but I>have yet to discover the right path. I tried>37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). Umm... Please review the laws of exponents.>I have also thought about>residue classes, but I just cant seem to make any connections. Could>anybody give me an idea?obtained by four squarings.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 I am working on a problem from Dummit & Footes book, Abstract Algebra.> The problem is tofind the remainder of 37^100 when divided by 29.> I have played around with 37 = 8 (mod 29) by raising it to powers, but I> have yet to discover the right path. I tried> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). I have also thought about> residue classes, but I just cant seem to make any connections. Could> anybody give me an idea?Start with 37^1 (mod 29) = 8 (mod 29). Then 37^2 (mod 29) = 37*8 (mod 29) = 296 (mod 29) = 6 (mod 29). Then 37^3 (mod 29) = 37*6 (mod 29) = ...and so on. Repeat until you have 37^100 mod 29. There is a shortcut to this: If you go on with the calculation, you will eventually get the same result again. For example, you mightfind that 37^2 (mod 29) = 37^10 (mod 29). (I am not saying these two are the same, but there are only 29 possible results, so at some point you must get the same result that you got before). If you found that 37^2 (mod 29) = 37^10 (mod 29), for example, then you would know that 37^18, 37^26, 37^34 and so on are the same again, so you couldfind 37^100 mod 29 quite quickly. I am working on a problem from Dummit & Footes book, Abstract Algebra.> The problem is tofind the remainder of 37^100 when divided by 29.>> I have played around with 37 = 8 (mod 29) by raising it to powers, but I> have yet to discover the right path. I tried>> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). I have also thought about> residue classes, but I just cant seem to make any connections. Could> anybody give me an idea?>> TIA>> LurchBeing not familiar with this stuff, I did some trial and errorwith a high precision calculator and this is what I found:Everything mod 29:37^100 = 8^100 = 2^300 = 2^(300-28) ??? = 2^(300-2*28) ??? = 2^(300-10*28) ??? = 2^(20) = 1048576 = 23Does anybody know whether this is a (special caseof some) theorem: 2^k = 2^(k-p+1) (mod p)?Dirk Vdm == couldnt see the connection :( And, yes Robert I do see my bonehead mistakewith the exponents. Sorry.Lurch> I am working on a problem from Dummit & Footes book, Abstract Algebra.> The problem is tofind the remainder of 37^100 when divided by 29.>> I have played around with 37 = 8 (mod 29) by raising it to powers, but I> have yet to discover the right path. I tried>> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). I have also thought about> residue classes, but I just cant seem to make any connections. Could> anybody give me an idea?>> TIA>> Lurch>> = Dirk Van de moortel > I am working on a problem from Dummit & Footes book, Abstract Algebra.> The problem is tofind the remainder of 37^100 when divided by 29.>> I have played around with 37 = 8 (mod 29) by raising it to powers, but I> have yet to discover the right path. I tried>> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). I have also thought about> residue classes, but I just cant seem to make any connections. Could> anybody give me an idea?>> TIA>> Lurch> Being not familiar with this stuff, I did some trial and error> with a high precision calculator and this is what I found:> Everything mod 29:> 37^100> = 8^100> = 2^300> = 2^(300-28) ???> = 2^(300-2*28) ???> = 2^(300-10*28) ???> = 2^(20)> = 1048576> = 23> Does anybody know whether this is a (special case> of some) theorem:> 2^k = 2^(k-p+1) (mod p)> ?> Dirk Vdm>Yes. For every prime p, x^p == x (mod p) for all integers x,so if GCD(x,p) = 1, then x^(p-1) == 1 (mod p).Then it follows for all primes p and positive integers x,y and k,that x^(y + k*(p-1)) == x^y (mod p)Thus, if p is prime and u == v (mod p-1)one gets x^u == x^v (mod p) for all integers x. Dirk (special case> of some) theorem:> 2^k = 2^(k-p+1) (mod p)> ?>> Dirk Vdm>> Yes.>> For every prime p, x^p == x (mod p) for all integers x,> so if GCD(x,p) = 1, then x^(p-1) == 1 (mod p).>> Then it follows for all primes p and positive integers x,y and k,> that x^(y + k*(p-1)) == x^y (mod p)> Thus, if p is prime and u == v (mod p-1)> one gets x^u == x^v (mod p) for all integers x.virtually *nothing* of it.Well, maybe I remember this: x = y (mod p) <= There is a whole number k such that: x-y = k*p.and thats where it ends.So I have been trying to prove that rst line of yours: For every prime p, x^p == x (mod p) for all integers xand I got stuck. Help!Any elementary on-line textbook (pdf or ps) you couldrecommend?Dirk Vdm Dirk Van de moortel>> case> of some) theorem:> 2^k = 2^(k-p+1) (mod p)> ?>> Dirk Vdm>> Yes.>> For every prime p, x^p == x (mod p) for all integers x,> so if GCD(x,p) = 1, then x^(p-1) == 1 (mod p).>> Then it follows for all primes p and positive integers x,y and k,> that x^(y + k*(p-1)) == x^y (mod p)> Thus, if p is prime and u == v (mod p-1)> one gets x^u == x^v (mod p) for all integers x.> virtually *nothing* of it.> Well, maybe I remember this:> x = y (mod p) <=> There is a whole number k such that: x-y = k*p.> and thats where it ends.> So I have been trying to prove that rst line of yours:> For every prime p, x^p == x (mod p) for all integers x> and I got stuck. Help!> Any elementary on-line textbook (pdf or ps) you could> recommend?> Dirk VdmMaybe Course 311 - Abstract Algebra, Part I: Topics in Number Theory:http://www.maths.tcd.ie/~dwilkins/Courses/311/ 311NumTh.pdfby D.R. Wilkinsor follow the links given in Annotated Web Links forKenneth H. Rosens Elementary Number Theory Book:http://www.aw-bc.com/rosen/resources.htmle.g.Fermats Little Theoremhttp://www.cut-the-knot.org/blue/Fermat.shtmlor search the homepages of number theorists:http://www.numbertheory.org/ntw/list.htmlMany of them have lecture notes online.Hugo Pfoertner =[...]> Any elementary on-line textbook (pdf or ps) you could> recommend? Dirk VdmOnline number theory lecture notes:http://www.numbertheory.org/ntw/lecture_notes.htmlHugo Pfoertner Part I: Topics in Number Theory:> http://www.maths.tcd.ie/~dwilkins/Courses/311/311NumTh.pdf> by D.R. WilkinsJust what I needed.Apparently this thing For every prime p, x^p == x (mod p) for all integers xis Fermats theorem. Not so trivial. no wonder I didntfinda simple proof :-)> or follow the links given in Annotated Web Links for> Kenneth H. Rosens Elementary Number Theory Book:> http://www.aw-bc.com/rosen/resources.html>> e.g.> Fermats Little Theorem> http://www.cut-the-knot.org/blue/Fermat.shtmlYep, had found http://www.cut-the-knot.org/blue/Modulo.shtmlThis was exactly what I was looking for.Had this stuff is almost 30 years ago - Good refresherfor the basics.>> or search the homepages of number theorists:> http://www.numbertheory.org/ntw/list.html>> Many of them have lecture notes online.>> Hugo PfoertnerDirk Vdm == So I have been trying to prove that rst line of yours:> For every prime p, x^p == x (mod p) for all integers x> and I got stuck. Help!The set {1,2,...,p-1} is a group of order p-1 under multiplication modp, so whenever x is not zero (mod p)x^(p-1) = 1 (mod p) (Lagranges theorem) whencex^p = x (mod p)which also holds for x=0 (mod p). =without the aid of a calculator/computer, or Fermats little theorem. Inthe book, all he discusses is modular arithmetic. FLT is not mentioned tilpage 97. I am on page 10.Even with FLT, I cant seem to get it. If a^p = a mod p, a is the remainderright? So, if, for example, I take 3^2 mod 2 the theorem says I should get3, but isnt it 1? I mean 9 mod 2 should be 1, right? Futhermore, if Itake 37^29 = 37 mod 29 and I try to raise this congruence by powers until Iget my 100, then my remainder gets larger and larger. So, if my remainderis supposed to be 23, I doubt that 37^x is going to be my answer. What theheck am I missing here? (Besides a brain)Lurch> I am working on a problem from Dummit & Footes book, Abstract Algebra.> The problem is tofind the remainder of 37^100 when divided by 29.>> I have played around with 37 = 8 (mod 29) by raising it to powers, but I> have yet to discover the right path. I tried>> 37^2*37^2*37^5*37^5 = 8^2*8^2*8^5*8^5 (mod 29). I have also thought about> residue classes, but I just cant seem to make any connections. Could> anybody give me an idea?>> TIA>> Lurch>> without the aid of a calculator/computer, or Fermats little theorem. In> the book, all he discusses is modular arithmetic. FLT is not mentioned til> page 97. I am on page 10.hm, maybe something like this?x = 37^100 (mod 29) = 8^100 (mod 29) casted out 29 = 64^50 (mod 29) = 6^50 (mod 29) casted out 2*29 = 58 = 36^25 (mod 29) = 7^25 (mod 29) casted out 29 = 7*7^24 (mod 29) = 7*49^12 (mod 29) = 7*20^12 (mod 29) casted out 29 = 7*400^6 (mod 29) = 7*23^6 (mod 29) = 7*529^3 (mod 29) = 7*7^3 (mod 29) casted out 29 = 49*49 (mod 29) = 20*20 (mod 29) = 400 (mod 29) = 23 (mod 29) casted out 29Dirk Vdm =fairly soon, but I was starting to get a little frustrated. On the otherhand, they say close only counts in horseshoes and handgrenades; so, whoknows? I think that your method will do the trick.Lurch>out> without the aid of a calculator/computer, or Fermats little theorem.In> the book, all he discusses is modular arithmetic. FLT is not mentionedtil> page 97. I am on page 10.>> hm, maybe something like this?> x = 37^100 (mod 29)> = 8^100 (mod 29) casted out 29> = 64^50 (mod 29)> = 6^50 (mod 29) casted out 2*29 = 58> = 36^25 (mod 29)> = 7^25 (mod 29) casted out 29> = 7*7^24 (mod 29)> = 7*49^12 (mod 29)> = 7*20^12 (mod 29) casted out 29> = 7*400^6 (mod 29)> = 7*23^6 (mod 29)> = 7*529^3 (mod 29)> = 7*7^3 (mod 29) casted out 29> = 49*49 (mod 29)> = 20*20 (mod 29)> = 400 (mod 29)> = 23 (mod 29) casted out 29>> Dirk Vdm>> fairly soon, but I was starting to get a little frustrated. On the other> hand, they say close only counts in horseshoes and handgrenades; so, who> knows? I think that your method will do the trick.I wouldnt have found it if I hadnt taken a nice refreshinglook this afternoon at http://www.cut-the-knot.org/blue/Modulo.shtmland specially at the solved elementary problems in http://www.cut-the-knot.org/blue/chinese.shtml (the casting out business)Heres a little proof I just found for this casting out p: x = a (mod p) = x = kp + a = x = kp+np + a-np = x = (k+n)p + a-np = x = a-np (mod p)and a more general case: x = a^m (mod p) = x = kp + a^m = x = kp + Poly[a,n,p,m]*p + (a-np)^m = x = (k+Poly[a,n,p,m])*p + (a-np)^m = x = (a-np)^m (mod p)hm, I think I have been wrong never having liked numbertheory ;-)Dirk Vdm > without the aid of a calculator/computer, or Fermats little theorem. In>> the book, all he discusses is modular arithmetic. FLT is not mentioned til>> page 97. I am on page 10.>>hm, maybe something like this?>x = 37^100 (mod 29)> = 8^100 (mod 29) casted out 29> = 64^50 (mod 29)> = 6^50 (mod 29) casted out 2*29 = 58> = 36^25 (mod 29)> = 7^25 (mod 29) casted out 29> = 7*7^24 (mod 29)> = 7*49^12 (mod 29)> = 7*20^12 (mod 29) casted out 29> = 7*400^6 (mod 29)> = 7*23^6 (mod 29)> = 7*529^3 (mod 29)> = 7*7^3 (mod 29) casted out 29> = 49*49 (mod 29)> = 20*20 (mod 29)> = 400 (mod 29)> = 23 (mod 29) casted out 29Or simpler: 37^100 (mod 29) = 8^100 (mod 29) = 2^300 (mod 29) = 2^20 (mod 29) casted out 28*10 by F = 32^4 (mod 29) = 3^4 (mod 29) = 3*27 (mod 29) = 3*-2 (mod 29) = -6 (mod 29) = 23 (mod 29).Each of these steps is quite easy to verify in ones head. -- Erick == Even with FLT, I cant seem to get it. If a^p = a mod p, a is the remainder> right? So, if, for example, I take 3^2 mod 2 the theorem says I should get> 3, but isnt it 1? I mean 9 mod 2 should be 1, right? Futhermore, if I> take 37^29 = 37 mod 29 and I try to raise this congruence by powers until I> get my 100, then my remainder gets larger and larger. So, if my remainder> is supposed to be 23, I doubt that 37^x is going to be my answer. What the> heck am I missing here? (Besides a brain)The remainder may always be taken as a non-negative integer less than the modulus. For modulus 2, that means zero or 1 (even or odd).Note that, according to one denition of congruences, 9 == 1 (mod 2) just means (9 - 1) is divisible by 2, which it is.In the following = is ordinary equality and == is congruenceSince 37 == 8 (mod 29), you know that 37^29 == 8^29 == 8 (mod 29).Further, x^(29-1) = x^28 == 1 (mod 29), for GCD(x,29) = 1.Consider that 100 = 3*28 + 16 == 16 (mod 28),so that now 37^100 == 37^ 16 == 8^16 (mod 29).Also 8^2 = 64 == 6 (mod 29)and 6^2 = 36 == 7 (mod 29) and 7^2 = 49 == 20 (mod 29)and 20^2 = 400 == 23 (mod 29 so 8^16 = (((8^2)^2)^2)^2 == 23 (mod 29).All done by hand without need for electronic aids. > I am working on a problem from Dummit & Footes book, Abstract Algebra.> The problem is tofind the remainder of 37^100 when divided by 29.>> I have played around with 37 = 8 (mod 29) by raising it to powers, but> I have yet to discover the right path. [...] Id like to gure this out> without the aid of a calculator/computer, or Fermats little theorem. > In the book, all he discusses is modular arithmetic. FLT isnt mentioned> til page 97. I am on page 10. [...]Mod 29: 37^100 = 8^100 = 2^300. To easily compute thislets search for a small power of 2 that equals +-1 (mod 29).We search for numbers equal +-1 (mod 29) that factor intopowers of 2 times small known powers of 2, such as 3 = 2^5.Mod 29: 1 = 6*5 = 6(-24) = -2^4 3^2 = -2^14 via 3 = 2^5so 2^300 = 2^(14*21+6) = (-1)^21 2^6 = -6 = 23.The search succeeds quickly however you do it, e.g. -1 = 4*7 = 4(6^2) = 2^4 3^2 1 = 8*11 = 8(-18) = -2^4 3^2 1 = 9*13 = 9(-16) = -2^4 3^2 -1 = 12*12 ...-Bill Dubuque without the aid of a calculator/computer, or Fermats little theorem. In>the book, all he discusses is modular arithmetic. FLT is not mentioned til>page 97. I am on page 10.>>Even with FLT, I cant seem to get it. If a^p = a mod p, a is the remainder>right? So, if, for example, I take 3^2 mod 2 the theorem says I should get>3, but isnt it 1? Yes, and yes. 3=1 (mod 2), after all, so its hardly surprising thatboth answers are correct, modulo 2.If you want to skip Fermats Little Theorem, then study how the powersof your number cycle moudlo 29. For example, if I wanted to gure outwhat the last digit of 3^{200}, what I would do is consider the powersof 3 modulo 10:3 = 3 (mod 10)3^2 = 9 (mod 10)3^3 = 7 (mod 10)3^4 = 1 (mod 10)3^5 = 3 (mod 10)At this point, it should become obvious that 3^a = 3^{a+4} (mod 10)for all a, so I just need to take the residue of 200 modulo 4; this is0, so 3^{200} = 3^0 = 1 (mod 10).>I mean 9 mod 2 should be 1, right? Futhermore, if I>take 37^29 = 37 mod 29 and I try to raise this congruence by powers until I>get my 100, then my remainder gets larger and larger.First, you would not do that; what you would do is note that 37^{28} =1 (mod 29), so that 37^{k*28} = 1 (mod 29) for all integers k. Thatmeans that37^{100} = 37^{84+16} = 37^{84}*37^{16} = 37^{3*28}*37^{16} =1*37^{16}=37^{16} (mod 29), so you just need to gure out how much37^{16} is.And second, every time you get a number larger than 29, YOU REDUCEMODULO 29, to get a smaller number and work with that; usually, yourbest bet would be a number between -13 and 15, to keep them small. what I accept as reality. --- Calvin (Calvin and Hobbes) just thought about something and hoped someone could help me explainwhy Im wrong in thinking it or why I could be right...I thought really simple, Take a sphere and begin walking on it (itnever ends). When I want to cut the knot I just do something radicaland go right through the sphere or something. So are we thinking thatsmall? This is somewhat metaphor I had on innite numbers...Anyone who can help me nd that red line again? >I just thought about something and hoped someone could help me explain>why Im wrong in thinking it or why I could be right...>I thought really simple, Take a sphere and begin walking on it (it>never ends). When I want to cut the knot I just do something radical>and go right through the sphere or something. So are we thinking that>small? This is somewhat metaphor I had on innite numbers...>>Anyone who can help mefind that red line again?So in 1 dimension you have innite numbers because of 1 dimensionalthinking.. In another you have hypernumbers that could resemble theinnite So in 1 dimension you have innite numbers because of 1 dimensional> thinking.. In another you have hypernumbers that could resemble the> inniteWell, that wouldnt really work, atleast if youre talking about thereal number system. Cantor proved that |R^n (thats R to the n, theset of real, ordered n-tuples, so the dimension is n) has the samecardinality as |R. That is, there is a function which maps everypoint in |R^n to a point in |R uniquely. So |R and |R^n have the samenumber of points.You may want to read about the Riemann sphere, however. Imagineplacing a sphere on the origin of the complex plane (or just |R^2, thestandard real plane). Imagine a line connecting the top of the sphereto a point on the plane. For any point on the plane, this line willintersect the sphere at a unique point. Now, what happens as thedistance from the origin of the plane gets very far? The line tendsto the tangent line at the top of the sphere! So it only intersectsthe sphere at one point. In complex analysis, we call this the pointat innity, and we call the plane with this point added theaugmented plane.Check outhttp://tinyurl.com/haxyThe rst picture is of the Riemann Sphere, the rest are of functionsmapped onto the Riemann sphere.Alex > So in 1 dimension you have innite numbers because of 1 dimensional>> thinking.. In another you have hypernumbers that could resemble the>> innite>>Well, that wouldnt really work, atleast if youre talking about the>real number system. Cantor proved that |R^n (thats R to the n, the>set of real, ordered n-tuples, so the dimension is n) has the same>cardinality as |R. That is, there is a function which maps every>point in |R^n to a point in |R uniquely. So |R and |R^n have the same>number of points.>>You may want to read about the Riemann sphere, however. Imagine>placing a sphere on the origin of the complex plane (or just |R^2, the>standard real plane). Imagine a line connecting the top of the sphere>to a point on the plane. For any point on the plane, this line will>intersect the sphere at a unique point. Now, what happens as the>distance from the origin of the plane gets very far? The line tends>to the tangent line at the top of the sphere! So it only intersects>the sphere at one point. In complex analysis, we call this the point>at innity, and we call the plane with this point added the>augmented plane.>>Check out>http://tinyurl.com/haxy>>The rst picture is of the Riemann Sphere, the rest are of functions>mapped onto the Riemann sphere.>>AlexI hope Ill get new inspiration from it :D Alex>Check out>http://tinyurl.com/haxy>>The rst picture is of the Riemann Sphere, the rest are of functions>mapped onto the Riemann sphere.>Wow. I never knew you could put a graphic in one table entry and a paragraph inthe other. I have a box function on my old Panasonic RK-P400C plotter that doesthat. Very cool. Makes for easy reading. Maybe a little too much whitespace.Yours,Doug Goncz, Replikon Research, Seven Corners, VA Fair use and Usenet distribution without restriction or feeCivil and criminal penalties for circumvention of any embedded encryption This thread is not an effective presentation any longer. I guess> I just wanted somebody to shoot this down and it hasnt happened yet.> some more impressive results.> -TimI had a few thoughts on the subject. Have you heard of the QUATERNIONSof William Rowan Hamilton?If your +, -, and * are none-other than the i, j, and k of Hamilton,then you have REDISCOVERED the Quaternions.Thats OK. I have often discovered something only tofind thatsomebody got there rst. nd that the great men of science thought alongthe same lines as I do - so on those occasions when I am second, Ifeel FLATTERED.I have ALSO found things that nobody ever found before me. That isbecause ones condence grows with every discovery - new or not.Do investigate the Quaternions: http://mathworld.wolfram.com/Quaternion.html Charles Douglas Wehner =I think this construction is more primitive than the quaternions.Im merely adding another sign to the real numbers and seeing whathappens.Quaternions appear to contain lots of real numbers and so cannot beequivalent.Have you ever tried graphing in a plane like this? + plus pole + + + + + + . . . . . . . p1 = - 7 + 4 + . + . . + . . 0 - - - - - - - - - - minus pole . * origin . p3 . * . . * . * . * . * . . . . . . . . . p2 = - 9 * 6 * * * * star poleThis is Y space as a plane. It is simpler than both cartesian 2Dand complex values in that it has just the three way branch instead ofa four way branch. As you can see parallelograms at angle 2pi/3resolve the entire plane symmetrically. A reduced Y value always hasat most a pair of magnitudes. Every position in the plane can beresolved.The dimensionality of Y is quite a conundrum. Becauseinformationally there are two magnitudes informationally it is twodimensional, yet the value is just one three-signed element in Y.The question why a pair? should be in the mind at this point.The answer is that in order to should require an equal magnitude for each poleto provide : - x + x * x = 0. In Y : y1 = - x + x is not zero! We need y1 * x to get to zero.If you believe that summation is superposition then I think you willbe convinced. The trouble is that we all think in context of realnumbers.Much of this thinking does not extend when more signs are used in theconstruction.I started this all in the context of the question Why R X R X R Xt?.This is the spacetime of classical physics which even string theoristsare not destroying. I consider Y X Y to be a competitor to R X R X R Xt.But also if you add yet another sign ( four signs ) then the extensionwould yield at most three magnitudes by the same law effect is looked upon asaccumulation, which then provides a basis for time. I think it is wiseto continue opening up the three-signed can of worms before moving onto the four-signed can. It is my hope that dynamics will be found in Ywhich will surprise us.> I had a few thoughts on the subject. Have you heard of the QUATERNIONS> of William Rowan Hamilton?> If your +, -, and * are none-other than the i, j, and k of Hamilton,> then you have REDISCOVERED the Quaternions.> Thats OK. I have often discovered something only tofind that> somebody got there rst. The effect on my mind is not science thought along> the same lines as I do - so on those occasions when I am second, I> feel FLATTERED.> I have ALSO found things that nobody ever found before me. That is> because ones condence grows with every discovery - new or not.> Do investigate the Quaternions:> http://mathworld.wolfram.com/Quaternion.html >> Charles Douglas Wehner == MISSING.erased it by mistake.just waiting until I had time for a fuller answer.I have to answer it here.The rst question of data going missing and being recovered camefrom this statement of mine:> The TRANSFINITE mathematics states that it becomes vanishingly small,> but if it could be scaled up by a special innity it would return.> Here, we can draw a NUMBER LINE -3,-2,-1,0,1,2,3and put the factorials above the positive part:1,1,2,60,1,2,3Tofind the factorial of 2 from that of 3 - which is 6 -we DIVIDE by 3. It gives 2.But we can always go back up again by multiplying by 3.Tofind the factorial of 1 from 2!, we divide by 2.We can also go back up.Tofind 0! from 1! we divide by 1, and can go back.Tofind (-1)! - note the brackets - we divide by 0.ANYTHING divided by zero is said to be INFINITE.Zero times innite is said to be INDETERMINATE.So we cannot go back up!!!!!However, at the console we dened zero by typing it FROM.We know therefore that (-1)! is 1 divided by CONSOLE-ZERO.We call this CONSOLE-INFINITY.As we still have console-zero on our number-line, we can still multiply console-innity by it.So we can go back up.(-2)! is -(console innity)(-3)! is +(1/2)(console-innity)&c.And so, by strict denition of the nature of the zeroes andof the innities, we do not have to lose information bydening it as indeterminate.This is one of my discoveries in the as-yet unpublishedfantasy maths - where fantasy means anything containingeasy-to-use numbers (FUN NUMBERS) and the FOLLIES zero andinnity.Another name is the TRANSFINITE MATHS.However, as I made my discoveries by one route it was pointedout to me by an eminent mathematician that Georg Cantor hadfound something very similar a hundred years ago by another route. The FOLLIES are almost the same as the ALEPHS of Cantor.I am not ashamed. We sought in Nature, and we BOTH found.I also stated:> The COMPLEX mathematics states that it shifts SIDEWAYS in complex> space, and so leaves the real world. Multiplication by -i1 would bring> it back.> Here we could consider a cosine. It starts at 1 and descends.However, I want you to imagine some micro-polynomial that, whengiven the number 1 computes the cosine a trace further - that is,0.999999999975625 or whatever.Think of a pico-polynomial that is even smaller in its steps.0.999999999999999999999999999999999Think of it being applied as the cosine passed through zero atthe point Pi/2But 2Cos(X) goes through zero just as 1Cos(X) and 3Cos(X) do.And we are not allowed special zeroes.As the waveform creeps through zero, how does it know how tore-emerge in such a way as to create a symmetrical cosine,instead of Cos(X) above the zero and 2Cos(X) below?In electronics engineering, we like to think of such a cosinebeing part of a PAIR. The cosine is REAL, and is the SIGNAL -but there is a hidden sine accompanying it. This we call thePHASE.The PHASE is hidden because it exists only in the IMAGINARYworld. However, it denes the SLOPE of the original cosine.Thus, if the cosine began at Y=1, and the units of the X-axisare RADIANS, the slope will be a downward slope of 1 as thecosine goes through zero. By means of its phase, a cosine can be allowed to pass throughzero without getting lost.So - with RESTRICTIONS - we can use complex maths.We use the Euler equation Exp(iX)=Cos(X)+iSin(X)However, i means CURRENT in electronics - so we use j.X is too vague, so we use omega-tOmega is the angular frequency. t is the time.So we write Exp(j.omega.t)=Cos(omega.t)+jSin(omega.t)Here, the jSin(omega.t) is actually the INVERTED slope.Yet it still denes the slope.Such complications are necessary because the simple mathematics will not always dene what we want.In the transnite case, we have no need for theimaginary numbers. In the complex maths we have noneed for the transnite.At least in theory.Perhaps there are problems that can only be solvedby BOTH.Then we have the mad maths of Boole. 1+1=1At rst sight, it is ridiculous.Then somebody explains that 1 is TRUE.He translates TRUE AND TRUE IS TRUE.Now he seems REALLY to have gone bananas.Yet without Boole, there would be no computer andno Internet for me to write this on.The Quaternions are sometimes used in elementary-I cannot ACCEPT a new mathematical system withoutseeing its benets.I cannot REJECT a new mathematical system forfear of rejecting something good.If I had time to delve and delve, I might knowwhether the THREE-SIGNED ARITHMETIC is usefulor not.But it is not mine, and I am busy. I will, however,if there is an adequate summary, look back now andthen to see if something important is emerging.I reserve judgement. == Im still largely confused by T and its operations. Could you give us a fewexhaustive examples from scratch? My guess is that it is probably isomorphicto something were already familiar with, in which case there will be lotsof useful things you could draw on that already are known to understand howT works. Im still largely confused by T and its operations. Could you give us a few> exhaustive examples from scratch? My guess is that it is probably isomorphic> to something were already familiar with, in which case there will be lots> of useful things you could draw on that already are known to understand how> T works.Whereas the real numbers (R)can be dened as a magnitude with twosigns( -, + ), T is a magnitude with three signs( -, +, * ).Examples of elements in T are: - 1.23 + 2.134 * 6.54Now consider summation in T.The sum of the example numbers reads: ( - 1.23 + 2.134 * 6.54 ).It is a mistake to say that - 1.23 * 6.54 = * 5.31.This style of too strict.If we use the law: Sum( t1, t2, t3 ) = Sum( Sum( t1, t2 ), t3 ) = Sum( Sum( t3, t1 ),t2 )we willfind the problem.Consider the following using the fraudulent style of 2.134 ( + 0.904 * 6.54 ) = ( * 5.31 + 2.134 ) ( * 5.636 ) = ( * 3.176 ). This is nonsense.The problem occurred at the ?Or should - 1 + 1 * 1 = 0?The latter is the correct selelection.For a magnitude x: In R - x + x = 0. In T - x + x * x = 0.Some values are not reducible. Thich leads me to describing a newspace Y because Sum( t1, t2 ) is not generally in T. So even thoughthe subject in this thread is T space it is only a starting step toget to Y space, which is general three-signed arithmetic.Y is a general sum of T which is always reducible to at most a pair ofthree-signed magnitudes. Now Sum( y1, y2 ) is always in Y.This leads to ( - 1.23 + 2.134 * 6.54 ) = ( + 0.904 * 5.31 ). where ( - 1.23 have time for right now. Arithmetical productsfollow much more easily. This is nonsense.Yes, now youre talking sense.Max de Macs == Product of Sums Equals Sum of Products---------------------------------------Here is the one law that may really break Y space: Product( y1, Sum( y2, y3 )) = Sum( Product( y1, y2 ), Product( y1,y3 ).Allow * to represent the sum of two elements in Y since it preservessign and follows symmetrically to the addition operator in reals.Allow arithmetical product to be as previously dened. These conceptsare quite symmetrical to the real number concept of opertations.The above law can be restated as: y1(y2 * y3) = y1y2 * y1y3.If we invoke this law then the rules for arithmetical product and sumthat I have laid out might break down.Let y1 = * 1.0 - 1.0, y2 = * 1.0 - 1.0, y3 = - 2.0 + 2.0 .Invoking: (y1)(y2*y3) = (y1)(y2)*(y1)(y3)(*1.0-1.0)(*1.0-1.0-2.0+2.0) =(*1.0-1.0)(*1.0-1.0)*(*1.0-1.0)(-2.0+2.0)(*1.0-1.0)(-2.0+1.0 ) = (*1.0-1.0-1.0+1.0)*(-2.0+2.0+2.0*2.0)-2.0+1.0+2.0*1.0 = - 1.0 + 2.0- 1.0 + 2.0 = - 1.0 + 2.0 .Oops! That one worked.Let y1 = * 1.0 - 2.0, y2 = * 1.0 - 1.0, y3 = - 2.0 + 2.0 .Invoking: (y1)(y2*y3) = (y1)(y2)*(y1)(y3)(*1.0-2.0)(*1.0-1.0-2.0+2.0) =(*1.0-2.0)(*1.0-1.0)*(*1.0-2.0)(-2.0+2.0)(*1.0-2.0)(-2.0+1.0 ) = (*1.0-1.0-2.0+2.0)*(-2.0+2.0+4.0*4.0)-2.0+1.0+4.0*2.0 = -2.0+1.0+4.0*2.0+ 3.0 = + 3.0 .Oops! That one worked too.Now prove that (y1)(y2*y3) = (y1)(y2)*(y1)(y3). Dene operators m(y), p(y), and s(y) which return the magnitude of yfor the signs minus, plus, and star respectively (e.g. m(-2.0*6.5) =2.0.y2 * y3 = - Sum(m(y2), m(y3)) + Sum(p(y2), p(y3)) * Sum(s(y2), s(y3)).(y1)(y2*y3) =( + m(y1)(Sum(m(y2),m(y3)) + p(y1)(Sum(s(y2),s(y3)) + s(y1)(Sum(p(y2),p(y3)) - m(y1)(Sum(s(y2),s(y3)) - p(y1)(Sum(p(y2),p(y3)) - s(y1)(Sum(m(y2),m(y3)) * m(y1)(Sum(p(y2),p(y3)) * p(y1)(Sum(m(y2),m(y3)) * s(y1)(Sum(s(y2),s(y3)))= [ distributing the product through the sum and gathering liketerms...]( + m(y1)m(y2) + p(y1)s(y2) + s(y1)p(y2) - m(y1)s(y2) - p(y1)p(y2) - s(y1)m(y2) * m(y1)p(y2) * s(y1)s(y2) * p(y1)m(y2) + m(y1)m(y3) + p(y1)s(y3) + s(y1)p(y3) - m(y1)s(y3) - p(y1)p(y3) - s(y1)m(y3) * m(y1)p(y3) * p(y1)m(y3) * s(y1)s(y3)).(y1)(y2) = [This matches the upper half of the equation above]( + m(y1)m(y2) + p(y1)s(y2) + s(y1)p(y2) - m(y1)s(y2) - p(y1)p(y2) - s(y1)m(y2) * m(y1)p(y2) * s(y1)s(y2)).(y1)(y3) = [This matches the lower half]( + m(y1)m(y3) + p(y1)s(y3) + s(y1)p(y3) - m(y1)s(y3) - p(y1)p(y3) - s(y1)m(y3) * m(y1)p(y3) * p(y1)m(y3) * s(y1)s(y3)).This is adequate proof that the law y1y2 * y1y3 = y1(y2*y3) is holding. Im still largely confused by T and its operations. Could you give us a few> exhaustive examples from scratch? My guess is that it is probably isomorphic> to something were already familiar with, in which case there will be lots> of useful things you could draw on that already are known to understand how> T works.Ive got a proof that aligns well with standard arithmetic which is apretty good example of some three-signed arithmetic. It seems to beholding up.Product of Sums Equals Sum of Products---------------------------------------Here is the one law that may really break Y space: Product( y1, Sum( y2, y3 )) = Sum( Product( y1, y2 ), Product( y1,y3 ).Allow * to represent the sum of two elements in Y since it preservessign and follows symmetrically to the addition operator in reals.Allow arithmetical product to be as previously dened. These conceptsare quite symmetrical to the real number concept of opertations.The above law can be restated as: y1(y2 * y3) = y1y2 * y1y3.If we invoke this law then the rules for arithmetical product and sumthat I have laid out might break down.Let y1 = * 1.0 - 1.0, y2 = * 1.0 - 1.0, y3 = - 2.0 + 2.0 .Invoking: (y1)(y2*y3) = (y1)(y2)*(y1)(y3)(*1.0-1.0)(*1.0-1.0-2.0+2.0) =(*1.0-1.0)(*1.0-1.0)*(*1.0-1.0)(-2.0+2.0)(*1.0-1.0)(-2.0+1.0 ) = (*1.0-1.0-1.0+1.0)*(-2.0+2.0+2.0*2.0)-2.0+1.0+2.0*1.0 = - 1.0 + 2.0- 1.0 + 2.0 = - 1.0 + 2.0 .Oops! That one worked.Let y1 = * 1.0 - 2.0, y2 = * 1.0 - 1.0, y3 = - 2.0 + 2.0 .Invoking: (y1)(y2*y3) = (y1)(y2)*(y1)(y3)(*1.0-2.0)(*1.0-1.0-2.0+2.0) =(*1.0-2.0)(*1.0-1.0)*(*1.0-2.0)(-2.0+2.0)(*1.0-2.0)(-2.0+1.0 ) = (*1.0-1.0-2.0+2.0)*(-2.0+2.0+4.0*4.0)-2.0+1.0+4.0*2.0 = -2.0+1.0+4.0*2.0+ 3.0 = + 3.0 .Oops! That one worked too.Now prove that (y1)(y2*y3) = (y1)(y2)*(y1)(y3). Dene operators m(y), p(y), and s(y) which return the magnitude of yfor the signs minus, plus, and star respectively (e.g. m(-2.0*6.5) =2.0.y2 * y3 = - Sum(m(y2), m(y3)) + Sum(p(y2), p(y3)) * Sum(s(y2), s(y3)).(y1)(y2*y3) =( + m(y1)(Sum(m(y2),m(y3)) + p(y1)(Sum(s(y2),s(y3)) + s(y1)(Sum(p(y2),p(y3)) - m(y1)(Sum(s(y2),s(y3)) - p(y1)(Sum(p(y2),p(y3)) - s(y1)(Sum(m(y2),m(y3)) * m(y1)(Sum(p(y2),p(y3)) * p(y1)(Sum(m(y2),m(y3)) * s(y1)(Sum(s(y2),s(y3)))= [ distributing the product through the sum and gathering liketerms...]( + m(y1)m(y2) + p(y1)s(y2) + s(y1)p(y2) - m(y1)s(y2) - p(y1)p(y2) - s(y1)m(y2) * m(y1)p(y2) * s(y1)s(y2) * p(y1)m(y2) + m(y1)m(y3) + p(y1)s(y3) + s(y1)p(y3) - m(y1)s(y3) - p(y1)p(y3) - s(y1)m(y3) * m(y1)p(y3) * p(y1)m(y3) * s(y1)s(y3)).(y1)(y2) = [This matches the upper half of the equation above]( + m(y1)m(y2) + p(y1)s(y2) + s(y1)p(y2) - m(y1)s(y2) - p(y1)p(y2) - s(y1)m(y2) * m(y1)p(y2) * s(y1)s(y2)).(y1)(y3) = [This matches the lower half]( + m(y1)m(y3) + p(y1)s(y3) + s(y1)p(y3) - m(y1)s(y3) - p(y1)p(y3) - s(y1)m(y3) * m(y1)p(y3) * p(y1)m(y3) * s(y1)s(y3)).This is adequate proof that the law y1y2 * y1y3 = y1(y2*y3) is holding. =Im looking at timesheets scheduling and optimisation task: a shopmanager would like to specify a number of employees working in anygiven time interval, and employees would like to stay within theirpreferable time constraints.I think such optimisation task should be quite common, but I justdont know the category of algorithms it falls to, or at least where Ican pick up bits of information in advance,Vlad Im looking at timesheets scheduling and optimisation task: a shop>manager would like to specify a number of employees working in any>given time interval, and employees would like to stay within their>preferable time constraints.>I think such optimisation task should be quite common, but I just>dont know the category of algorithms it falls to, or at least where I>can pick up bits of information on this subject.Yes, this is quite common. The usual formulations involve integerlinear programming. The general eld in which this is done is operations research, and you might try the sci.op-research newsgroup.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 == Im looking at timesheets scheduling and optimisation task: a shop> manager would like to specify a number of employees working in any> given time interval, and employees would like to stay within their> preferable time constraints.> I think such optimisation task should be quite common, but I just> dont know the category of algorithms it falls to, or at least where I> can pick up bits of information on this subject.Dr. Israel mentioned that the general eld is Operations Research.It may also be of interest that some complex examples of timetablingproblems are solved with Evolutionary or Genetic Algorithms, and therehas been discussion within the past year of this in comp.ai.genetic.xanthian.-- =There are several plug-ins for Excel which will deal with this as long asyour problem is not too large.>Im looking at timesheets scheduling and optimisation task: a shop>manager would like to specify a number of employees working in any>given time interval, and employees would like to stay within their>preferable time constraints.>>I think such optimisation task should be quite common, but I just>dont know the category of algorithms it falls to, or at least where I>can pick up bits of information on this subject.>> Yes, this is quite common. The usual formulations involve integer> linear programming. The general eld in which this is done is> operations research, and you might try the sci.op-research newsgroup.>> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2 =1. What is considered to be the toughest eld of mathematics at thegraduate level?2. Which schools are considered to be the best schools for graduatelevel mathematics? == I guess that is going to be how one denes toughest.As for top schools, it will probably come as no surprise:These schools are always in the top 20:Harvard, Princeton, M.I.T., Duke, Cal @ Berkeley, U of Michigan, U of Wisc@Madison, U of Minnesota@ Twin Cities, etc...> 1. What is considered to be the toughest eld of mathematics at the> graduate level?> 2. Which schools are considered to be the best schools for graduate> level mathematics? =Dont know about toughest. If you ask which one is hottest, it will have to be algebraic geometry, which is continuously expanding and absorbing other elds of mathematics.> 1. What is considered to be the toughest eld of mathematics at the> graduate level?> 2. Which schools are considered to be the best schools for graduate> level mathematics? 1. What is considered to be the toughest eld of mathematics at the> graduate level?assumption of easiest to toughestall others >> stats >> abstractHerc =okay, how about we dene toughest as in the toughest to get an A in.I mean a signicant number of people will make As in calculus anddifferential eqns etc. but are there certain courses out there thatare extremely difcult to make As in? 1. What is considered to be the toughest eld of mathematics at the> graduate level?There is no such thing. Its a personal thing.> 2. Which schools are considered to be the best schools for graduate> level mathematics?Again, for graduate level there is no such thing. What counts is theadvisor and the critical mass (the number) of the faculty working inthe eld you are interested in. Jan Bielawski Dont know about toughest. If you ask which one is hottest, it will have > to be algebraic geometry, which is continuously expanding and absorbing > other elds =I think getting an A would primarily depend on your professor/instructor.Acheiving an A in a maths course doesnt always mean you have mastered thesubject, nor does getting a B or C mean you havent mastered the subject.Like it or not, no matter what course you take at university, subjectivityreigns supreme in the grading arena. Albeit, some subjects are worse thanothers. I think the term toughest may prove difcult to dene in thiscontext. Any subject within the realm of Advanced maths is difcult andrequires a lot of work to understand it thoroughly.I think toughest may turn out to be a matter of personal opinion.Everyone thinks and learns differently; so, some people may catch on to onesubject faster than another subject.LurchLurch> 1. What is considered to be the toughest eld of mathematics at the> graduate level?> 2. Which schools are considered to be the best schools for graduate> level mathematics? okay, how about we dene toughest as in the toughest to get an A in.>I mean a signicant number of people will make As in calculus and>differential eqns etc. but are there certain courses out there that>are extremely difcult to make As in?That will certainly differ enormously from school to school, andfrom instructor to instructor within a school. What a particularpersonfinds tough depends mainly on that persons level of preparationand abilities.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 okay, how about we dene toughest as in the toughest to get an A in.> I mean a signicant number of people will make As in calculus and> differential eqns etc. but are there certain courses out there that> are extremely difcult to make As in?Math majors often have the hardest time in the rst course where theyreally learn to write proofs. Here at Ohio State that could be eitherIntroductory Analysis (aka Advanced Calculus) or Abstract Algebra,depending on which they take rst. Postponing them as long aspossible, which some students are wont to do, is NOT a good idea.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ == Got this puzzle the other day, but cant gure it out. Impossible?First, let a friend constructs a right-angled triangle with *integer*sides a and b (the shorter sides, that is)Then he throws away the integers, but hands you the area, A.Question: Can youfind two integers, a and b, to recreate aright-angled triangle with area A ??? Got this puzzle the other day, but cant gure it out. Impossible?> First, let a friend constructs a right-angled triangle with *integer*> sides a and b (the shorter sides, that is)> Then he throws away the integers, but hands you the area, A.> Question: Can youfind two integers, a and b, to recreate a> right-angled triangle with area A ???Sure. If all else fails, an approach of trial and error will work.You know that A = 1/2 ab, so ab=2A. Start counting up until youfind integers a and b that have product 2A. You may notfind what your friend started with, but you willfind something.For Example: Given area=6if a=1, b=12if a=2, b=6if a=3, b=4This gives you three possible choices for a and b.Your solution wont be unique unless A=1/2 or A=p/2 where p is prime.-- Will Twentyman Got this puzzle the other day, but cant gure it out. Impossible?> First, let a friend constructs a right-angled triangle with *integer*> sides a and b (the shorter sides, that is)> Then he throws away the integers, but hands you the area, A.> Question: Can youfind two integers, a and b, to recreate a> right-angled triangle with area A ???How long are you willing to wait?-- Dave SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal ruling.> First, let a friend constructs a right-angled triangle with *integer*> sides a and b (the shorter sides, that is)>> Then he throws away the integers, but hands you the area, A.>> Question: Can youfind two integers, a and b, to recreate a> right-angled triangle with area A ???Are you sure the problem was stated this way?You can always take a=1 and b=2AArea = a*b/2 = 1*2A/2 = ADirk Vdm =Pythagorean Triangles (right triangles whose sides are all integers) haveinteger area.We can sometimes patch two PTs together, either additively or subtractively,and get a non-right triangle whose sides and area are all integers. Forexample, the PT (5,12,13) and the PT (9,12,15) can be aligned along their 12side to make either the triangle (13,14,15) or the triangle (4,13,15).Prove that, given a triangle D whose sides and area are integers, either Dis a PT or there is some positive integer multiple of D which can be patchedfrom two PTs, as described above. =Will Self escribi.97 en Triangles (right triangles whose sides are all integers)> have integer area.>> We can sometimes patch two PTs together, either additively or> subtractively, and get a non-right triangle whose sides and area are> all integers. For example, the PT (5,12,13) and the PT (9,12,15) can> be aligned along their 12 side to make either the triangle (13,14,15)> or the triangle (4,13,15).>> Prove that, given a triangle D whose sides and area are integers,> either D is a PT or there is some positive integer multiple of D> which can be patched from two PTs, as described above.If the triangle isnt rectangle and its sides are integers, by cosine lawthe three cosines are rationals. If also the area is integer, the threeheights are rationals and then also the three sines. Therefore, the tangentsof the three angles are rationals.Then any height, interior to triangle, divides it in two right triangles,whose acute angles have rational tangents, being its hypotenuses integers.Multiplying the triangle sides by the gcd od the denominators of thetangents of its two sub-triangle acute angles, we get two PythagoreanTriangles. If the height is exterior, the triangle id the difference of tworight triangles with rational sides and, multiplying by a suitable factor asbefore, integers.-- Ignacio Larrosa == Triangles (right triangles whose sides are all integers)> have integer area.>> We can sometimes patch two PTs together, either additively or> subtractively, and get a non-right triangle whose sides and area are> all integers. For example, the PT (5,12,13) and the PT (9,12,15) can> be aligned along their 12 side to make either the triangle (13,14,15)> or the triangle (4,13,15).>> Prove that, given a triangle D whose sides and area are integers,> either D is a PT or there is some positive integer multiple of D> which can be patched from two PTs, as described above.>> If the triangle isnt rectangle and its sides are integers, by cosine law> the three cosines are rationals. If also the area is integer, the three> heights are rationals and then also the three sines. Therefore, thetangents> of the three angles are rationals.>> Then any height, interior to triangle, divides it in two right triangles,> whose acute angles have rational tangents, being its hypotenuses integers.> Multiplying the triangle sides by the gcd od the denominators of the> tangents of its two sub-triangle acute angles, we get two Pythagorean> Triangles. If the height is exterior, the triangle id the difference oftwo> right triangles with rational sides and, multiplying by a suitable factoras> before, integers.> -- >> Ignacio Larrosa Ca.96estro> A Coru.96a (Espa.96a)> ilarrosaQUITARMAYUSCULAS@mundo-r.com>Golly, that was fast. Well done, Ignacio.BTW for readers, when Ignacio says If the triangle isnt rectanglehe means If the triangle isnt a right triangle. Just to dispel anyconfusion. (Spanish would be a much better International Language :-) =both of which are integers. Of course given time one canfind the twointegers that A came from initially--something the original writer did notrequest.>> Got this puzzle the other day, but cant gure it out. Impossible?>> First, let a friend constructs a right-angled triangle with *integer*> sides a and b (the shorter sides, that is)>> Then he throws away the integers, but hands you the area, A.>> Question: Can youfind two integers, a and b, to recreate a> right-angled triangle with area A ???>> How long are you willing to wait?>> == A hemispherical bubble is placed on a spherical bubble of radius 1. Asmaller hemispherical bubble is then placed on the rst one. This processis continued until n chambers, including the sphere, are formed. Usemathematical induction to prove that the maximum height of any bubble towerwith n chambers is 1 + sqrt(n).The solution of this problem requires a little mathematical discovery. A hemispherical bubble is placed on a spherical bubble of radius 1. A> smaller hemispherical bubble is then placed on the rst one. This process> is continued until n chambers, including the sphere, are formed. Use> mathematical induction to prove that the maximum height of any bubbletower> with n chambers is 1 + sqrt(n).>> The solution of this problem requires a little mathematical discovery.This is a good problem, and the formulation in terms of bubbles isnicely phrased, an elegant touch. Also, the little mathematical discoverymay come as a pleasant surprise. A hemispherical bubble is placed on a spherical bubble of radius 1. A> smaller hemispherical bubble is then placed on the rst one. This process> is continued until n chambers, including the sphere, are formed. Use> mathematical induction to prove that the maximum height of any bubble tower> with n chambers is 1 + sqrt(n).>> The solution of this problem requires a little mathematical discovery.>Discovery: Reduce the problem to a unit circle with ever smallersemi-circles being place upon the next larger. =computer program that allows to plot the bubble tower and see for yourselfthe effects of the decrease in angle theta. Also I can distribute theformula for the height (not the maximum height) of the tower in *.bmp le computer program that allows to plot the bubble tower and see for yourself> the effects of the decrease in angle theta. Also I can distribute the> formula for the height (not the maximum height) of the tower in *.bmp le>Too, bumpy when the only thing that needs be done is to nd f for which:Given a circle of radius r and a semicircle of diameter d exterior to thecircle and with endpoints on the circle, the distance from the center ofthe circle to the middle or top of the semicircle arc = f(r,d).Once that formula is establish, itd elucidate if the bubbletower diverges or not and how accurate 1 + sqr n is. =You think I didnt do that? You want to know the formula for Height oftower of n chambers?H(n)=1 + sin(theta) * (1-cos(theta)^(n-1)) / (1-cos(theta)) + cos(theta)^(n-1)How is that? Converges? Derivative? Try it.>> Everyone who is trying to solve this problem please be aware that I> computer program that allows to plot the bubble tower and see foryourself> the effects of the decrease in angle theta. Also I can distribute the> formula for the height (not the maximum height) of the tower in *.bmple>> Too, bumpy when the only thing that needs be done is tofind f for which:>> Given a circle of radius r and a semicircle of diameter d exterior to the> circle and with endpoints on the circle, the distance from the center of> the circle to the middle or top of the semicircle arc = f(r,d).>> Once that formula is establish, itd elucidate if the bubble> tower diverges or not and how accurate 1 + sqr n is. == You think I didnt do that? You want to know the formula for Height of> tower of n chambers?>> H(n)=1 + sin(theta) * (1-cos(theta)^(n-1)) / (1-cos(theta)) + cos(theta)> ^(n-1)>Huh? Didnt you say height is 1 + sqr n?> How is that? Converges? Derivative? Try it.>Whats theta? lim H(n) = 1 when cos t > 0 = oo when cos t < 0 = 1 + sin t when cos t = 0>> Given a circle of radius r and a semicircle of diameter d exterior to the> circle and with endpoints on the circle, the distance from the center of> the circle to the middle or top of the semicircle arc = f(r,d).>> Once that formula is establish, itd elucidate if the bubble> tower diverges or not and how accurate 1 + sqr n is. Given a circle of radius r and a semicircle of diameter d exterior to the> circle and with endpoints on the circle, the distance from the center of> the circle to the middle or top of the semicircle arc = f(r,d).>> Once that formula is establish, itd elucidate if the bubble> tower diverges or not and how accurate 1 + sqr n is.The height of the center of the base circle of raduis 1 is is 1.Let h_i be the rise from of the center of ith semicircle/circle to the center of the next higher semicircle, Let r_i be the radius of the ith semicircle/circle, with r_n = h_n being the radius of the top semicircle.Clearly, 1 = _r1 > r_2> ...r_n > 0 for a maximum overall height.Then (r_i)^2 = (h_i)^2 + (r_{i+1})^2, for 1 <= i <= n-1,with r_1 = 1.Then 1 = r_1^2 = (h_1)^2 + ... + (h_n)^2Total height = T = 1 + h_1 + ... + h_n subject to (h_1)^2 + ... + (h_n)^2-1 = 0.Let F = 1 + h_1 + ... + h_n + lambda((h_1)^2 + ... + (h_n)^2 - 1)Take partials wrt each h_i and wrt lambda, and set each equal to zero.It is quickly obvious that all the h_1 must be equal to some h. =send you a picture or a program that would describe you the problem ingreater detail.You willfind it a lot more interesting since so far I have not seen aperson who would solve this problem by the means of single variablecalculus. Nonetheless, this problem is given to students in their rstcalculus course (see the remarks about the origins of the problem inprevious letters), when they are just about to nish with simpledifferentiation. How is that?This problem is a mathematical discovery for a person who solves it, it isintense experience, not comparable with something else. Like George Polyasaid once, ..if you solve it [problem] by you own means you may enjoy thesense of discovery.I really wish you continue to work on this problem, give me you validapologize for may be extremely ofcial or serious tone, but your lettersreect your much deeper knowledge of the subject than mine. Collaborationis, nonetheless, the method of any scientic community.>> You think I didnt do that? You want to know the formula for Height of> tower of n chambers?>> H(n)=1 + sin(theta) * (1-cos(theta)^(n-1)) / (1-cos(theta)) + cos(theta)> ^(n-1)>> Huh? Didnt you say height is 1 + sqr n?>> How is that? Converges? Derivative? Try it.>> Whats theta? lim H(n) = 1 when cos t > 0> = oo when cos t < 0> = 1 + sin t when cos t = 0> Given a circle of radius r and a semicircle of diameter d exterior tothe> circle and with endpoints on the circle, the distance from the centerof> the circle to the middle or top of the semicircle arc = f(r,d).>> Once that formula is establish, itd elucidate if the bubble> tower diverges or not and how accurate 1 + sqr n is.> =The problem we are talking about comes from James Stewarts textbookCalculus 5ed, pg. 313, ex.#22.>> A hemispherical bubble is placed on a spherical bubble of radius 1. A> smaller hemispherical bubble is then placed on the rst one. Thisprocess> is continued until n chambers, including the sphere, are formed. Use> mathematical induction to prove that the maximum height of any bubble> tower> with n chambers is 1 + sqrt(n).>> The solution of this problem requires a little mathematical discovery.>> This is a good problem, and the formulation in terms of bubbles is> nicely phrased, an elegant touch. Also, the little mathematicaldiscovery> may come as a pleasant surprise.>> == You will find it a lot more interesting since so far I have not seen >a person who would solve this problem by the means of single >variable calculus. Nonetheless, this problem is given to students in >their rst calculus course (see the remarks about the origins of the >problem in previous letters), when they are just about to nish with >simple differentiation. How is that?Impressive. Just what were the topics covered by the end of the book. >I really wish you continue to work on this problem, give me you >conclusion. I apologize for may be extremely ofcial or serious >tone, but your letters reect your much deeper knowledge of the >subject than mine. Collaboration is, nonetheless, the method of any >scientic community.Virgil did a nice job of solving the problem. As for collaboration,youve yet to explain your function H(n) and theta. So now you want asolution based upon single variable calculus? How could that be possible? >The problem we are talking about comes from James Stewarts textbook >Calculus 5ed, pg. 313, ex.#22.Dont have a copy.> You think I didnt do that? You want to know the formula for Heightof> tower of n chambers?>> H(n)=1 + sin(theta) * (1-cos(theta)^(n-1)) / (1-cos(theta)) +cos(theta)> ^(n-1)> Huh? Didnt you say height is 1 + sqr n?>> How is that? Converges? Derivative? Try it.>> Whats theta? lim H(n) = 1 when cos t > 0> = oo when cos t < 0> = 1 + sin t when cos t = 0>> Given a circle of radius r and a semicircle of diameter d exterior tothe> circle and with endpoints on the circle, the distance from the center of> the circle to the middle or top of the semicircle arc = f(r,d).>The height of the center of the base circle of raduis 1 is is 1.Let h_i be the rise from of the center of ith semicircle/circle tothe center of the next higher semicircle,Let r_i be the radius of the ith semicircle/circle,with r_n = h_n being the radius of the top semicircle.Clearly, 1 = r_1 > r_2> ...r_n > 0 for a maximum overall height.Then (r_i)^2 = (h_i)^2 + (r_{i+1})^2, for 1 <= i <= n-1,with r_1 = 1.Then 1 = r_1^2 = (h_1)^2 + ... + (h_n)^2Total height = T = 1 + h_1 + ... + h_nsubject to (h_1)^2 + ... + (h_n)^2 - 1 = 0.Let F = 1 + h_1 + ... + h_n + lambda((h_1)^2 + ... + (h_n)^2 - 1)f_j = 1 + 2.lambda.h_j = 0; h_j = -1/2lambdaTake partials wrt each h_i and wrt lambda, and set each equal to zero.It is quickly obvious that all the h_1 must be equal to some h.from 1 = (h_1)^2 + ... + (h_n)^2 = n*h^2,find h = 1/sqrt(n)then max(T) = 1 + n*h = 1 + sqrt(n).It is clear that the only critical point for F must represent amaximim total height.---- Youve often asked questions here about trigonometric diophantine > equations. Have you looked at the literature? Here are some papers > that you will enjoy. They should help you answer your questions. > thank you for all these informations.Nicola =Exhibit A: http://tinyurl.com/fuf8 she looks exactly like Laurie HoldenExhibit B: http://tinyurl.com/fuf2 government has spied on me so them clearlyin between the release dates ofThe Truman Show : 1998 : Jim CarreyMajestic : 2002 : Jim Carrey and Laurie HoldenShall this evidence be allowed to proceed to court?And quote your degree if you say yes.Herc-- _ == Exhibit A: http://tinyurl.com/fuf8 she looks exactly like Laurie Holden> Exhibit B: http://tinyurl.com/fuf2 government these 2 posts place them clearly> in between the release dates of> I am sorry that I cant comment about your question, however I foundthese links that may interest you: http://www.healthyplace.com/Communities/Thought_Disorders/ Site/paranoid_schizophrenia.htmhttp://www.healthyplace.com/ Communities/Thought_Disorders/site/schizophrenia_ overview.htmPlease, will you seek evaluation from a competent mental healthprofessional? I am worried about you.> The Truman Show : 1998 : Jim Carrey> Majestic : 2002 : Jim Carrey and Laurie Holden> Shall this evidence be allowed to proceed to court?> And quote your degree if you say yes.> Herc> --> __> / /> / / > / / / > / / / > / /__/__ > /________ > ___________/> www.A1Sites.com == Oh come on sci.math thats why im posting here to get away fromdocile as cheese comments like this one.if i was a court judge and these ulrs were presented Id say :I need motive, opportunity, and something that puts him at the sceneTHESE urls put me at the scene of being the truman, TRUE OR FALSE?just check the highlighted phrases in the two urls PLEASE sci.math,or the truman is doomed to listen to 1000s of atheists spiels about being on camera syndrome.Herc> Exhibit A: http://tinyurl.com/fuf8 she looks exactly like Laurie Holden Exhibit B: http://tinyurl.com/fuf2 government has spied on posts place them clearly> in between the release dates of I am sorry that I cant comment about your question, however I found> these links that may interest you:>> http://www.healthyplace.com/Communities/Thought_Disorders/ Site/paranoid_schizophrenia.htm>> http://www.healthyplace.com/Communities/Thought_Disorders/ site/schizophrenia_overview.htm>> Please, will you seek evaluation from a competent mental health> professional? I am worried about you.> The Truman Show : 1998 : Jim Carrey Majestic : 2002 : Jim Carrey and Laurie Holden>> Shall this evidence be allowed to proceed to court?> And quote your degree if you say yes.>> Herc> --> __> / /> / / > / / / > / / / > / /__/__ > /________ > ___________/> www.A1Sites.com =Im having trouble guring these out:1) Is the set of all functions from the boolean set {0,1} to thenatural numbers countable or uncountable?2) Let P(A) denote the power set for some set A.Given a set B, a subset A of P(B) is called an antichain if no elementof A is a subset of any other element of A. Does P(N) contain anuncountable antichain? =on sci.math:> Im having trouble guring these out:> 1) Is the set of all functions from the boolean set {0,1} to the> natural numbers countable or uncountable?Hmm, Im not an expert in this by any means, more like a beginner,but I think that the set youre after can be expressed as theCartesian product between the set of all functions from {0} to N,and the set of all functions from {1} to N. As the sets {0} and {1}are singletons, those sets of functions have the same cardinality asN itself. Therefore the Cartesian product has... what cardinality?> 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?Im too lazy to try to gure out this one.-- /-- == Im having trouble guring these out:> 1) Is the set of all functions from the boolean set {0,1} to the> natural numbers countable or uncountable?> 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?In this question it makes no difference if we substitute N byany countably innite set. So lets replace N by Q, the setof all rationals. Can you see an uncountable antichain in P(Q)?-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen on sci.math:>> Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the>> natural numbers countable or uncountable?>Hmm, Im not an expert in this by any means, more like a beginner,>but I think that the set youre after can be expressed as the>Cartesian product between the set of all functions from {0} to N,>and the set of all functions from {1} to N. Not quite. A function from {0,1} to N is a set of the form{(0,n), (1,m)} for some n and m in N, not necessarily distinct,so the set of all such functions is {{(0,n), (1,m)} : n,m in N},a set of sets of ordered pairs.The sets of functions from {0} and {1} to N {{(0,n)} : n in N}and {{(1,n)} : n in N}, respectively, and their Cartesian productis {({(0,n)}, {(1,m)}) : n,m in N}, a set of ordered pairs ofsingletons of ordered pairs. Theyre denitely not the sameset. They are, however, the same size, and as you suggested inthe bit that I snipped below, theyre the same size as N x N. Inparticular, it should be easy to write down a bijection between{{(0,n), (1,m)} : n,m in N} and N x N, thereby showing that theset in question is countable.[...]>> 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?Yes. I could simply tell you how to construct one, but youlllearn more if you work at it a bit yourself. Heres a hinttowards one possible solution:find an uncountable antichain inP(Q x Q), where Q is the set of rational numbers. Since Q x Q iscountably innite, this automatically gives you an uncountableantichain in P(N). You may also want to consider that for anyreal numbers c and d with 0 <= c < d, the set of points in Q x Qwith polar coordinates (r, t) satisfying the conditions r > 0 andc < t < d is non-empty.Brian Robin Chapman,didnt complete his address so it can be viewed directly by news brouser.http://www.maths.ex.ac.uk/~rjc/rjc.html> The League of Gentlemen>Gentelmen arent corrupted by odors, smells and scents? Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the> natural numbers countable or uncountable?>> 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?>Havent sci.mathers done about enough homework for this guy? Now he demandsproof, not just hints. Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the> natural numbers countable or uncountable?>> 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?> Havent sci.mathers done about enough homework for this guy? Now he demands> proof, not just hints.>Tho 1 is easy, I dont get 2, even with the hint to use the bijectionbetween Q and N. Also I wonder if OP meant N^{0,1}, and not {0,1}^N.Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,while |{0,1}^N| = 2^Aleph_0 = Aleph_1. > Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the>> natural numbers countable or uncountable?>> 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?>> Havent sci.mathers done about enough homework for this guy? Now he demands>> proof, not just hints.> Tho 1 is easy, I dont get 2, even with the hint to use the bijection> between Q and N. Also I wonder if OP meant N^{0,1}, and not {0,1}^N.> Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,> while |{0,1}^N| = 2^Aleph_0 = Aleph_1.Think a bit about Dedekind cuts. They dont quite ll the bill as is,but a slight modication of the idea will work.By the way, 2^Aleph_0 is not equal to Aleph_1, unless you are assumingCH == Im having trouble guring these out:>> 1) Is the set of all functions from the boolean set {0,1} to the>> natural numbers countable or uncountable?> 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?[...]>Tho 1 is easy, I dont get 2, even with the hint to use the bijection>between Q and N. How about my hint to look at Q x Q and sectors from the origin?Its also easy to do by transnite recursion, but I didntexpect Stuck to be familiar with that. Use N x N instead of N;your antichain is going to be a family F of functions from N to Nsuch that if f and g are distinct members of F, there is ann(f,g) in N such that either f(n) > g(n) for all n >= n(f,g) -- fdominates g -- or g(n) > f(n) for all n >= n(f,g) -- g dominatesf. Start with F_0 = the family of constant functions. Given acountable family A of functions from N to N, its easy toconstruct a new function from N to N that dominates every memberof A.>Also I wonder if OP meant N^{0,1}, and not {0,1}^N.>Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,>while |{0,1}^N| = 2^Aleph_0 = Aleph_1.That last step is invalid: the statement that 2^Aleph_0 = Aleph_1is the Continuum Hypothesis, which is independent of ZFC.Brian == 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?>> Tho 1 is easy, I dont get 2, even with the hint to use the bijection> between Q and N. Also I wonder if OP meant N^{0,1}, and not {0,1}^N.>> Think a bit about Dedekind cuts. They dont quite ll the bill as is,> but a slight modication of the idea will work.>{ (r,r+1) / Q | r in RQ }/ intersect> By the way, 2^Aleph_0 is not equal to Aleph_1, unless you are assuming> CH.>c, senior. <3f18c949.11042797@enews.newsguy.com 2) Let P(A) denote the power set for some set A.>> Given a set B, a subset A of P(B) is called an antichain if no element>> of A is a subset of any other element of A. Does P(N) contain an>> uncountable antichain?>>Tho 1 is easy, I dont get 2, even with the hint to use the bijection>between Q and N.>> How about my hint to look at Q x Q and sectors from the origin?Too complicated. How about { (r,r+1) / Q | r in RQ } ?/ intersect> Its also easy to do by transnite recursion, but I didnt> expect Stuck to be familiar with that. Use N x N instead of N;> your antichain is going to be a family F of functions from N to N> such that if f and g are distinct members of F, there is an> n(f,g) in N such that either f(n) > g(n) for all n >= n(f,g) -- f> dominates g -- or g(n) > f(n) for all n >= n(f,g) -- g dominates> f. Start with F_0 = the family of constant functions. Given a> countable family A of functions from N to N, its easy to> construct a new function from N to N that dominates every member> of A.> <3f18c949.11042797@enews.newsguy.com> == Im having trouble guring these out:::>> 1) Is the set of all functions from the boolean set {0,1} to the:>> natural numbers countable or uncountable?::>> 2) Let P(A) denote the power set for some set A.:>> Given a set B, a subset A of P(B) is called an antichain if no element:>> of A is a subset of any other element of A. Does P(N) contain an:>> uncountable antichain?:::[...]::>Tho 1 is easy, I dont get 2, even with the hint to use the bijection:>between Q and N.::How about my hint to look at Q x Q and sectors from the origin?:Its also easy to do by transnite recursion, but I didnt:expect Stuck to be familiar with that. Use N x N instead of N;:your antichain is going to be a family F of functions from N to N:such that if f and g are distinct members of F, there is an:n(f,g) in N such that either f(n) > g(n) for all n >= n(f,g) -- f:dominates g -- or g(n) > f(n) for all n >= n(f,g) -- g dominates:f. Start with F_0 = the family of constant functions. Given a:countable family A of functions from N to N, its easy to:construct a new function from N to N that dominates every member:of A.::>Also I wonder if OP meant N^{0,1}, and not {0,1}^N.:>Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,:>while |{0,1}^N| = 2^Aleph_0 = Aleph_1.::That last step is invalid: the statement that 2^Aleph_0 = Aleph_1:is the Continuum Hypothesis, which is independent of ZFC.Maybe the problem isnt being solved in ZFC though. Perhaps ZFCH was beingused or == Im having trouble guring these out:>:>:>> 1) Is the set of all functions from the boolean set {0,1} to the>:>> natural numbers countable or uncountable?>:>:>> 2) Let P(A) denote the power set for some set A.>:>> Given a set B, a subset A of P(B) is called an antichain if no element>:>> of A is a subset of any other element of A. Does P(N) contain an>:>> uncountable antichain?>:>:>:[...]>:>:>Tho 1 is easy, I dont get 2, even with the hint to use the bijection>:>between Q and N.>:>:How about my hint to look at Q x Q and sectors from the origin?>:Its also easy to do by transnite recursion, but I didnt>:expect Stuck to be familiar with that. Use N x N instead of N;>:your antichain is going to be a family F of functions from N to N>:such that if f and g are distinct members of F, there is an>:n(f,g) in N such that either f(n) > g(n) for all n >= n(f,g) -- f>:dominates g -- or g(n) > f(n) for all n >= n(f,g) -- g dominates>:f. Start with F_0 = the family of constant functions. Given a>:countable family A of functions from N to N, its easy to>:construct a new function from N to N that dominates every member>:of A.>:>:>Also I wonder if OP meant N^{0,1}, and not {0,1}^N.>:>Whence |N^{0,1}| = (Aleph_0)^2 = Aleph_0,>:>while |{0,1}^N| = 2^Aleph_0 = Aleph_1.>:>:That last step is invalid: the statement that 2^Aleph_0 = Aleph_1>:is the Continuum Hypothesis, which is independent of ZFC.>>Maybe the problem isnt being solved in ZFC though. Perhaps ZFCH was being>used or ZFG.Irrelevant to the problem -- only uncountability was required --and unlikely in any case.Brian 2) Let P(A) denote the power set for some set A.> Given a set B, a subset A of P(B) is called an antichain if no element> of A is a subset of any other element of A. Does P(N) contain an> uncountable antichain?>>Tho 1 is easy, I dont get 2, even with the hint to use the bijection>>between Q and N.>> How about my hint to look at Q x Q and sectors from the origin?>Too complicated. Depends on what you happen to see rst. Not intrinsically morecomplicated than the one below, in my opinion. In any case, Ithought that since my hint was a little broader than Robins, itmight point you in the right direction.>How about> { (r,r+1) / Q | r in RQ } ?>/ intersectBrian Im having trouble guring these out:>>1) Is the set of all functions from the boolean set {0,1} to the>natural numbers countable or uncountable?Each function f:{0,1} --> N may be expressed as: ((0,a),(1,b)), wherea=f(0) and b=f(1). We can establish a bijection between these and NxN as: ((0,a),(1,b)) <> (a,b)Since NxN is countable, the given set of functions must be countable.>Please prove your conclusions.Why does a phrase like this always sound like its taken directly offof an exercise sheet? Should I feel guilty of helping somebody withtheir homework? Or should they feel stupid for taking the advice ofsomebody who unked out of an undergraduate math program (due tonot doing homework)?-- Michael F. Stemper#include Give a man a sh, and you feed him for a day. Teach him how to sh,and you can sell him equipment. >I am an undergraduate mathematics student and i by the time i graduate>i will have taken 5-6 undergraduate/graduate split classes. Basically, im learning the same as the graduate student minus a>presentation or more difcult assignments. Is it typical procedure to>not be aloud to transfer any of these classes to graduate studies? I>mean, im in a situation where either i have to take the same course>all over again as a graduate student or be aloud to transfer thecredits?..how does it work in your experience ??> Whats _allowed_ varies from place to place. Here at OSU for example> there are required classes for the graduate program and also a certain> number of hours required. If youve already taken a required course> as an undergrad youre not required to take that course again to> full the required-course part, but youre also not allowed to use> the course twice in the required-number-of-hours part - if you used> it as part of the undergrad requirements then you _cannot_ use> it as part of the n hours required in the graduate program.> It sounds like youre not being allowed to use these courses> and you think you should be. I have to ask: I see the words> minus a presentation or more difcult assignments. If youd> actually done the same work required of the grad students> that would be one thing, but given that you have _not_ done> what the grad students in the course are required to do> why _would_ you think that you should get grad credit? == , I think there should be a minimal number of transferablecredits. I know in other programs if you get an A in aundergraduate course you can transfer a maximum of two courses overto grad school. I think this is a reasonable request, since in myexperience with taking these split grad/undergrad courses 90% of thecourse is exactly the same. Spending an extra 5-6 hours a week on anassignment or maybe a total of 20 hours extra to do a presentationshows little difference between the two courses. =imbibe@mindspring.com (David James Polewka) whines:> Thats all the economy is, people and money!The more people (aka power) and money you have, the more useful thingsyou can do to justify your own ideology. Or else, you canl only watchand force yourself to accept others to do whatever they like. Theres less suffering with SARS!> You are an idiot!> No people mean no consumers and no money. > U.S economy is deterioating and isnt going to get any better because of SARS.A few hundred people wont make much of a difference. Earth has beendeclared SARS free. Im not saying it is, mind, but there havent beenany new cases in some time. =Lan Gang - Local Area Network!! :) christ non locality theory following !!(im getting good at these)I have numerous effects of delayed choice catalogued at www.adamskingdom.complus 50 datapoints that prove people live their whole lives just so theirname reects what they rst write or do to me! If anyone can pass ona serious suggestion to get these quantum physics claims examined, takea smart person a couple hours at the site to verify the quantum effects, thengive me a lead. If you can even seriously suggest thatchrist non locality theory following (Chris LAN gang) to Chris as him beinga datapoint of his own theory, HawkKing being the smartest man as anotherexample and that I have detailed the by name by nature theory as part ofthe macro quantum effects, then maybe Ill be taken seriously rather than100s of newsgroups laugh for 2 years because their names were chosenHercJohn Archibald Wheeler has some very interesting ideas with == In his delayed-choice thought experiment, Wheeler suggests that asingle photon emitted from a distant quasar (far right) cansimultaneously follow two paths to Earth, even if those paths areseparated by many light-years. Here one photon travels past twodifferent galaxies, with both routes deected by the gravitationalpull of the galaxies. Stranger still, Wheeler theorizes, theobservations astronomers make on Earth today decide the path thephoton took billions of years ago.--------------------------------------------------------- -----------------------The CTMU(conspansive duality-telic recursion) of Chris Langan alsoexplains this type of possible feedback.If the locality principle is not going to be thrown into the trashheap, then a viable option is that space is something analogous tohomogeneously distributed probability density functions(a perfectuid?) i.e. increasing density gradients, giving the observedthermodynamic arrow of time. The observed cosmic expansion is arelative one! A perspective effect from our local vantage point. Ashrinking object gives the illusion of receding motion. Increasing*refractive* density gradients give the appearence of adoppler-red-shift. Space increases density as matter is re-sized.Spacetime then remembers the input! A quantum measurement is made,the action principle demands the shortest distance between two pointsbe taken, whatever that may be. There is no instantaneous action at adistance!Interesting... Either a creator with innite god-like intelligencecreated this universe as a clockworks mechanism, or the creator andthe mechanism are one and the same.A universal quantum computer?It seems that in order to continue with the idea of a physicallyinnite multiverse instantaneous communication at a distance mustbe accepted as an absurd axiom. But we must remember, Newtonsclassical reality was superceded by Albert Einsteins relativity!There is no action at a distance!Instantaneous action at a distance was shown to be unecessary, byEinsteins theories of special and general relativity. How can aninnite multiverse be in accordance with probability theory if all ofthe separate universes are not in a type of corrspondence with eachother? Who observes the entire multiverse?So it can be shown that the multiverse DOES carry to muchmetaphysical, and heavy conceptual baggage. Goodbye multiverse. Itappears that the best solution is that our universe is self contained.The locality principle is not violated.and backwards in time for brief periods below the Planck time.10^(-43) sec.The closed loop histories explanation of quantum uctuations.another. Hawking explains that these closed loop histories areconrmed by one interpretation of the Casmir effect.This is allowed by the Heisenberg uncertainty principle for briefperiods smaller than the Planck time.http://clinton4.nara.gov/Initiative...m/ shawking.htmlquote:------------------------------------------ --------------------------------------faster than light and even paths that go back in time. Before anyonerushes out to patent a time machine let me say that in normalcircumstances at least, one can not use this for time travel. Howeverpaths that go back in time are not just like angels dancing on a pin.They have real observational consequences. Even what we think of astime. That is they move forward in time on one side of the loop andbackwards in time on the other side. These closed loops are said to beOne way is to have a pair of metal plates close together. The effectof the plates is to reduce slightly the number of closed loops in theregion between the plates relative to the number outside. There arethus more closed loops hitting the outside edges of the plates andbouncing off than there are hitting the inside edges. One wouldtherefore expect there to be a small force pushing the platestogether. This force, which was rst predicted by the Turkishphysicist Hendrick Casimir, has been observed experimentally. So we----------------------------------------------------------- ---------------------Interesting...THE UNREASONABLE EFFECTIVENSS OF MATHEMATICS IN THE NATURAL SCIENCESEugene Wignerhttp://nedwww.ipac.caltech.edu/leve...ner/ Wigner.htmlhttp://www-gap.dcs.st-and.ac.uk/~hi...ons/ Wigner.htmlquote:-------------------------------------------- ------------------------------------The enormous usefulness of mathematics in the natural sciences issomething bordering on the mysterious--------------------------------------------------- -----------------------------Why does mathematics correspond to the real world?It appears that some type of instantaneous correlation orcommunication between regions of spacetime is happening, or it couldbe that spacetime has a type of memory analogous to a computerhardwarePerhaps there is a type of residual imaging going on at thesub-microscopic quantum level? What if the so called compactieddimensions are fractal in nature?A holographic 3+1 space-time?Russell E. Riersonanalog57@yahoo.com John Archibald Wheeler has some very interesting ideas with feedback:>> http://www.discover.com/june_02/featuniverse.html> quote:John Archibald Wheeler, high priest of quantum mysteries, suspects that reality exists not becausebe just what we learn about the world, he says. It may be what makes the world.Compare this to my quote :Scientists make principles from observations of their environment, it isnot mathematically sound for them to discount theories just because theydont t current theory. The line between fantasy and reality is blurred.Reaching a complete understanding of the universe involves theoriesof information, not matter. Our knowledge of events, plus facts thatwww.adamskingdom.com [Introduction] Study Time! The line between fantasy and reality is blurred.Yes, schizophrenia works that way. Get help. >The line between fantasy and reality is blurred.>> Yes, schizophrenia works that way. Get help.>psyched elephant??? F*CK OFF BIRD10 times youve used the same line on me, not since .=Hawk=. in alt.congsuch bird brain ame antics.Theres 100 other topics in the thread on advanced physics, you know therst computer text talking program was a dumb AI that responded to recognised words,Eliza, was programmed with the personality of a psych therapist.Herc John Archibald Wheeler has some very interesting ideas with feedback:> http://www.discover.com/june_02/featuniverse.html>> snip> One has to remember that Wheeler is 91 years old and his mind depends on a brain which is part of a failing body.> Contemplate the implications on science of destroying> causality as we know it. The fact of the matter is that> observation can affect observation but not the facts of> path selection which have previously become established.>> while very interesting, physics is not my thing. as> a result, i am often both confused and uncertain.>> with that in mind, ill ask how does that last sentence> above relate to:> In 1984 physicists at the University of Maryland> set up a tabletop version of the delayed-choice> scenario. Using a light source and an arrangement> of mirrors to provide a number of possible photon> routes, the physicists were able to show that the> paths the photons took were not xed until the> physicists made their measurements, even though> those measurements were made after the photons had> already left the light source and begun their circuit> through the course of mirrors.>Computer scientists interpretation:Its the classic quantum experiment usually known as the double slit experiment.When a light source is forced through a narrow slit photons interact atthe boundary and the light spreads out, like a torch beam is wider thanthe angle of the globe enclosure!Now imagine the slit is like a pebble dropping in a pool, and the wavestravel to the side and hit in concentric circles. Now drop two pebblesadjacent to each other and the pattern on the side of the pool changesto a uctuating ripple, like a sine wave higher in the middle.The two slits in a barrier allow light to travel through and spead onto a wallmuch like the two wave fronts in the pool, and the light source indeed forms adiffraction pattern on the wall, like the ripples. So when there is one slita smooth bell curve is featured, when there are 2 slits, a diffraction patternalternating bright dark! With me so far? Then, reduce the power of thetorch so that only a single photon of light is emmited each time, and sometimesit will go through slit 1, sometimes it will go through slit 2, and a photographicplate records the pattern after a lot of trials. What will the pattern be onthe wall, the interference pattern from 2 slits? or the bell curve pattern fromonly one slit having a photon go through at a time? (actually 2 bell curvessuperimposed but added together not subtractive forming the pattern)It still makes an interference pattern! Its as though the one photon went throughboth slits at the same time. This is the basis of Schroedingers Cat, where a catin a sealed box gets poisened from a mechanism depending on a single randomphoton. Can the cat be alive and dead at the same time?Delayed choice is a different extension still at micro scales but longer time, wherethe measurement of the photons path is in galaxy units. But thats all theory, themajor gap in physics today might be the function of when a quantum state mapsto its real / chosen / observed counterpart, and to keep the universe conventional mostmeasurable effects probably only manifest at micro scales.reality and fantacy are distinct, huhHerc>> even though...>> ?>> The arrow of time will not be violated. If someone discovers> time travel they can never come back to tell us about it.> Of course if they discovered it in the past.....>> I also disagree with the following pasted from the> end of the URL:>> Does Wheeler think that physicists might one day> have a similarly clear understanding of the origin> of the universe?>> Absolutely, he says. Absolutely.>> For every answer, or even likely answer, we have discovered> multiple additional new questions. Consider a nite space> bounded by an innitely long periphery as the appropriate> model. You know whats in the space, but you cant actually> dene it completely because theres not enough time to draw> the detail of the perimeter. One must content oneself with> generalizations to some acceptable degree of resolution> while realizing that whats in the undiscoverable detail is just as fascinating, and perhaps even moreso.> perhaps higher orders are involved. recognition occurs> without quantum examination.>> true, one can make error in recognition, but life itself> depends on a very high degree of accuracy.>> rgrds,> The line between fantasy and reality is blurred.>> Yes, schizophrenia works that way. Get help.>>psyched elephant??? F*CK OFF BIRD>>10 times youve used the same line on me, not since .=Hawk=. in alt.cong>such bird brain ame antics.Its not a ame, its an attempt to help you. Youre clearly mentallyill, and I strongly suggest that you seek professional help, for yourown benet. == thank Goodness,that we must apparently believe in the Copenhagen interpretation,for things to work out as you suggest -- and > I have numerous effects of delayed choice catalogued at www.adamskingdom.com> plus 50 datapoints that prove people live their whole lives just so their> name reects what they rst write or do to me! If anyone can pass on > pull of the galaxies. Stranger still, Wheeler theorizes, the> observations astronomers make on Earth today decide the path the> photon took billions of years ago. > and backwards in time for brief periods below the Planck time.> 10^(-43) sec. > The enormous usefulness of mathematics in the natural sciences is> something bordering on the mysterious--Dec.2000 WAND Chairman Paul ONeill, reelectedto Board. Newsish?http://www.rand.org/publications/randreview/issues/rr .12.00/http://members.tripod.com/~american_almanac =There is an update to the site hosting the settlement to the P-NPquestion at www.cqr.info/tcne/tcne.shtml. The update includes the following: 1. An explanation in response to a (largely irrelevant) questionconcerning the proof for NP = coNP. (Casual Notes)interest in the P-NP question. (Correspondence (1)) 3. An open letter to the author of the Why Me? le.(Correspondence (1)) The open letter is also placed at the end of thismessage.name is replaced with Xxxxxxx. Together with the message was a proof,a slightly broader version than the one published at www.cqr.info. When you follow the link to the Why Me? le publication, pleasescroll down a little as the actual text (relevant to reader, if you ever take the (NP=coNP) proof to some one, yourcolleague or professor, or if you give it in an exam for your studentsto offer critique on, never use it to test out the mathauthority/capability but the basic human decency of being truthful.titled P, NP, coNP) about the settlement at cqr.info is no longeraccessible. Therefore, this post can not be attached to the thread(and is posted separately). There appears to be a gap in the ********************** (The questionably desirable thing called respect gave me theimpression that I could benet from your reply, should there be any,to my private message, but I learnt the hard way that you decided tobenet the public (with the publication) instead, while keeping meunaware. I am not too sure if the public really beneted, just as Idoubt if the person, who expressedly in response to the mentionedpublication questioned the meaningfulness of it, would ever reallyunderstand the Why Me? le. In fact, I hope that in the entire world,only you yourself alone can understand it. The mentioned publication of yours seems to have quite a lot packedin it. With the same amount of ink, I believe, you might haveenlightened any ignorant soul should you have written sincerely to theperson directly. Despite the non-singular logical inconsistency inthe mentioned publication and despite my strong doubt about thecorrectness of your understanding of, say, Cantors conclusion, Iprefer to stay focused on this one issue: Is it a valid proof? Since you can see clearly in a few seconds, can you share with uswhy it is invalid (not much is surely such a reference)? Whichdenition used is ill-dened? Which proof step is wrong? Whichassumption, if any, is false? Or even, which concept is beyondelementary mathematics? Since you even took time to expound onAlbertian Pythagorasism, could you also philanthropically spend alittle time pointing out the fallacy (that you see clearly in a fewseconds) in the proof for NP = coNP? I cowardly ask these questionswith the sure hope that you can courageously answer them all in strongsupport of the mentioned publication of yours. Whether that, or anything else, is the theorem of the century,possesses little signicance for me. I do not consume my life on suchpetty stuff. But I am not afraid of sending it to a grad student,taking your kind advice. Please let me know the grad students whowill, regardless of the existence of any, happily tell me my errors ofmy ways. But just to avoid introducing not much again with a lot ofwriting, I think Id better run So&so, Because you are an expert in complexity, and because I am such a despicable liar (as I do not know you and do not really know if you are an expert), but just because lying attery and attering lie may get me somewhere... How does that sound? Should I also redundantly add that being anExpert or being not an Expert as an addressee can turn a non-proof toa proof or vice versa? Judging from this opening line, am I alsoqualied now to advise people on how, what, who and whose way towrite? Please kindly advise. == on..partially because i do not know where to begin and partiallybecause i believe i dont even fully understand the problems...and iwas wondering if any of you would be kind enough to show me what todo? THANK YOU!!!1. Suppose that (x_0,y_0)=(0,0) and for every positive interger n,f(1/n,0)=1 and f(0.1/n)=-1. Does lim (x,y)-->(x_0,y_0) exist? Proveyour answer.2. Let g(x,y)= (sin(x-y))^2/(abs(x)+abs(y)), where abs(x)=absolutevalue of x. Prove that lim (x.y)-->(0,0) g(x,y)=0 using either theabs(sin(s+t)) f(1/n,0)=1 and f(0,1/n)=-1. Does lim (x,y)-->(x_0,y_0) exist? Prove> your answer.No. Take any disk that is centered at the origin.No matter how small the radius of the disk is,within the disk, along x-axis (y = 0) there are valuesof the function equal to 1, while, within the disk, alongthe y-axis there are value of the function equal to -1.Since a limit is unique, there isnt one.> 2. Let g(x,y)= (sin(x-y))^2/(abs(x)+abs(y)), where abs(x)=absolute> value of x. Prove that lim (x.y)-->(0,0) g(x,y)=0 using either the> abs(sin(s+t)) to.When (x,y) is not (0, 0), we have:abs(sin(x+(-y))(0,0)(abs(x)+abs(y)) = 0,by Squeeze Theorem, so does lim g(x,y).> 3. Prove or disprove: If f_x(x_0,y_0) and f_y(x_0,y_0) both exist,> then f is continuous at (x_0,y_0).Consider the raised cross:f(x,y) = 1 along y = 0 and along x = 0and f(x,y) = 0 everywhere else.Both partial dervatives are 0 at (x, y) = (0, 0),but f is not continuous at (0, 0).> 4. Suppose f(x,y) is differentiable for all (x,y), f(x,y)=17 on the> unit circle x^2+y^2=1 and grad f is never zero on the unit circle.> For any real number K, find a unit vector parallel to grad> f(cos(k),sin(k))....grad f stands for the gradient of f.(For example, consider an inverted cone with vertex at the origin,Its level curves are circles but the gradient is never 0.)In this case, the unit circle is a level curve of f.But the gradient is perpendicular to level curves.The point (cos(k), sin(k)) is on the unit circlex^2+y^2 = 1, as can be checked.The gradient (also in the plane) is perpendicularto this unit circle. Note that the dot product ofthe vectors [cos(k), sin(k)] and [-sin(k), cos(k)] is zero.So these last two vectors are perpendicular.Therefore, the vector [-sin(k), cos(k)] is parallelto the gradient at (cos(k), sin(k)), on..partially because i do not know where to begin and partially> because i believe i dont even fully understand the problems...and i> was wondering if any of you would be kind enough to show me what to> do? THANK YOU!!!> 1. Suppose that (x_0,y_0)=(0,0) and for every positive interger n,> f(1/n,0)=1 and f(0.1/n)=-1.Is that second one f(0.1/n,0) or f(0,0.1/n)? Heres my hint on the assumption that its f(0.1/n, 0):The values 1/n get arbitrarily close to 0. Pick any smallinterval around 0, and there are innitely many 1/n closerthan that.The values 0.1/n have the same property.> Does lim (x,y)-->(x_0,y_0) exist? Prove> your answer.Something missing again. The limit of what?Assuming you mean the limit of f(x,y) go to the epsilon-deltadenition of limit of a function. Can you conne f(x,y)within an arbitrarily small range by conning (x,y)close to (0,0)? Do the wiggles get smaller and smalleras you get closer to the origin?> 2. Let g(x,y)= (sin(x-y))^2/(abs(x)+abs(y)), where abs(x)=absolute> value of x. Prove that lim (x.y)-->(0,0) g(x,y)=0 using either the> abs(sin(s+t)) to.That means that f(x,y)/(abs(x) + abs(y)) <= f(x,y)/(abs(x+y)).It also means (but youd have to show this) thatf(x,y)/(abs(x) + abs(y)) <= f(x,y)/(abs(x-y)). Yournumerator is a function of (x-y) so you could havef1(x-y)/abs(x-y). The two-D limit becomes a one-D limit.Your g(x,y) is positive and this gives you an upper bound onit. Does that upper bound go to zero? If it does, g(x,y)is sandwiched. You may have to invoke a sandwich theoremto make it rigorous.> 3. Prove or disprove: If f_x(x_0,y_0) and f_y(x_0,y_0) both exist,> then f is continuous at (x_0,y_0) (f_x(x_0,y_0) means the partial> derivative of f a(x_0,y_0)...the problem does not state that the> partials are continuous which leads me to believe that the statement> is false...i just do not see how to disprove it.)Good instincts. A disproof is a counterexample. f_x tellsyou how it varies along one direction. f_y tells youhow it varies along another direction. That leavesyou an innity of in-between directions. Can youthink up a function that looks nice along the axesat x_0,y_0 but not in other directions? If you canpicture it in your mind, you can then cook up amathematical representation of it.> 4. Suppose f(x,y) is differentiable for all (x,y), f(x,y)=17 on the> unit circle x^2+y^2=1 and grad f is never zero on the unit circle. For> any real number K, find a unit vector parallel to grad> f(cos(k),sin(k))....grad f stands for the gradient of f. (This> problem....i do not even understand...and im pretty sure the textbook> did not make a typo or an error or anything like that...well, i> probably just do not understand it but isnt it contradicting what its> saying? it says f(x,y)=17 on the unit circle x^2+y^2=1 yet grad f is> never zero on the unit circle. -_-)Two ways to think of whats going on:Since f(x,y) = 17 on the unit circle, if you did a contourplot of f(x,y), the unit circle would be a contour. Thegradient crosses contour lines at right angles. Imaginea circular hill. The contour lines, the level lines, areconcentric circles around the peak. But the gradientis a direction straight uphill.Another way to think of it, algebraically instead ofgeometrically: Youre bothered by f(x,y) being aconstant somewhere but the gradient being nonzero.The gradient tells you the direction in whichf(x,y) changes most rapidly. In other directions itchanges less rapidly, or not at all. Algebraically,the rate of change in a direction d is given bygrad(f) dot d, where dot means the vector dot product.This is zero when d is orthogonal to grad(f).So what that tells you, same as the geometric argument,is that lines along the unit circle are orthogonal tothe gradient direction.The point (cos(k), sin(k)) is a point on the unit circle atangle k. Draw the picture. Which direction are tangentlines at that point? Which direction is the gradient?Ive given you the crucial clue twice now to be ableto answer that question. You need to translate that clueinto algebra. - Randy i think so...=T >on..partially because i do not know where to begin and partially>>because i believe i dont even fully understand the problems...and i>>was wondering if any of you would be kind enough to show me what to>>do? THANK YOU!!!>>1. Suppose that (x_0,y_0)=(0,0) and for every positive interger n,>>f(1/n,0)=1 and f(0.1/n)=-1.>Is that second one f(0.1/n,0) or f(0,0.1/n)? >No, its f(0,1/n). The . is next to the , on the keyboard.This is easily solved by looking at the denition of limit.Jon Miller 1. Suppose that (x_0,y_0)=(0,0) and for every positive interger n,> f(1/n,0)=1 and f(0.1/n)=-1.> Is that second one f(0.1/n,0) or f(0,0.1/n)? oops i obviously types this too fast....i meant to say that the secondone is f(0,1/n)> Something missing again. The limit of what?> (x,y)-->(x_0,y_0)f(x,y) 1. Suppose that (x_0,y_0)=(0,0) and for every positive interger n,> f(1/n,0)=1 and f(0.1/n)=-1.> Is that second one f(0.1/n,0) or f(0,0.1/n)? > oops i obviously types this too fast....i meant to say that the second> one is f(0,1/n)> Something missing again. The limit of what?> (x,y)-->(x_0,y_0)f(x,y)Youve gotten several answers. Look at the epsilon-deltadenition of limit (x->x0) f(x,y). Do you understand whatepsilon-delta is all about? - Randy > Wouldnt it be cool if the SI could dene the kilogram> as some number of the same cesium-133 photons that> are currently uses to dene the second and the meter ?>> The SI clock measures time by counting these photons .> Is there an SI Radar Gun that measures length> by recording the travel time of a reected light beam ?> What about an SI device that could use light to measure mass ...> How would it work ?>> There are several alternate denitions for the kilogram in the works:> http://www.wikipedia.org/wiki/Kilogram>> All of those denitions related to using parallel wires of neglible cross> section and innite length are sure interesting(;^)>>An artfact like that probably wouldnt change in a million years(;^? > There are several alternate denitions for the kilogram in the works:> http://www.wikipedia.org/wiki/Kilogram>> All of those denitions related to using parallel wires of neglible cross section and innite length are sure interesting(;^)>> An artfact like that probably wouldnt change in a million years(;^?I wasnt going to reply since anyone who checked the link wouldrealize that your ramblings are solely the product of a deranged mindbut I also have a slim hope you might realize it. > There are several alternate denitions for the kilogram in theworks:> http://www.wikipedia.org/wiki/Kilogram>> All of those denitions related to using parallel wires of negliblecross section and innite length are sure interesting(;^)>> An artfact like that probably wouldnt change in a million years(;^?>> I wasnt going to reply since anyone who checked the link would> realize that your ramblings are solely the product of a deranged mind> but I also have a slim hope you might realize it.Click on amp. in your wikipedia and see what Im rambling about Mickey. =In sci.math, Donald G. Shead<3rHSa.11298$u94.3251000409@newssvr10. news.prodigy.com>:>> There are several alternate denitions for the kilogram in the> works:>> http://www.wikipedia.org/wiki/Kilogram>> All of those denitions related to using parallel wires of neglible> cross>> section and innite length are sure interesting(;^)>> An artfact like that probably wouldnt change in a million years(;^?>> I wasnt going to reply since anyone who checked the link would>> realize that your ramblings are solely the product of a deranged mind>> but I also have a slim hope you might realize it.> Click on amp. in your wikipedia and see what Im rambling about Mickey.> This is a disambiguation page; that is, one that just points to other pages that might otherwise have the same name. If you followed a link here, you might want to go back and x that link to point to the appropriate specic page.Well assume Ampere. In physics, the ampere (symbol: A, often informally abbreviated to amp) is the SI base unit used to measure electrical currents. By denition, 1 ampere is that constant current which, if maintained in two straight parallel conductors of innite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2x10^-7 newton per meter of length. The ampere is named after Andr.8e-Marie Amp.8fre, one of the main discoverers of electromagnetism. The unit of electric charge, the coulomb, is dened in terms of the ampere: 1 coulomb is the amount of electric charge carried in a current of 1 ampere owing for 1 second.(The wires would have to be of negligible cross section inorder to show any effect at all. 1 Newton is about theforce of 100 grams in the Earths gravitational eld.2x10^-7 Newtons would be roughly equivalent to thegravitational force produced by 20 nanograms. Of courseone could move the wires closer together and push morecurrent through them. :-) )-- #191, ewill3@earthlink.net -- who prefers an ampmeter (ammeter?) himselfIts still legal to go .sigless. =:>: minimizes the sum of the squares of the (vertical) errors. Is:>: that really your goal when you choose best? Another goal will:>: give another answer. For example, if you decide that the best:>: line is the one which minimizes the largest of the (vertical):>: errors, then in this case I believe the best line is:>:>Thats an interesting criterion. So, if all the points lie on a line:>except for one outlier, the best line is -- I think -- halfway:>between the outlier and the rest. Anti-robust regression, anyone?: Well, youve got to choose a function f from among a family of: options (in this case, the linear functions f(x)=ax+b) and for any: such choice you will have errors e_i = y_i - f(x_i). You want to: make all the errors small, but how do you balance several objectives: like that? You can amalgamate the vector-valued function (e_1, ..., e_n): into one real-valued objective function E to be minimized.: The usual choice is: E = ( sum (e_i)^2 )^(1/2): which is just the L^2 norm on this vector space. A non-mathematician: might rst think to use the L^1 norm, and there are L^p norms for: all p. All are reasonable. I was merely suggesting the L^infty norm.As a mathematician, you can measure the distance between your data and your tted values any way you like. But your original point wasthat, for a given practical problem, not all metrics are reasonable (no quotes necessary). In virtually every real problem Ive encounteredin which the L^2 norm is inappropriate, its inappropriate at least inpart because it gives too much weight to outliers. (All L^p norms arebad in the sense that one anomalous data point can have an arbitraryinuence on the t).In the case of L-innity regression, I think it is hard even tocook up articial examples for which it would be appropriate (see below).: Certainly there are times when the L^infty norm better measures what: youre trying to do. You could for example be trying to adjust some : environmental variables for a crowd, knowing that no combination of them : will make everyone happy. Choosing conditions which make everyone a little: bit unhappy is not so bad; choosing conditions which happen to be: lethal for somebody is considered really poor form! So you might: want in that case to minimize the maximum error.I dont completely understand the setup here, in which theerrors in the t correspond to unhappiness, but lets assume that makes sense. The L^innity norm says that, underany circumstances, you should sacrice everyones happiness as much as possible to make the unhappiest person a tiny bit happier. That doesnt seem right. Mike =Greek equivalent of 11th grade. We are doing vectors, and I have aquestion that nobody seems to be able to help with, so I post it herehoping to an answer.This is my rst post on sci.math and despite of my brieooking-around, I failed tofind what the Usenet style forrepresenting vectors is. So please excuse my ignorance and let a and bbe vectors and k E R.My Algebra textbook refers to the following proporties of the dotproduct of a vector :(k*a)*b = a(k*b)=k*(a*b)While solving any excercise I couldfind, I stumbled across thefollowing True-False question :(k*a)*b=(a*b)*kI chose True on the basis that (a*b) E R and since k E R too, theirmultiplication is probably OK. (k*a) E V and b E V, so their dotproduct [(k*a)*b] E R , and must be equal to (a*b)*k.I also tried to verify this theoretically, not just by intuition :Let vector a = (x_1,y_1)Let vector b = (x_2,y_2)Let k E R.(k*a)*b= (k*x_1,k*y_1)*(x_2,y_2) = k*x_1*x_2 + k*y_1*y_2 = k*(x_1*x_2+ y_1*y_2) =(x_1*x_2,y_1*y_2)*k= (a*b)*kThis verication of my guess seems to be alright to my inexperienceeye, but this property fails to appear not only in the textbook, butin any other book Ive read concerning vectors ( Id be glad toreference, but they are books available only here and are helpingbooks for the school - not English Algebra textbooks ) .Does this property exist ? It is unusual to have the coefcient afterthe dot product of two vectors, but it seems to be alright since thedot product is enclosed in parentheses. Or, if not, what am I doingwrong?Vassilis Pandis My Algebra textbook refers to the following proporties of the dot> product of a vector :>> (k*a)*b = a(k*b)=k*(a*b)>You need to be clear about the dot product and scalar multiplication.For example using * for dot product and . for scalar multiplication. (k.a)*b = k.(a*b) = a*(k.b)or (ka)*b = k(a*b) = a*(kb)> While solving any excercise I couldfind, I stumbled across the> following True-False question :>> (k*a)*b=(a*b)*k>> I chose True on the basis that (a*b) E R and since k E R too, their> multiplication is probably OK. (k*a) E V and b E V, so their dot> product [(k*a)*b] E R , and must be equal to (a*b)*k.>> I also tried to verify this theoretically, not just by intuition :>> Let vector a = (x_1,y_1)> Let vector b = (x_2,y_2)> Let k E R.>> (k*a)*b= (k*x_1,k*y_1)*(x_2,y_2) = k*x_1*x_2 + k*y_1*y_2 = k*(x_1*x_2> + y_1*y_2) =(x_1*x_2,y_1*y_2)*k= (a*b)*k>> This verication of my guess seems to be alright to my inexperience> eye, but this property fails to appear not only in the textbook, but> in any other book Ive read concerning vectors ( Id be glad toIndeed, youve a notational problem.a*(b*c) doesnt exist nor does k*a.What exists is k.(b*c) and k.a.You wont see a*(b*c) = (a*b)*c, look fork.(a*c) = (k.a)*c or k(a*c) = ka * c> Does this property exist ? It is unusual to have the coefcient after> the dot product of two vectors, but it seems to be alright since the> dot product is enclosed in parentheses. Or, if not, what am I doing> wrong?>No not as you write it. Otherwise yes when written correctly. Byconvention scalar multiplication places the scalar rst and to dootherwise makes for bothersome reading. What youre doing wrong is notdistiguishing between vector dot multiplication and scalar multiplicationwith a vector. =I would like a proof of the following theorem (originally due toWeierstrass):Given any function f, continuous in some interval [a,b], and any e >0, there exists some polynomial P, such that |f(x) - P(x)| < e, for any x in [a,b].--Thanos I would like a proof of the following theorem (originally due to> Weierstrass):>> Given any function f, continuous in some interval [a,b], and any e >> 0, there exists some polynomial P, such that>> |f(x) - P(x)| < e,> for any x in [a,b].> --ThanosHere is the one using Bernstein polynomials:http://apollonius.math.nthu.edu.tw/d2/num/ uniform-approximation/bernstein.html =I am an intern at NASA and am working on a certain problem where Iwant to use weighted least squares, except I want> to use NxN covariances > for data in R^{N} instead of scalar weigths.> For instance, I have a several points in 2D, with> each point having its > own 2x2 covariance matrix. The covariance matracies> vary from one point > to the next. > I would like to t a line to this data, using> weighted least squares > where the covariance matracies are used in tting> the data as weights. I > also would like to get a covariance for the parameters of> the nal line > (specically the slope of the line -- or normal to> the line from the > origin -- and the distance from the origin to the> line).> Lastly, I assume, that I can use the same algorithm> for higher dimensions, > for example tting a plane to 3D (x,y,z) data which> have a 3x3 covariance =These are absolute fodder for my by name by nature theory!Ill just outline some obvious ones before people start thinkingIm on some he sees patterns thing1st post to uk.org.mensa in year 2000>peoples names determine what scientic discovery or art>they will uncover> Major contribution to the rst practical nuclear reactor: Roberta Woods (> Fermis student and collaborator)endless fuel ~ woods> Inventor of Nuclear medical tagging:> Rosyln Yalow. ( Nobel Prize in Medicine)label ~ yellow>> Discover of DNA structure: Roslynn Franklinsequence ~ line>> Inventor of Computer Programming: Ada Lovelace>> Inventor of high level computer language:> Grace Hopperhopper ~ moth (the rst computer bug she also found)> Inventor of Agile Signal coding: Heddi Lamour>> First discoverer of Chaos in electronics and its mathematical theory: Mary> Cartright. She also invented nonlinear electronics.>> First to apply abstract algebra ( which she invented to number theory) Emmy> Noether.> error correcting codes without which your PC couldnt work.>physics ~ no ether>> First Solution of time evolution for Einsteins equations: Madam Choquet>> Major inventor of Quantum Gravity (using path integrals): Cecile Dewitt>> Discoverer of Radium and Polonium ( Winner of Nobel Prizes in Physics and> Chemistry): Marie Curie>> Shell Model of Atomic Nucleus: ( Nobel Prize):> Maria Mayer>> Wu> ( Stonybrook)>> Mathematical theory of transonic gas dynamics: Lesley Sibner, Kathleen> Morawitz, Penny Smith>> Theory of dimensions past arbitrary obstacles: Penny Smith>> Supercritical Wing theory: Lesley Sibner>> Inventor of the D-Wavlet ( responsible for trainingless voice recognition, and> modern digital tv) and Macarthur winner Ingrid Daubeiches.>> Fundamental work in Genetics and member of the national academy of sciences> Lynne Margulis.>> Jumping genes ( Non-nuclear genetics) and a Nobel Prize. Also a woman from> Cold Spring Harbor labs.jump ~ spring>> What about the environmental movement: Every hear of Rachel Carson?>> I could go on and on and on.>> best> Penny>>(Example, Women have contributed almost>nothing to the technological or democratic society we have today in the US.>> Ever hear of Rosa Parks, OR Emma Goldman, or Elizabeth Candy Stanton?> Or Wilsons wife--who was defacto president of the USA for years after her> husbands stroke. Or Eleanor Roosevelt, who was responsible for many of the> ideas of the New Deal.> As to abolition of slavery, ever read Uncle Toms Cabin?>> Women have been severely discriminated and excluded, but we have made quite a> few contributions.> On the other hand, our greatest contributions have YET to come.>> p.s. You may have heard of Salk and Sabin, but the rst person to isolate the> polio virus was a woman.>> You may have heard of Hubbles expanding universe but the method he used> to conrm it ( and the idea to look) was invented by Cecila Payne ( Member> National Academy of Sciences).>> Harvard has quite a tradition of unpaid world class women astronomers. They> must have been inspired by Catherine Herschel ( the woman who did all the math> for her husband--including the discovery of> Uranus, and is credited with dozens of independent discoveries.)>> etc. __ / / / / / / / / / / / /__/__ /________ ___________/Proof of numerology at www.adamskingdom.comname one billionaire?what did Lady Di do?Tiger :: golf :: ?star wars program was introduced by which president ?Nic Cage stars in what kind of movies ?Who is the smartest man, Haw.... ? Discover of DNA structure: Roslynn Franklin>> sequence ~ lineActually that ones quite notable, RosLYNN FrankLINLINE LINE double helix !!Hercgo on , prove numerology, take the test at www.adamskingdom.com10% of James Randi million in it for you! ok 15% Discover of DNA structure: Roslynn Franklin> sequence ~ line> Actually that ones quite notable, RosLYNN FrankLIN>> LINE LINE double helix !!The only problem is her name was Rosalind, not Roslynn.--Mark with me your analysis.> I read through all your questions and confusions.> My answers to the confusions> 1. This statement is originated from The Merchant of Venice by> William Shakespear.> 2. I dont think I miss something important for the reasoning, what I> did is simply by changing some phrases, such as picture, gold> box(silver, lead), to necklace, box A(B,C). By doing that, nothing> harms getting the conclusion.> 3. The statement is There are three boxes, A, B and C, there is only> one necklace in any one of the boxes, which tells there is three and> only three boxes(A,B,C), one and only one necklace in either A, or B,> or C at one time; By given three propositions/conditions on three> boxes respectively, anyone can tell in which box the necklace> mentioned in the statement is?> My analysis> 1. When I rst looked at the statement I had the same impression as> most of you do, it seems to omit some condition which leads to> conclusion, such as, of three propositions, two true and one false, or> two false and one ture.> 2. But when I worked on it further, I found the statement conveyed me> a very important idea, that is uniqeness, with it I dont think the> statement omit any crutial for the conclusion.> 3. I consider, since the statement tells the necklace is in either A,> or B, or C, then propositions A, B and C can neither be all ture nor> all false.> There must be one box containing the necklace mentioned, I> suppose(if not stated Box A, B or C, letters A, B and C mean> proposition A, B and C thereafter):> if its in Box A-> A(F), B(T), C(T)> if its in Box B-> A(T), B(F), C(T)> if its in Box C-> A(T), B(F), C(F)> SO you can see, in the rst two possibilities, two T one F; the> last one shows the difference, two F one T. My conclusion is, the> necklace is in Box C.> 4. If you dont nd it persuasive, let me show my other reasoning,> suppose, A(T)-> B(F), C(F), then must be in Box C> A(F)-> B(T), C(T), then must be in Box A> B(T)-> A(F), C(T), then must be in BOx A> B(F)-> A(T), c(F), then must be in Box A> C(T)-> either A or C is ture, we cannot get the> conclusion> C(F)-> A(T), B(F), then must be in Box C> since, we know the necklace must be in Box A, or B, or C at one> time from these three propositions, C cannot be true, if it is true,> we dont know the necklace is in Box A or B, and it is contradictial> to the statement.> so, C must be false, that is to say, The necklace is not this> box is false. The conclusion is, the necklace is in Box C.>> I dont know whether you agree with me of the above, and you are again> welcome to tell me of your opinion.> best> cat>I agree box C is a weak conclusion from your rst argument, but the secondargument you seem to be selecting what propositions to consider and I suspectthere are equally valid arguments for each box like this.You should spend an hour at www.adamskingdom.com [paranormal test],50 simple questions that depend on selecting an option IF YOU HAD TOselect an option, just like this, which would give a denite answer.Not only that, if you guess 25 out of 50 right, (4 multi choice) and can see thatthat should be impossible it proves Im paranormal (from a simple test how about that?)and if you can further scream and jump up and down upon passing an impossibletest to get James Randi to peruse it youve earnt $100,000. (or anyone here)[quantum experiment] explains the test.Hercsample 10 questions, get 5 right and were on track, you can see by the dateand use rigged.------------------------------------------------------ --------------------------///////////all you moon-huggers!-----------------Greg Neillscribe2bRoundtableJohn L------------------------------------------------------------ --------------------Ive spent basically my whole life at 28 degrees lat. and have gotten quitegood at telling the time of day by the sun. In the past couple years Ivebeen to England three times and was bafed by the sun there - even knowingto expect it. It was wild there in February actually being able to feel theday getting longer. That was different.Also different was the snow we got today. The Space Center had enough thatit piled up on cars.-----------------CNoteMitch DicksonTim KozuskoJ.y.n.x----------------------------------------------- ---------------------------------The evidence based on metallurgical analysis of fractured surfaces(produced by Geller) indicates that a paranormal inuence must have been operative in theformation of the fractures. Dr Wilbur Franklin (Physics Department,Kent State University - U.S.A.) We have observed certain phenomena with...-----------------G=EMC^2 GlazierCNotemalcolm burtonGreg Neill-------------------------------------------------------- ------------------------> hey Im the one solving dillemmas round here, contribute> something useful yourself. its been well discussed b4 that> odd binaries are allowed here.No.> formal rules such as attachments> only in binaries named groups are easily programmed into> news servers yet they dont actually exist do they?>> now which part of too bad didnt you get?> HercWelcome to the brackish depths of my killle, Sparky.*plonk* ...problem solved.-----------------Rich ShewmakerIanGreg Neillscribe2b------------------------------------------------ plastic was held by the magnet with a small metal supercial insertthe bottom of the tear drop was the same proportions as the sphere, and inall was much heavierI was considering the possibility of the teardrop wabbling and causingturbulence on its rst moments of fall, but wouldnt that have showed up asa sphere being held by a magnet?>> the magnetic eld will remain after the voltage is cut> and different shapes and materials will drop sooner>> the weight (also size) will inuence the fall, a teardrop> might be faster than a sphere in general but lighter> material will fall slower in atmosphere in general aswell>> maybe your teardrop doesnt maintain vertical at low speeds>> Herc>> Someone HAS to know whats going on.> If it helps, its not anything wrong with our particular apparatus because> similar phenomenon happened with other groups on different workstations.> Im doing a rst year university physics lab on free fall.. the apparatus> is simple. An electromagnet positioned above two, moveable, photogates. When> the voltage is cut, the object falls and the time interval between the two gates is measured on a timer accurate to 0.1 ms. We used two different> objects to examine their free fall, we used a plastic sphere and a> streamlined (i.e. looks like a teardrop) steel object.> We hypothesized that the plastic sphere would feel greater drag during its> fall, and would thus, take longer to fall.> However, when we performed the lab, we found that the plastic sphere was> consistently falling approximately 5-15 ms faster than the steel (for> heights of 50-200 cm).> When we tted the two sets of data our confusion grew as the plot for the> plastic sphere had a cubic term, representing the 1/3 the drag coefcient,> while the steel did not. (veried by chi-squared results).>> Can anyone explain this phenomenon?-----------------ShanxWandaTheKidraven1----------- ------------------------------------------------------------- --------|-|erc:The picture looks great.Interestingly, the ill-logicof the basic design congurationis more visible in the sketchthan in photos. The problemswould be that, numerous strongpoints in the fuselage are neededto support the heavy componentsthat are exterior to the main tankand to resist and distribute highlocal stresses that are due to thehigh forces that are applied to thestructural components.A vertical tower provides for theinline stacking of components. Thesymmetrical design of the USSR rocketsystem would probably have far lessstructural problems.Neat sketch, however.The unit spacing worked just ne._________________________________>|-|erc:>>Please advise us regarding the font setting that>>you used in making the picture. We can then select>>the same font in order to see the picture as you>>intended.> this is the original ascii art, I altered it for my default outlook font,> but if you are adjusting font this one is better (any monospaced) :>> /> / > / > / > / > ^ /__________ ^> / | | / > /___| / |/___> | || / || |> | || /____ || |> | || // || |> | || //______ || |> | || | || | || |> |___|| | || | ||___|> | || | || | || |> | || | || | || |> | || | || | || |> | || | || | || |> | ||/| || ||| |> | |/ | || | | |> | / | || | |> | / | || | |> | / |___||___| |> |/ | | |> / |___||___| > (______|____||____|______)> /___ /_/||/_ /___>> // // > //// //// > //// // //// // > /// // // // //// //// / > / / ///// / //// // / // > / /// // // /// /// /// / // > / // / /// //// / //// // // > /// /// //// // // /////// > / // / ///// ///// / /// // / / > /// /// / /// / // //// // // / / > // // ///// /// // //// ///// // / > ////// //// // // /// // ////// // > /// / ///// / /// // // ///// /// / > -Jas-----------------RustRalph HertleMercury481PlanetaryMatrix------------------------------ --------------------------------------------------> thats really good, if you have a spare 30 secs download it , hollywood> doesnt do much better. I will humbly accept the compliment even though I am never humble.http:www.sworld.org/artiv/fs.mpg is the next scene. I was very surprised with how good it looked with almost no effort. Of coursecomments on what is unrealistic are appreciated as I want to improve it.-----------------Tim KozuskoMatt GiwerChrisApostate------------------------------------------- -------------------------------------Yep, I took the data from both sources for 1 bar-----------------SomeoneLawrence & BobbieOdysseusmalcolm burton------------------------------------------------------- -------------------------> First of all I need a volunteer from the audience,>> give me the name of any rec newsgroup!>> Dont be shy, say you there, any rec newsgroup, step> right up and see real magic!>> HercYoure doing it wrong.Please immediately purchase a copy of Jim Cellinis DVD and view it ve timesin a row. (Potty breaks are allowed; its a long video.)Itll be a good start for you. The rest of us are praying, burning whitecandles, chanting, thinking positive thoughts, and/or invoking any number ofChopra Quantum Whatevers to help bring about the release of volume two sometimereal soon now.-----------------RustJohn LGreg EvansBen Sauvin------------------------------------------------------- -------------------------Randi will test you when you properly apply to be tested. Sign up here:http://www.randi.org/research/challenge.html------------ -----Rich ShewmakerCNoteWandaRust-------------------------------------- ------------------------------------------It really all depends on the situation.-----------------ShanxSee You In Hell My Friend.SomeoneGreg Neill-------------------------------------------------------- ------------------------If ever I actually found myself in that situation, Id hold it upright,with the intent of attacking my assailants knife hand.-----------------cliff86RustShanxNormDePloom------------ ------------------------------------------------------------- -------ANSWERS10 Scribe2b scribes11 Tim Kozusko mount koziosko12 CNote see note13 Greg Neill nil14 Shanx show then space16 Matt Giwer give away17 Someone one bar18 John L loo19 Rich Shewmaker rich showmaker20 See You In Hell My Friend. depends21 Rust attacks metal =hi, HericI am just a beginner to Logic. I dont understand your purpose of thequotes and what they mean.Furthermore, I cannot open your link.Again, anyone interested in this thread is welcome to refer to my lastpost, I want to know about views from differet all your questions and confusions.> My answers to the confusions> 1. This statement is originated from The Merchant of Venice by> William Shakespear.> 2. I dont think I miss something important for the reasoning, what I> did is simply by changing some phrases, such as picture, gold box(silver, lead), to necklace, box A(B,C). By doing that, nothing> harms getting the conclusion.> 3. The statement is There are three boxes, A, B and C, there is only> one necklace in any one of the boxes, which tells there is three and> only three boxes(A,B,C), one and only one necklace in either A, or B,> or C at one time; By given three propositions/conditions on three> boxes respectively, anyone can tell in which box the necklace> mentioned in the statement is?> My analysis> 1. When I rst looked at the statement I had the same impression as> most of you do, it seems to omit some condition which leads to> conclusion, such as, of three propositions, two true and one false, or> two false and one ture.> 2. But when I worked on it further, I found the statement conveyed me> a very important idea, that is uniqeness, with it I dont think the> statement omit any crutial for the conclusion.> 3. I consider, since the statement tells the necklace is in either A,> or B, or C, then propositions A, B and C can neither be all ture nor all false.> There must be one box containing the necklace mentioned, I> suppose(if not stated Box A, B or C, letters A, B and C mean> proposition A, B and C thereafter):> if its in Box A-> A(F), B(T), C(T)> if its in Box B-> A(T), B(F), C(T)> if its in Box C-> A(T), B(F), C(F)> SO you can see, in the rst two possibilities, two T one F; the> last one shows the difference, two F one T. My conclusion is, the> necklace is in Box C.> 4. If you dontfind it persuasive, let me show my other reasoning,> suppose, A(T)-> B(F), C(F), then must be in Box C> A(F)-> B(T), C(T), then must be in Box A> B(T)-> A(F), C(T), then must be in BOx A> B(F)-> A(T), c(F), then must be in Box A> C(T)-> either A or C is ture, we cannot get the> conclusion> C(F)-> A(T), B(F), then must be in Box C> since, we know the necklace must be in Box A, or B, or C at one> time from these three propositions, C cannot be true, if it is true,> we dont know the necklace is in Box A or B, and it is contradictial> to the statement.> so, C must be false, that is to say, The necklace is not this> box is false. The conclusion is, the necklace is in Box C.>> I dont know whether you agree with me of the above, and you are again> welcome to tell me of your opinion.> best> cat>> I agree box C is a weak conclusion from your rst argument, but the second> argument you seem to be selecting what propositions to consider and I suspect> there are equally valid arguments for each box like this.> You should spend an hour at www.adamskingdom.com [paranormal test],> 50 simple questions that depend on selecting an option IF YOU HAD TO> select an option, just like this, which would give a denite answer.> Not only that, if you guess 25 out of 50 right, (4 multi choice) and can see that> that should be impossible it proves Im paranormal (from a simple test how about that?)> and if you can further scream and jump up and down upon passing an impossible> test to get James Randi to peruse it youve earnt $100,000. (or anyone here)> [quantum experiment] explains the test.> Herc> sample 10 questions, get 5 right and were on track, you can see by the rigged.> ------------------------------------------------------------- -------------------> ///////////> all you moon-huggers!> -----------------> Greg Neill> scribe2b> Roundtable> John L> ------------------------------------------------------------- -------------------> Ive spent basically my whole life at 28 degrees lat. and have gotten quite> good at telling the time of day by the sun. In the past couple years Ive> been to England three times and was bafed by the sun there - even knowing> to expect it. It was wild there in February actually being able to feel the> day getting longer. That was different.> Also different was the snow we got today. The Space Center had enough that> it piled up on cars.> -----------------> CNote> Mitch Dickson> Tim Kozusko> J.y.n.x> ------------------------------------------------------------- -------------------> The evidence based on metallurgical analysis of fractured surfaces> (produced by Geller) indicates that a paranormal> inuence must have been operative in the> formation of the fractures.> Dr Wilbur Franklin (Physics Department,> Kent State University - U.S.A.)> We have observed certain phenomena with> ... -----------------> G=EMC^2 Glazier> CNote> malcolm burton> Greg Neill> ------------------------------------------------------------- -------------------> hey Im the one solving dillemmas round here, contribute> something useful yourself. its been well discussed b4 that> odd binaries are allowed here.> No.> formal rules such as attachments> only in binaries named groups are easily programmed into> news servers yet they dont actually exist do they?>> now which part of too bad didnt you get?> Herc> Welcome to the brackish depths of my killle, Sparky.> *plonk* ...problem solved.> -----------------> Rich Shewmaker> Ian> Greg Neill> scribe2b> ------------------------------------------------------------- plastic was held by the magnet with a small metal supercial insert> the bottom of the tear drop was the same proportions as the sphere, and in> all was much heavier> I was considering the possibility of the teardrop wabbling and causing> turbulence on its rst moments of fall, but wouldnt > some ideas :>> how is a plastic sphere being held by a magnet?>> the magnetic eld will remain after the voltage is cut> and different shapes and materials will drop sooner> the weight (also size) will inuence the fall, a teardrop> might be faster than a sphere in general but lighter> material will fall slower in atmosphere in general aswell>> maybe your teardrop doesnt maintain vertical at low speeds>> Herc>> Someone HAS to know whats going on.> If it helps, its not anything wrong with our particular apparatus because> similar phenomenon happened with other groups on different workstations.> Im doing a rst year university physics lab on free fall.. the apparatus> is simple. An electromagnet positioned above two, moveable, photogates. When> the voltage is cut, the object falls and the time interval between the two gates is measured on a timer accurate to 0.1 ms. We used two different> objects to examine their free fall, we used a plastic sphere and a> streamlined (i.e. looks like a teardrop) steel object.> We hypothesized that the plastic sphere would feel greater drag during its> fall, and would thus, take longer to fall.> However, when we performed the lab, we found that the plastic sphere was> consistently falling approximately 5-15 ms faster than the steel (for> heights of 50-200 cm).> When we tted the two sets of data our confusion grew as the plot for the> plastic sphere had a cubic term, representing the 1/3 the drag coefcient,> while the steel did not. (veried by chi-squared results).>> Can anyone explain this phenomenon?> -----------------> Shanx> Wanda> TheKid> raven1> ------------------------------------------------------------- -------------------> |-|erc:> The picture looks great. Interestingly, the ill-logic> of the basic design conguration> is more visible in the sketch> than in photos. The problems> would be that, numerous strong> points in the fuselage are needed> to support the heavy components> that are exterior to the main tank> and to resist and distribute high> local stresses that are due to the> high forces that are applied to the> structural components.> A vertical tower provides for the> inline stacking of components. The> symmetrical design of the USSR rocket> system would probably have far less> structural problems.> Neat sketch, however. The unit spacing worked just ne.> _________________________________>|-|erc:>Please advise us regarding the font setting that>>you used in making the picture. We can then select>>the same font in order to see the picture as you>>intended.>> this is the original ascii art, I altered it for my default outlook font,> but if you are adjusting font this one is better (any monospaced) :>> /> / > / > / > / > ^ /__________ ^> / | | / > /___| / |/___> | || / || |> | || /____ || |> | || // || |> | || //______ || |> | || | || | || |> |___|| | || | ||___|> | || | || | || |> | || | || | || |> | || | || | || |> | || | || | || |> | ||/| || ||| |> | |/ | || | | |> | / | || | |> | / | || | |> | / |___||___| |> |/ | | |> / |___||___| > (______|____||____|______)> /___ /_/||/_ /___>> // // > //// //// > //// // //// // > /// // // // //// //// / > / / ///// / //// // / // > / /// // // /// /// /// / // > / // / /// //// / //// // // > /// /// //// // // /////// > / // / ///// ///// / /// // / / > /// /// / /// / // //// // // / / > // // ///// /// // //// ///// // / > ////// //// // // /// // ////// // > /// / ///// / /// // // ///// /// / > -Jas> -----------------> Rust> Ralph Hertle> Mercury481> PlanetaryMatrix> ------------------------------------------------------------- -------------------> thats really good, if you have a spare 30 secs download it , hollywood> doesnt do much better.> I will humbly accept the compliment even though I am never humble.> http:www.sworld.org/artiv/fs.mpg is the next scene.> I was very surprised with how good it looked with almost no effort. Of course> comments on what is unrealistic are appreciated as I want to improve it.> -----------------> Tim Kozusko> Matt Giwer> Chris> Apostate ------------------------------------------------------------- -------------------> Yep, I took the data from both sources for 1 bar> -----------------> Someone> Lawrence & Bobbie> Odysseus> malcolm burton> ------------------------------------------------------------- -------------------> First of all I need a volunteer from the audience,>> give me the name of any rec newsgroup!>> Dont be shy, say you there, any rec newsgroup, step> right up and see real magic!>> Herc Youre doing it wrong.> Please immediately purchase a copy of Jim Cellinis DVD and view it ve times> in a row. (Potty breaks are allowed; its a long video.)> Itll be a good start for you. The rest of us are praying, burning white> candles, chanting, thinking positive thoughts, and/or invoking any number of> Chopra Quantum Whatevers to help bring about the release of volume two sometime> real soon now.> -----------------> Rust> John L> Greg Evans> Ben Sauvin> ------------------------------------------------------------- -------------------> Randi will test you when you properly apply to be tested. Sign up here:> http://www.randi.org/research/challenge.html> -----------------> Rich Shewmaker> CNote> Wanda> Rust> ------------------------------------------------------------- -------------------> It really all depends on the situation.> -----------------> Shanx> See You In Hell My Friend.> Someone> Greg Neill> ------------------------------------------------------------- -------------------> If ever I actually found myself in that situation, Id hold it upright,> with the intent of attacking my assailants knife hand.> -----------------> cliff86> Rust> Shanx> NormDePloom> ------------------------------------------------------------- -------------------> ANSWERS> 10 Scribe2b scribes> 11 Tim Kozusko mount koziosko> 12 CNote see note> 13 Greg display hurtle, hurtle into space> 16 Matt Giwer give away> 17 Someone one bar> 18 John L loo> 19 Rich Shewmaker rich showmaker> 20 See You In Hell My Friend. depends> 21 Rust attacks metal hi, Heric> I am just a beginner to Logic. I dont understand your purpose of the> quotes and what they mean.> Furthermore, I cannot open your link.> Again, anyone interested in this thread is welcome to refer to my last> post, I want to know about views from differet people.> the statement conveyed me> a very important idea, that is uniqeness, with it I dont think the> statement omit any crutial for the conclusion.> 3. I consider, since the statement tells the necklace is in either A,> or B, or C, then propositions A, B and C can neither be all ture nor> all false.> There must be one box containing the necklace mentioned, I> suppose(if not stated Box A, B or C, letters A, B and C mean> proposition A, B and C thereafter):> if its in Box A-> A(F), B(T), C(T)> if its in Box B-> A(T), B(F), C(T)> if its in Box C-> A(T), B(F), C(F) SO you can see, in the rst two possibilities, two T one F; the> last one shows the difference, two F one T. My conclusion is, the> necklace is in Box C.> I agree box C is a weak conclusion from your rst argument,I may have to retract that, uniqueness by itself is insufcient, and box B hasa unique label, hence the fact that box C entails the most number of falsestatements is not uniquely unique! There are 2 answers, immediately Bby being the odd one out, then C by uniqueness in quantity of conclusions,which leaves out A, making A the odd one out at a meta level!! Kind of like thoseIQ tests where one option is symmetric and one option is self similar and dependingon how the marker of the IQ test catagorises things changes the answer.On box A, it is written the necklace is not in this boxon box B, it is written the necklace is in box Aon box C, it is written the necklace is not in this boxMy test www.adamskingdom.com is just a different example of subjectivematerial that still has an explicit answer, site should be up, not directly relatedto your prob but my life depends on someone doing it.Herc =Divine test for you :*****************************> TTHTTTTHTTTTTTHTHHHTHHTTTHHHTHHHTHTHHHHTTHTHTTTTTH.>>Now if I can only get 1267650600228229401496703205376>people to read this and try it, one of them should>be convinced Im God.Ah, the reverse pyramid scam. Excellent use of it,by the way.I learned of this scam from a short story, Adam HadThree Brothers, by R. A. Lafferty. I recommendit, it is quite funny.-- -john***********************OK, youre a lecturer, I removed your name from the above post http://tinyurl.com/h9wdYou print it out together with 9 other posts to you (actually 9 replies to you to keep itconsistent, you didnt specically say Id be rich, the reply suggested it AND his namewas gamble which is just as good).THEN hand it out as a test to your students and they should select this post as being yours.To make it easier for skeptics to follow select ANY other 9 posts, students identify yourpost VS students identify your post to me. Have the questions :1/ Have you read any of usenet this year? [y/n]2/ Have you read any of sci.math or sci.logic this year? [y/n]3/ Which of the following posts do you think is a reply to David C Ullrich (quoted) [1/2/../10]HercOne sample point please!ps if youre after a sequence of words mortals wouldnt knowhttp://tinyurl.com/fuf8 predict Laurie Holden as the next Truman costar 2000http://tinyurl.com/gutr predict the Shuttle Tragedy date using word preminition one year beforehttp://tinyurl.com/h6uz sci.skeptic marking my 100% accurate psychic test calls me a private investigator =: There is plenty of arrogance in this thread thus far (including this: post), but let me assure you I detect none in the rst post. Since the original post, and the direct answers to it, are off the server,I will supply the answer, although no doubt I will only be echoing whathas already been said.It is true that quad- is a prex meaning four. So why wouldntquadratic mean the fourth power, triatic the third power, and biatic thesecond power?arithmetical terms. They were thought of in geometric terms. After all, wedo call them squares, and a square is a geometric shape.In fact, a square has four sides - and one old name for a square is a*quadrate*. So, quadratic means of or pertaining to squares. Also notethat the corners of squares are right angles, which split the circle intofour parts, or *quadrants*.A quadratic equation is also known as a second-degree equation, and acubic equation (of or pertaining to _cubes_, again geometric shapes) as athird-degree equation.John Savard Stan Browns post had an s in it (lower case), and Brian Chandlers> arguably showed disdain for mathematics. Neither of them was the> original poster; the original post came from Singapore, and neither> Browns or Chandlers did.I didnt think Brian Chandler was showing disdain for mathematics.It looked to me like he was joking and making an allusion to James Harris,who keepsfinding nonsensical (and nonexistent) gaps in mathematics.(One of the more amusing is James insistence that the ring of algebraicintegers is incomplete because hes found algebraic integers thatdont t the denition of algebraic integers! Its rather like saying,Ive found a tree thats yellow, has four wheels, and has School Buspainted on the sides, so the denition of tree must be incomplete!)-- Wayne Brown | When your tails in a crack, you improvisefwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock > Stan Browns post had an s in it (lower case), and Brian Chandlers>> arguably showed disdain for mathematics. Neither of them was the>> original poster; the original post came from Singapore, and neither>> Browns or Chandlers did.> I didnt think Brian Chandler was showing disdain for mathematics.> It looked to me like he was joking and making an allusion to James Harris,> who keepsfinding nonsensical (and nonexistent) gaps in mathematics.Oh, that might well be. Im an American, you know; we dontdo irony. > x^2 - (1+i)x + i = 0>> b^2 = (1+i)^2 = 2i> b^2-4ac = -2i> sqr(b^2-4ac) = 1-i>> x1 = [1+i + (1-i)]/2 = 1> x2 = [1+i - (1-i)]/2 = iThis equation works irrespective of what i is assumed to be. For example,take i=7, then the quadratic x^2 - 8x + 7does indeed have the solutions 1 and 7 as indicated above. Yet theintermediate steps in the working fail: (1+i)^2 = 2idoes NOT hold for i=7. So how come the right answer is still obtained?!?!? > x^2 - (1+i)x + i = 0>> b^2 = (1+i)^2 = 2i>> b^2-4ac = -2i>> sqr(b^2-4ac) = 1-i>> x1 = [1+i + (1-i)]/2 = 1>> x2 = [1+i - (1-i)]/2 = i> This equation works irrespective of what i is assumed to be. For example,> take i=7, then the quadratic> x^2 - 8x + 7> does indeed have the solutions 1 and 7 as indicated above. Yet the> intermediate steps in the working fail:> (1+i)^2 = 2i> does NOT hold for i=7. So how come the right answer is still obtained?!?!?False proofs of true things are permitted to have false steps in them.That doesnt invalidate the truth of the original statement.The discriminant = i^2 - 2*i + 1 = (1-i)^2 irrespective of i.Therefore its true if you plug in a certain numeric value for i. Thatcannot invalidate the truth of the statement for other different alues,even though the intermediate values in you calculation are now different.Phil > x^2 - (1+i)x + i = 0>> b^2 = (1+i)^2 = 2iSkip this step and let i be any constant. b^2-4ac = -2i(1+i)^2 - 4i = (1-i)^2 just by algebra(Note: if i = sqr -1, then (1-i)^2 = -2i)> sqr(b^2-4ac) = 1-iThus you got the same sqr just using algebra instead of the value of i.> x1 = [1+i + (1-i)]/2 = 1> x2 = [1+i - (1-i)]/2 = i>> This equation works irrespective of what i is assumed to be.Interesting observation. = I have development algorithm to solve the bin-packing problem usingthe simulated annealing. But I have found some unstable situations. Final solution should be stable after enough running time. But sometimesthe nal solutions cost function arrives at minimum, and sometimes thenal solution stays at a local minimum. Why? How to adjust the parameter of simulated annealing for a stablesolution?Jun Yang-- Serwis Usenet w portalu Gazeta.pl -> http://www.gazeta.pl/usenet/ I have development algorithm to solve the bin-packing problem using> the simulated annealing. But I have found some unstable situations.> Final solution should be stable after enough running time. But sometimes> the nal solutions cost function arrives at minimum, and sometimes the> nal solution stays at a local minimum.> Why? How to adjust the parameter of simulated annealing for a stable> solution?>> Jun Yang> Anything else would be a sensational result. Tofind a proven globaloptimum the only method is an exhaustive search through all possiblearrangements. It is known that there exists no _efcient_ algorithmto reach a solution that differs by less than 22% from the globalTo get more help it would be useful if you give more details aboutyour problem and the SA algorithm used.Hugo Pfoertnerhttp://www.pfoertner.org/ I have development algorithm to solve the bin-packing problem using> the simulated annealing. But I have found some unstable situations.>> Final solution should be stable after enough running time. But sometimes> the nal solutions cost function arrives at minimum, and sometimes the> nal solution stays at a local minimum.>> Why? How to adjust the parameter of simulated annealing for a stable> solution?Instead of running longer, cool the annealing slower.Thatll give you a better chance offinding a better local minimum.With better local minimums youll have better chance offinding, as theremay not be a unique minimum, a _minimal_ point. I have development algorithm to solve the bin-packing problem using> the simulated annealing. But I have found some unstable situations.>> Final solution should be stable after enough running time. But sometimes> the nal solutions cost function arrives at minimum, and sometimes the> nal solution stays at a local minimum.>> Why? How to adjust the parameter of simulated annealing for a stable> solution?> Instead of running longer, cool the annealing slower.> Thatll give you a better chance offinding a better local minimum.> With better local minimums youll have better chance offinding, as there> may not be a unique minimum, a _minimal_ point.diminishing results, do a set of runs and if there is a set of minimums thenits probable thats the solution.Herc > While surfing the Web, I stumbled upon the>> following site:>> http://www.mathpages.com/> Oooh, for:http://www.mathpages.com/home/quotes.htmxanthian.--