mm-47 === >Suppose that T1 and T2 are shift operators on R. Is it provable that>if there exists function K such that K o T1 = T2 o K then K is linear.>Hmm. This says that there exist constants a and b such that> K(x+a) = K(x) + b>for all x. This certainly does not imply that K is linear, (or even>affine, which is no doubt what you meant). For example,>the restriction of K to the interval [0,a) could be any function>whatever.If there are a and b so that for all x K(x+a) = K(x) + bthen K(x) - bx/a is periodic with period a. Thus, K is the sum of alinear function and a periodic function. This confirms the statementthat K can be anything on [0,a), but K on that interval plus one morepoint uniquely determines K.Rob Johnson take out the trash before replying === This post presents a topic from intuitionistic logic.The following statements provide some context for the longer excerptthat follows. All excerpts are from The Blackwell Guide toPhilosophical Logic.1)This shows that even a relatively simple theory likeequality is incomparably richer than the classical theory.2)The theory of equality has the usual axioms: reflexivity,symmetry, transitivity. One can strengthen the theory inmany ways, for example, the theory of stable equality isgiven byEQ^st = EQ + Axy(--x=y -> x=y)And the decidable theory of equality is axiomatized byEQ^dec = EQ + Axy(x=y / -x=y)[Aside: In deference to John Correy's treatment of identity andnon-self-identicals, I would note that treating this as an inclusivedisjunction admits the possibility of a superimposed interpretation.]3)Intuitionistic predicate logic itself is, just like itsclassical counterpart, undecidable. Fragments are,however, decidable. For example, the class ofprenex formulas is decidable, and, as a corollary,not every formula has a prenex normal form inIQC. On the other hand, although monadic predicatelogic is decidable in classical logic, Kripke showedit is undecidable in intuitionistic logic; see alsoOrekov et al. Likewise, Lifschitz showed that thetheory of equality is undecidable.The following is an excerpt about apartness. With respect to thestatements above, one thing to note is that stable equality is obtainedunder the apartness axioms.In intuitionistic mathematics, there is also a strongnotion of inequality: apartness, #, as mentioned above.This was introduced by Brouwer (1919) and axiomatizedby Heyting (1925). The axioms of AP are given byEQ and the following listAxyx'y'(x#y / x=x' / y=y' -> x'#y')Axy(x#y -> y#x)Axy(-x#y <-> x=y)Axyz(x#y -> x#z / y#z)The gluing technique will now be used to show thatAP has the disjunction and existence properties.---Aside:(DP) |- A / B => |- A or |- B(EP) |- ExA(x) => |- A(t) for a closed term t'|-' is interperted with respect to intuitionistic deduction(DP) is presented with a reductio ad absurdum argument;but, the author notes that there are proof-theoreticjustifications that invoke no such conflict with intuitionisticprinciples.---Let AP |- A / B and assume ~(AP |- A) and~(AP |- B) ['~' is classical exclusion negation].Then, by the strong completeness theorem, thereare models K_1 and K_2 of AP such that~(K_1 ||- A) and ~(K_2 ||- B). Consider thedisjoint union of K_1 and K_2 and place theone-point world k_0 below it. That is to say,designate points _a_ and _b_ in K_1 and K_2which are identified with the point k_0. The newmodel obviously satisfies the axioms of AP.Hence k_0 ||- A / B and so k_0 ||- A ork_0 ||- B. Both are impossible on the grounds ofthe choice of K_1 and K_2. Contradiction.Hence AP |- A or AP |- B.For EP, it is convenient to assume that the theoryhas a number of constants, say{c_i | _i_ e I}Now, let AP |- ExA(x) and ~(AP |- A(c_i)) for all_i_. Then for each _i_ there is a model K_i with~(K_i ||- A(c_i)). As above, the models K_i canbe glued by means of a bottom world K^* witha domain consisting of just the elements c_i. Nonon-trivial atoms are forced in k_0 (i.e., only thetrivial identities c_i = c_i). The identification of thec_j with elements in the models K_i is obvious.Again, it is easy to check that the new model satisfiesAP. Hence k_0 ||- ExA(x), i.e., k_0 ||- A(c_i) forsome _i_. But, then also, K_i ||- A(c_i), contradiction.Hence AP |- A(c_i) for some _i_.The gluing operation thus demonstrates that thereare interesting operations in Kripke model theorythat make no sense in traditional model theory.The apartness axioms have consequences for theequality relations. In particular, stable equality isobtained:--x=y -> x=yFor,-x#y <-> x=ysox=y <-> ---x#y <-> --x=yIndeed, the equality fragment itself is axiomatizedby an infinite set of quasi-stability axioms. Put-(x [=_0] y) =df -x=y-(x [=_(n+1)] y) =df Az(-(z [=_n] x) / -(z [=_n] y)For these 'approximations to apartness,' formulatequasi-stability axioms:S_n =df Axy(-( x [=_n] y) -> x=y)The S_n axiomatize the equality fragment of AP. Tobe precise: AP is conservative overEQ + {S_n | n >= 0}This shows that even a relatively simple theory likeequality is incomparably richer than the classical theory.Apartness and linear order are closely connected.The theory LO of linear order has axioms:Axyz(x x z -(x x#yOne can also usex x=y)Stability of equality never excited me.|In intuitionistic mathematics, there is also a strong|notion of inequality: apartness, #, as mentioned above.One might mention that for real numbers, r#s is defined to meanthe existence of a positive integer n such that |r-s|>1/n. Theequality r=s is equivalent to the negative of that, |r-s|<1/nfor every n. The negation, ~r=s, is in a sense very close toequivalent to r#s. If you prove ~r=s for specific r and s, there'sa metatheorem which guarantees you can prove r#s too. Markov's ruleentails ~r=s -> r#s, and the Markov school used this (although Iguess they noted when they did). Once you've concluded that ~r=s,it's tempting to say that it's just a matter of computing r and sprecisely enough to find the level of precision where they differ,to determine that r#s.Nevertheless, it's considered inappropriate in standard intuitionismshows that the nonexistence of an n for which |r-s|>1/n leads to acontradiction; it isn't taken as exhibiting the n.Mainly, I would say it's just better not to muck about with weirdlittle statements like ~r=s anyway. It's rare enough that I want tostate the negation of an equation like ~r=s that in my own notes Ijust write the usual inequality symbol r =/= s instead of r#s forapartness, and write ~r=s if I really mean the negation of equality.|This was introduced by Brouwer (1919) and axiomatized|by Heyting (1925). The axioms of AP are given by|EQ and the following list||Axyx'y'(x#y / x=x' / y=y' -> x'#y')||Axy(x#y -> y#x)||Axy(-x#y <-> x=y)||Axyz(x#y -> x#z / y#z)I like sometimes to deal with relations that satisfy all the axiomsof this definition except with Axy(~x#y <-> x=y) replaced byAxy(x=y->~x#y) (or in other words, Ax(-x#x)), and don't necessarilysatisfy Axy(~x#y->x=y).I'm not sure whether the condition Axy(~x#y->x=y) really should beincluded in the definition of apartness, actually. All it serves isto tie equality to apartness, basically by defining it to be thenegation of apartness. The first axiom you list (which would usuallybe considered to follow just by the rules of equality) follows fromthe others with Axy(x=y->~x#y) as follows. If x#y, then either x#x'or x'#y. If x'#y then x'#y' or y'#y, hence y#y'. So if x#y, theneither x'#y' or x#x' or y#y'. But by assumption, x=x' and y=y', soneither x#x' nor y#y', hence x'#y'.Here's an example of a case where one has a natural # relation thatsatisfies my weaker definition. I don't remember who discovered itfirst, but I independently rediscovered it later. Define multisetsof a set S to be equivalence classes of functions from other sets Tunder the equivalence that f1:T1->S and f2:T2->S are equivalent ifthere's a one-to-one correspondence g:T1->T2 such that f1 = f2 o g.Call a multiset a finite multiset if the domain is finite.Defining multisets as equivalence classes amounts to defining theequality relation for them. But is there an apartness relation forfinite multisets of a set S which itself has an apartness relation?The first discoverer was using the definition above, and stated hisresult as being that there is no apartness relation in general.But for finite multisets of a set with apartness, if we weaken thedefinition as I just described, there is one.As a preliminary step, it might help to prove a simple lemmageneralizing the last axiom above to finite multisets. If I haveelements x1,...,xn and y1,...,ym of a set with apartness, and ifx_i # y_j for every i=1,...,n and j=1,...,m, and z is some furtherelement, then either z#x1, z#x2, ..., z#xn, or z#y1, z#y2,..., z#ym.This is just applying the distributive law (A or B) and (A or C)<-> A or (B and C) a finite number of times, using the fact that bythe last axiom for each i and j, either x_i#z or z#y_j.[By the way, the finite distributive law (A or B1) and (A or B2)and ... and (A or Bn) <-> A or (B1 and B2 and ... and Bn) isconstructive. The infinite version {for every i>=0 (A or B_i)}<-> {A or for every i>=0 B_i} is not. In the finite case, we caneffectively get for each i=1,...,n the fact that A is true, or thefact that B_i is true, and when we're done, we've either found thatA is true, or for each one we've found that B_i is true. But that'snot an effective procedure in the infinite case.]So if we have x1,...,xn and y1,...,ym which are collectively apartfrom each other, any new z is apart from all of one of them, whichmeans it can be appended to the other one while preserving thesituation. By induction, if we have z1,...,zk, then there's apartition of z1,...,zk so that the x's together with one of thepartitions are collectively apart from the y's together with theother of the partitions.The case of n=m=1 is maybe the most interesting. If among someelements u1,...,ur in a set with apartness we find some apartnessrelation u1#u2, then we can partition the u1,...,ur into two, sothat u1 is in one partition, u2 is in the other one, and the elementsof one partition are all apart from the elements of the otherpartition.Okay, now back to finite multisets. If two multisets have differentnumbers of elements, then they are of course distinct, so we justneed to consider finite multisets defined by functions from a fixed{1,...,n} to S. Now let me motivate my # relation. Given twomultisets {x1,...,xn} and {y1,...,yn}, in order to have a distinctionbetween them, it's only natural that we should have a distinctionbetween some two of the elements. Hence there exists some way ofpartitioning x1,...,xn,y1,...,yn into submultisets that arecollectively apart from each other. I define {x1,...,xn}#{y1,...,yn}to mean that we can partition them in such a way that one of thepartitions has a different number of elements from the x's as fromthe y's. Equivalently, I define it to mean that there are subsetsi_1,...,i_k and j_1,...,j_l of {1,...,n} where k+l>n, with theproperty that for every s=1,...,k and t=1,...,l, we havex_i_s#y_j_l. (k+l=n+1 is enough.)For example, an unordered pair {x1,x2} is apart from anotherunordered pair {y1,y2} if either x1#y1 and x1#y2, or x2#y1 and x2#y2,or y1#x1 and y1#x2, or y2#x1 and y2#x2.The # relation satisfies all the axioms except ~x#y -> x=y. It'sobviously symmetric and preserved under substitution of equals. Iftwo multisets are equal, they are not #. If {x1,...,xn}#{y1,...,yn}and {z1,...,zn} is some other multiset, then the apartness relationsbetween x_i_1,x_i_2,...,x_i_k and y_j_1,...,y_j_l can be extended bythe lemma to include the z1,...,zn. Now either enough z's are putwith those x's to make {z1,...,zn}#{y1,...,yn}, or enough z's areput with those y's to make {x1,...,xn}#{z1,...,zn}. If >n-l of thez's are put on the left, that's enough to make the former true. Ifnot, then there are at least l of the z's on the right, which makes{x1,...,xn}#{z1,...,zn} true. Hence the last axiom also holds.The negation of #, however, is not constructively equivalent toequality for multisets. Equality is really a pretty strict relation;one has to be able to say which element corresponds to which. Ingeneral, for unordered pairs of real numbers, {x,-x} cannot be #{|x|,-|x|}. But in order to prove that {x,-x}={|x|,-|x|}, we haveto have either that x=|x| and -x=-|x|, or that x=-|x| and -x=|x|.The first case is equivalent to x>=0 and the second case isequivalent to x<=0. (For constructive purposes one defines x>=yfor real numbers x and y to mean that in the sequence of rationalapproximations to x and to y, the approximation to within 1/n of xis >= the approximation to within 1/n of y, minus 2/n. There isn'ta constructive equivalence between x>=0 and x>0 or x=0. There isan equivalence between x>=0 and not x<0.) That x>=0 or x<=0 foreach real number x is nonconstructive.That x>=0 or x<=0 for each real number is equivalent to the statementthat if s0,s1,... is a sequences of bits, i.e. elements of {0,1},then either it isn't the case that the first 1 is at an odd position,or it isn't the case that the first 1 is at an even position.Nonconstructively speaking, either (a) all the sequence s is 0, or(b) the first 1 occurs at an even index, or (c) the first 1 occurs atan odd index. In cases (a) and (c), one can say that for each even nsuch that s_n=1, there exists an odd m |- A or |- B||(EP) |- ExA(x) => |- A(t) for a closed term t||'|-' is interperted with respect to intuitionistic deduction||(DP) is presented with a reductio ad absurdum argument;|but, the author notes that there are proof-theoretic|justifications that invoke no such conflict with intuitionistic|principles.|---I would approach the use of Kripke models with some caution. Kripkemodel theory and topos theory are ways for people to keep assumingthe law of excluded middle, but simulate to some degree what it'slike not to assume it.So I'd take seriously all the distinctions between differenttypes of validity until I knew that there was a known equivalence.I suppose the sentences valid in one-node Kripke models are the sameas the ones that are just plain valid. Usually, though, peopleworking with Kripke model theory are assuming the law of excludedmiddle, which implies that all the sentences implied by it are validin one-node Kripke models. There are Kripke models in which thelaw of excluded middle is definitely not valid, but only someconstructivists actually assume that the law of excluded middle isnot valid, as opposed to failing to assume that it is valid.I saw a proof assuming the law of excluded middle that for eachstatement S in second order logic, there's a correspondingfirst-order statement S' such that S holds in all models if and onlyif S' holds in all Kripke models. So the set of statements valid forall Kripke models is somewhat complex.The situation is complicated by the fact that the moststraightforward ways of asserting a completeness theorem areincorrect for intuitionist logic. There's a formal system, yes, andit's got an associated recursively enumerable set of theorems. Butthe classical proof I mentioned in the last paragraph shows that theset of first-order sentences valid in all Kripke models is way toologically complex to be recursively enumerable. I've seen mentionedcertain completeness theorems, but I wouldn't say the situationseemed simple.[...]|The apartness axioms have consequences for the|equality relations. In particular, stable equality is|obtained:||--x=y -> x=y||For,||-x#y <-> x=y||so||x=y <-> ---x#y <-> --x=yYes, stability is just the same as saying that equality is negative,i.e. that equality is equivalent to the negation of something. Idon't know what good that is. The thing it's a negation of doesn'thave to be an apartness, and as far as I know there's no guaranteethat there is an apartness.Here's a demonstration that ~~X->X is equivalent to saying X is(equivalent to) the negation of something. (I write negationswith ~ instead of -.)First, a simple lemma. Any statement implies its double negation:X->~~X. If we assume X, assuming ~X leads to a contradiction, fromwhich we can infer ~~X.Triple negation is the same as single negation (a theorem ofBrouwer). If ~X, then by the lemma applied to ~X, we get ~~~X too.together with ~~~X that's a contradiction. Hence, on the assumptionof ~~~X, we get ~X.A statement X satisfying the stability condition ~~X->X is equivalentto ~~X because X->~~X is automatic. So if it satisfies stability,then it's equivalent to the negation of something (namely, ~X).Conversely if X is equivalent to ~Y, then ~~X is equivalent to~~~Y which implies ~Y, hence X.[Some stuff about the theory of linear orders omitted.]Keith Ramsay === : This post presents a topic from intuitionistic logic.'Nuff said.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === at 09:29 PM, maneesh@drunkenbastards.com (maneesh) said:>To elaborate, I often found it confusing to associate particular>properties, algebraic objects and other such things with the name>they are given.Learning a new language is always confusing.>Surely it only adds a level of complexity to a field of study which>does not need one!Surely not. The problem is that the concept is new, not that there issomething wrong with the name.>What are your opinions on having things called Schurian, Jacobian,>or Gaussian?It's shorter than descriptive names, and no less enlightening. >Do you feel that names of things in mathematics should greater>reflect information about the said things? Should the word red be red? Names are arbitrary. >Do you feel it retards mathematical progress (on a personal, or >collective level)? No.>What is the worst named object you can think of?Ideal.None of what you have written addresses the real problems inMathematical nomenclature; the large number of synonyms and theinconsistency in the notation.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === ----------------------------- <^> <()> <^> ----------------------------->What is the worst named object you can think of?> Ideal.>why ideal?I would say 'abstract' has the most number of meanings in mathematics.function abstraction, outline, define proposition, approximate, ....Herc X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === at 10:55 AM, Saab Siddiqui said:>when i show this to other people they say the same thing. no offense>ment here but really people always claim such but bring no proof.>what other text have parallels like this?Others have already answered your question. Basically, given anysufficiently large text the odds are astronomically in favor of suchcoincidences. It's just a question of picking out the ones thatsupport your case and ignoring the ones that don't.>and bible codes and panim dont count because it is different sort >of pattern.Why? What sort of pattern would count? Why should anybody botherproducing more examples if you're just going to say That one doesn'tcount - it's different?>i mean where words duplicate like in quran.Words duplicate in the Tanakh, and even in novels.>what is your faith?Judaism.>but what holy texts?Tanakh. Talmud.>could you show me similar parallels?You've already rejected them. Jewish mysticism often makes use ofGemmatria, which is basically what you are doing. It's not Science andit's certainly not Mathematics.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > at 10:55 AM, Saab Siddiqui said:>i mean where words duplicate like in quran.>Words duplicate in the Tanakh, and even in novels.I happened to have the text of Moby Dick, and did some word counts.COFFIN, SHARKS, MAN'S, REST and BONE all occur 51 times.STERN, INDIAN, SPEAK, SLOWLY, and SAVAGE all occur 52 times.KING, ROPE and IVORY all occur 56 times.BOOK, SHORT, and LIVE all occur 60 times.ORDER, REASON and LORD all occur 64 timesFISHERY and WORK both occur 65 timesWHALEMEN and MYSELF both occur 69 timesFIRE and HARPOON both occur 76 timesMIND and SOUL both occur 80 timesCERTAIN, GOING, LEG and DEATH all occur 89 timesBECAUSE, LEVIATHAN and DEAD all occur 92 timesIf that doesn't convince you that Moby Dick was divinely inspired,I suppose nothing will.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > If that doesn't convince you that Moby Dick> was divinely inspired, I suppose nothing will.LOLDouble-LOLDouble-LOL, squared.(BTW: The Pythagoras confusion waits for some enlightenment)Rainer Rosenthalr.rosenthal@web.de === I have a problem. I don't see how the author finds the closed formexpression for the polynomials P_n(lambda) in the book by Akhiezer --'The Classical Moment Problem' at page 3. I have scanned the first 5pages (although one only needs 1,2,3 to get the backgroundinformation):http://hem.bredband.net/lukhor/moment/ Happy new year! === > I have a problem. I don't see how the author finds the closed form> expression for the polynomials P_n(lambda) in the book by Akhiezer --> 'The Classical Moment Problem' at page 3. I have scanned the first 5> pages (although one only needs 1,2,3 to get the background> information):> http://hem.bredband.net/lukhor/moment/> Happy new year!First, a quick summary for those who don't have the relevant pagesin front of them. Consider the space of polynomials R(lambda) = p_0 + p_1 lamba + p_2 lambda^2 + ... + p_n lambda^n.We consider a linear functional S defined by a sequence of numberss_0, s_1, ..., defined by S[R] = s_0 p_0 + s_1 p_1 + ....where the sum is finite because the polynomial has only a finite numberof non-zero coefficients. Now we can define a bilinear form, givenpolynomials R_1 and R_2, as S[ R_1 R_2 ]; that is, we muliply thepolynomials in the usual sense, then apply the linear functionalS[]. The coefficients s_0,..., are choosen so that the form ispositive definite.Now the author wishes to choose an orthogonal basis, and givesthe following explicit formulae: P_0 (lambda) = 1 P_n (lambda) = 1/ sqrt{ D_{n-1} D_{n} } times | s_0 s_1 ... s_n | | s_1 s_2 ... s_{n+1} | | ... | | s_{n-1} s_{n} ... s_{2n-1} | | 1 lambda ... lambda^n |where D_{m} is defined as the determinant of the matrix | s_0 s_1 ... s_m | | s_1 s_2 ... s_{m+1} | | ... | | s_{m} s_{m+1} ... s_{2m} |.Now the author notes that P_n is of degree 'n' (as is clear byexpanding the matrix using the bottom row), so that toprove orthogonality (leaving aside normalization for the moment) itis sufficient to prove that P_n is orthogonal to 1, lambda, lambda^2,..., lambda^{n-1}. Agreed?Now what happens when we multiply P_n by lambda^m? This isgiven on page 4 [let us call this equation#1], P_n (lambda) lambda^m = 1/ sqrt{ D_{n-1} D_{n} } times | s_0 s_1 ... s_n | | s_1 s_2 ... s_{n+1} | | ... | | s_{n-1} s_{n} ... s_{2n-1} | | lambda^m lambda^{m+1} ... lambda^{m+n} |Note: if we expand the determinant along using the last row, wewould get P_n(lambda) lambda^n = a polynominal in termsof lambda^{m}...lambda^{m+n} where the coefficients arethe minors of the above matix (with appropriate signs).The author then asks the reader to apply the functional S[] toboth sides of it, so that the left side becomes the inner productof P_n(lambda) and lambda^m, so that we obtain [this will beour equation#2]S[ P_n (lambda) lambda^m ] = 1/ sqrt{ D_{n-1} D_{n} } times | s_0 s_1 ... s_n | | s_1 s_2 ... s_{n+1} | | ... | | s_{n-1} s_{n} ... s_{2n-1} | | s_{m} s_{m+1} ... s_{m+n} |.To see this equalility, consider the expansion of this matrixand the previous matrix (from equation#1 above) using the lastrow--the minors are the same in both instances (and contain onlys_i's, no lambda's!) and we simply have replaced lambda^m withs_m, lambda^{m+1} with s_{m+1}, etc. Now, if m at 05:28 AM, victorfrankenstein2@juno.com (Benjamin) said:>Is Apostol from the baby boom generation or is he a generation Xer?>purchased my copy in the early 1960s.>-- > Shmuel (Seymour J.) Metz, SysProg and JOAT>not reply to spamtrap@library.lspace.orgI thought so.....Apostol must be from the WWII generation if he got his degrees in the mid to late 40's. So, he didn't grow up on the new math stuff. I wonder if he got to be in the Battle of the Bulge or if he was in Iwo Jima or the Battle of Midway. Hmmm...or do you think he was in the Bataan Death March? So, the best calculus books seem to be those of Apostol, Courant, and Spivak.Does anyone know of the best biostatistics book? It seems all the ones I've read have sucked eggs. . . . === Apostol is great, uses Liebniz' method. I've also heard thatJ.Bernoulli's text is excellent (see www.wlym.com .-) > So, the best calculus books seem to be those of Apostol, Courant, and Spivak.--Give the Gift of Trickier Dick Cheeny -- out of office at last!http://laroucehin2004.com === How many digits of pi have you memorized? I did 100. The world record === >How many digits of pi have you memorized?25, when I was ten or eleven. I had a book about maththat in a kind of decorative way had pi written to 25decimal places with ... at the top of a page. I didn'tbother to go past that point. Probably I would have ifmy friend Tom, who memorized somewhat fewer, hadgotten to 25.A few years later, another kid in the same school systemwas identified as being gifted mathematically for havingmemorized thirty-some digits of pi.Keith Ramsay === > How many digits of pi have you memorized? I did 100. The world record === >How many digits of pi have you memorized? I did 100. The world recordI have no idea if it's true or just an urban legend but: Reporter: Dr Einstein, how many digits of pi have you memorized? Dr E: Four, I think. R: Four! But sir, many people have memorized dozens, evenhundreds of digits. Dr E: Yes, but I know where to look it up if I need more. === >How many digits of pi have you memorized? I did 100. The world record> I have no idea if it's true or just an urban legend but:> Reporter: Dr Einstein, how many digits of pi have you memorized?> Dr E: Four, I think.> R: Four! But sir, many people have memorized dozens, even> hundreds of digits.> Dr E: Yes, but I know where to look it up if I need more.I took a course once that gave this story.Dr E was asked why he didn't carry a Franklin Planner to write down all ofhis wonderful thoughts and ideas.Dr E replied: I have had two great thoughts in my life and remember themboth.Not sure if it is true. === How many digits of pi have you memorized? I did 100. The world record> I have no idea if it's true or just an urban legend but:> Reporter: Dr Einstein, how many digits of pi have you memorized?> Dr E: Four, I think.> R: Four! But sir, many people have memorized dozens, even> hundreds of digits.> Dr E: Yes, but I know where to look it up if I need more.> Did Einstein really memorize only four?? === : Did Einstein really memorize only four??Why would that be suprising? Memorizing them is essentially useless.Justin === > How many digits of pi have you memorized? I did 100. The world recordEach word in this corresponds to a digit of pi. O means zero.Why, π! Stop, π! Weird anomalies do behave badly!You, madly conjured, imperfect, strange, numerical,Why do you maintain this facade?In finite time you are barbaric!You do wonders, mesmerize minds!O, do elements numerous have a beautiful meaning- A system isolating all mysteries, solutions for puzzles, chaos, aO snafu apparent in O Universal Concept from believing lies?That there, obstinate in you, O Strange Constant, A Divine Sign O exists is unlikely unlessIs O revealed Something Brilliant, negating belief!In formulas, O, you show yourself in Greek and math as a π forever--O hidden wonders absconded, infinite, in a tiny constant, O, sneakily, rather?Never, I say! === > How many digits of pi have you memorized? I did 100. The world record> Each word in this corresponds to a digit of pi. O means zero.> Why, π! Stop, π! Weird anomalies do behave badly!Sorry, something seems to have gone wrong with my newsreader. What doesπ! stand for please? The context requires a single letter, but none ofa, I or O make sense.--Paul V. S. TownsendInterchange the alphabetic elements to reply === > How many digits of pi have you memorized? I did 100. The world record> Each word in this corresponds to a digit of pi. O means zero.> Why, π! Stop, π! Weird anomalies do behave badly!> Sorry, something seems to have gone wrong with my newsreader. What does> π! stand for please? The context requires a single letter, but none of> a, I or O make sense.It was a pi symbol. Go to http://members.aol.com/loosetooth/poem.htmlto see the poem. === >How many digits of pi have you memorized? I did 100. The world recordI've memorized 100,000 digits. They're all 3. Of course I haven'tmemorized exactly where they occur.(Oh, you meant _consecutive_ digits? Never mind...)************************David C. Ullrich === >How many digits of pi have you memorized? I did 100. The world record>I've memorized 100,000 digits. They're all 3. Of course I haven't>memorized exactly where they occur.I've memorized infinitely many digits, and they're all the same!Of course I haven't memorized either exactly where they occur,or even which digit they all are. Lee Rudolph === >How many digits of pi have you memorized? I did 100. The world record> I've memorized 100,000 digits. They're all 3. Of course I haven't> memorized exactly where they occur.> (Oh, you meant _consecutive_ digits? Never mind...)LOL! === : I've memorized 100,000 digits. They're all 3. Of course I haven't: memorized exactly where they occur.: (Oh, you meant _consecutive_ digits? Never mind...)Well, in that vein, I've memorized all of them. They are 0,1,2,3,4,5,6,7,8 and 9. If only I could get the order right...Justin === ----------------------------- <^> <()> <^> -----------------------------> : I've memorized 100,000 digits. They're all 3. Of course I haven't> : memorized exactly where they occur.> : (Oh, you meant _consecutive_ digits? Never mind...)> Well, in that vein, I've memorized all of them.> They are 0,1,2,3,4,5,6,7,8 and 9.> If only I could get the order right...>Reminds me of the time I drove though the city of Melbourne and gotevery green light!Herchad to wait for some... === > : I've memorized 100,000 digits. They're all 3. Of course I haven't> : memorized exactly where they occur.> : (Oh, you meant _consecutive_ digits? Never mind...)> Well, in that vein, I've memorized all of them.> They are 0,1,2,3,4,5,6,7,8 and 9.> If only I could get the order right...That does not quite work out as well as David's example since there is anunending number of non repeating decimals and thus would imply that you haveinfinite storage capability, which even the Universe does not have. === >: I've memorized 100,000 digits. They're all 3. Of course I haven't>: memorized exactly where they occur.>: (Oh, you meant _consecutive_ digits? Never mind...)>Well, in that vein, I've memorized all of them. >They are 0,1,2,3,4,5,6,7,8 and 9. That's very impressive. I always forget the 8.>If only I could get the order right...>Justin************************David C. Ullrich === 3.14159265358979323846264338327950288419716939937510....It's a long story how it happened...(well, it's actually very short, but as if anyone would like to listen to it and believe it!!!) === >: I've memorized 100,000 digits. They're all 3. Of course I haven't>: memorized exactly where they occur.>: (Oh, you meant _consecutive_ digits? Never mind...)>Well, in that vein, I've memorized all of them. >They are 0,1,2,3,4,5,6,7,8 and 9. >That's very impressive. I always forget the 8. I used to work with a guy who had spent a coupleof months of concentrated software development time whichmeant that he had to think in octal about eight hours aday, six days a week (that doesn't include the time spentdreaming while sleeping). Then he balanced his checkbook.After a couple hundred heart palpitations, he verified thathe hadn't written the checks using octal./BAH === > I used to work with a guy who had spent a couple> of months of concentrated software development time which> meant that he had to think in octal about eight hours a> day, six days a week (that doesn't include the time spent> dreaming while sleeping). Then he balanced his checkbook.> After a couple hundred heart palpitations, he verified that> he hadn't written the checks using octal.Me too, caused a few riots keeping score for darts at the pub in theevenings. No, sorry, treble 17 isn't 55 is it : )--Paul V. S. TownsendInterchange the alphabetic elements to reply === > I used to work with a guy who had spent a couple> of months of concentrated software development time which> meant that he had to think in octal about eight hours a> day, six days a week (that doesn't include the time spent> dreaming while sleeping). Then he balanced his checkbook.> After a couple hundred heart palpitations, he verified that> he hadn't written the checks using octal.>Me too, caused a few riots keeping score for darts at the pub in the>evenings. No, sorry, treble 17 isn't 55 is it : )Unless you played darts with like-minded friends. They wouldn't havebatted an eye but would have done the conversion automatically./BAH === I learned 500, but now, I have probably forgotten most ofthem(It was some years ago)./Anders> How many digits of pi have you memorized? I did 100. Theworld record === I memorized 27 digits when I was 12, and since then I've not forgottenthem nor have I memorized any more.What might be more interesting is the number of digits someone may havememorized such that any particular digit can be recalled simply from itsplace. As someone said, what's digit 77? If you know 100 digits and canrecall each one by place, that's impressive! =)Justin: How many digits of pi have you memorized? I did 100. The world record === Hey, I need a drink, alcoholic of course, after the heavy lectures on quantummechanics...3 1 4 1 5 9 2 6 5 3 5 8 2 7 9....There is some chance I mishandled the quote.G C === > Hey, I need a drink, alcoholic of course, after the heavy lectures onquantum> mechanics...> 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9....> There is some chance I mishandled the quote.I can remember that there was a doggerel verse of seven or eight lines,which (by the same scheme) gave pi to 30 decimals, but I can only rememberthe first line: Now I, even I, would celebrate.Back in my student days, some students tried to devise mnemonics for pi ande. One mnemonic for e neatly avoided the charge that such sentences werecomplicated, tedious or full of obscurities: In seeking a mnemonic, we composed a sentense of sensible words.Not enough entries were received for the first prize of a bottle of wine tobe awarded, but one of the students received a Mars bar as a consolationprize for this one for pi: Buy a Mars a month. Chocolate is seldom cheap but costly.--Paul V. S. TownsendInterchange the alphabetic elements to reply === > I can remember that there was a doggerel verse of seven or eight lines,> which (by the same scheme) gave pi to 30 decimals, but I can only remember> the first line: Now I, even I, would celebrate.Found it, ascribed to a certain Adam C. Orr of Chicago in1906.Now I, even I, would celebrateIn rhymes unapt, the greatImmortal Syracusan rivalled nevermoreWho in his wondrous lorePassed on before,Left men his guidance howTo circles mensurate.(Should it be inept in line 2? Who is the guy from upstate New Yorkreferred to in line 3?)--Paul V. S. TownsendInterchange the alphabetic elements to reply === : Now I, even I, would celebrate: In rhymes unapt, the great...: (Should it be inept in line 2? Who is the guy from upstate New York: referred to in line 3?)Depends upon whether you want to say unapt or inept. The rhyme *is* awfully unapt. I would prefer:-----------------------------Can I givea digit numericalof number whichcan ratiodiameter-perimeter divulge?beautiful 'tis, we say.-----------------------------Justin === > Hey, I need a drink, alcoholic of course, after the heavy lectures on>quantum> mechanics...> 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9....> There is some chance I mishandled the quote.>I can remember that there was a doggerel verse of seven or eight lines,>which (by the same scheme) gave pi to 30 decimals, but I can only remember>the first line: Now I, even I, would celebrate.>Back in my student days, some students tried to devise mnemonics for pi and>e. One mnemonic for e neatly avoided the charge that such sentences were>complicated, tedious or full of obscurities:> In seeking a mnemonic, we composed a sentense of sensible words.>Not enough entries were received for the first prize of a bottle of wine to>be awarded, but one of the students received a Mars bar as a consolation>prize for this one for pi:> Buy a Mars a month. Chocolate is seldom cheap but costly.Good grief. It's easier just to remember the numbers rather thanhave call a conversion routine for each word. That's a waste ofbrainpower plus you have to remember where you were in thesentence./BAH === In sci.math, jmfbahciv@aol.com<3fef0112$0$4765$61fed72c@ news.rcn.com>:> Hey, I need a drink, alcoholic of course, after the heavy lectures on>quantum> mechanics...> 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9....> There is some chance I mishandled the quote.>I can remember that there was a doggerel verse of seven or eight lines,>which (by the same scheme) gave pi to 30 decimals, but I can only remember>the first line: Now I, even I, would celebrate.>Back in my student days, some students tried to devise mnemonics for pi > and>e. One mnemonic for e neatly avoided the charge that such sentences were>complicated, tedious or full of obscurities:> In seeking a mnemonic, we composed a sentense of sensible words.>Not enough entries were received for the first prize of a bottle of wine > to>be awarded, but one of the students received a Mars bar as a consolation>prize for this one for pi:> Buy a Mars a month. Chocolate is seldom cheap but costly.> Good grief. It's easier just to remember the numbers rather than> have call a conversion routine for each word. That's a waste of> brainpower plus you have to remember where you were in the> sentence.3.14159,Oh the digits are sublime,2653589,More and more come all the time,7932384,C'mon man let's do some more,62648338,Wow! This pi thing sure is great!:-)> /BAH> -- #191, ewill3@earthlink.net -- insert random weird poetry hereIt's still legal to go .sigless. === >Hey, I need a drink, alcoholic of course, after the heavy lectures on quantum>mechanics...>3 1 4 1 5 9 2 6 5 3 5 8 2 7 9....>There is some chance I mishandled the quote.The third-to-last digit above should be a 9, not a 2. Could it be that itwas supposed to have been heavy lectures regarding quantum mechanics? -- Erick === ----------------------------- <^> <()> <^> ----------------------------->Hey, I need a drink, alcoholic of course, after the heavy lectures on quantum>mechanics...>3 1 4 1 5 9 2 6 5 3 5 8 2 7 9....>There is some chance I mishandled the quote.> The third-to-last digit above should be a 9, not a 2. Could it be that it> was supposed to have been heavy lectures regarding quantum mechanics?>you mean 3 1 4 1 5 9 2 6 5 3 5 8 9 7 9....I thought 3 1 4 1 5 9 2 6 5 3 5 8 2 7 9 7... there's a ACBC around thereHerc === ----------------------------- <^> <()> <^> -----------------------------> How many digits of pi have you memorized? I did 100. The world record3.1415926536not sure with the rounding I'll pass the Dalek Pyramid test with theelectrocuting floor of 10 X 10 tiles, each row only the pi digit is safe.What's digit 77?Herc === In sci.math, |-|erc ----------------------------- <^> <()> <^> -----------------------------> How many digits of pi have you memorized? I did 100. The world record> 3.1415926536> not sure with the rounding I'll pass the Dalek Pyramid test with the> electrocuting floor of 10 X 10 tiles, each row only the pi digit is safe.> What's digit 77?> Herc> I sure hope you've got your sums right. -- Teegan Jovanka (Janet Fielding):-)-- #191, ewill3@earthlink.net -- who?It's still legal to go .sigless. === > How many digits of pi have you memorized? I did 100. The world recordI think you've got a real shot at the record. You're only 42,095 behind. === I did one symbol. Does that count? No actually I could never do morethan 12 digits before running out of interest. === Something like this:Hey, I need a drink, alcoholic of course.....I forgot the rest; was it:Hey, I need a drink, alcoholic of course, after the (lectures on quantummechanics...)Count the letters in each word:3 1 4 1 5 9 2 6 5 3 ...(.....?)G C === the ratio of the diameter of sphereto its circumference is greater than three! on teh other hand,the ratio of the area of the equatorial circle of a sphereto its spherical surface is 1/4.> How many digits of pi have you memorized? I did 100. The world record--ils duces d'Enron!http://laroucehin2004.com === > How many digits of pi have you memorized? I did 100. The world record3.1415926535Now... check this out.Assume that the Earth is a sphere. One can calculate the circumferenceusing 3.14159 to within an order of a meter. Each extra decimal reduces theamount that you are off by an order of magnitude.Now, let us assume that the galaxy is a disk 150000 LY in diameter (75000 LYradius)This is a radius of ~ 7E20 meters.With pi to 100 digits this will give the circumference to within some 1E-80meters of exact.Do you need to know pi to that precision?No. === In sci.math, Michael Varney:> How many digits of pi have you memorized? I did 100. The world record> 3.1415926535> Now... check this out.> Assume that the Earth is a sphere. One can calculate the circumference> using 3.14159 to within an order of a meter. Each extra decimal reduces the> amount that you are off by an order of magnitude.> Now, let us assume that the galaxy is a disk 150000 LY in diameter (75000 LY> radius)> This is a radius of ~ 7E20 meters.> With pi to 100 digits this will give the circumference to within some 1E-80> meters of exact.> Do you need to know pi to that precision?> No.> Probably not, but it's handy for checking out new processor units. :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > How many digits of pi have you memorized? I did 100. The world record3.141592Six digits.And I think it's enough :)Spider === >The world recordAlways avoid the use of goto! === >The world record> Always avoid the use of goto!*groan* === Spider> How many digits of pi have you memorized? I did 100. The world record> 3.141592> Six digits.> And I think it's enough :)> SpiderThe digits are the number of letters in the words of that well-known saying,How I need a drink, alcoholic of course, after the ....LH === > ----------------------------- <^> <()><^> -----------------------------> Skeptic organisations are not interested in the scientificinvestigation> of> the paranormal.> Skeptic organisations are interested in shutting down businesses thatmake> paranormal claims.> It doesn't matter what evidence or results the business is getting,simply> having> a paranormal claim is enough for the 1000s of skeptic organisations to> take> action to close them down.> Thousands? Name 60 please. Or did you make that up?> Atleast one in every state of Australia, pop < 20 million> http://www.skeptics.com.au/about/contact.htm> State & Regional Branches> New South Wales :> Victoria :> South Australia :> ACT (Australian Capital Territory):> Western Australia :> Northern Territory :> Queensland :> Gold Coast (Queensland)> Gold Fields (Ballarat, Victoria) :> Borderline (Mitta Mitta, Albury, Wodonga):> Hunter Valley (NSW) :> TasmaniaThat's less than thousands... Rather a lot really. Guess it was made up. === ----------------------------- <^> <()> <^> -----------------------------> ----------------------------- <^> <()> <^> -----------------------------> Skeptic organisations are not interested in the scientific> investigation> of> the paranormal.> Skeptic organisations are interested in shutting down businesses that> make> paranormal claims.> It doesn't matter what evidence or results the business is getting,> simply> having> a paranormal claim is enough for the 1000s of skeptic organisations to> take> action to close them down.> Thousands? Name 60 please. Or did you make that up?> Atleast one in every state of Australia, pop < 20 million> http://www.skeptics.com.au/about/contact.htm> State & Regional Branches> New South Wales :> Victoria :> South Australia :> ACT (Australian Capital Territory):> Western Australia :> Northern Territory :> Queensland :> Gold Coast (Queensland)> Gold Fields (Ballarat, Victoria) :> Borderline (Mitta Mitta, Albury, Wodonga):> Hunter Valley (NSW) :> Tasmania> That's less than thousands... Rather a lot really. Guess it was made up.>1 skeptic organisation per million capita in our country.extrapolate to 1 billion in western civilisation.Herc === In sci.math, Dik T. Winter:> In sci.math, Dik T. Winter> :> ...> A lot of conclusions are drawn from this factorisation, it is however> the consequence of another factorisation. Set y = 49x, we get:> P(y) = 125 y^3 - 375 y^2 - 360 y + 1078 => = (5 a_1(y) + 7)(5 a_2(y) + 7)(5 a_3(y) + 7)> where the a's are roots of:> a^3 + 3(-1 + y)a^2 - (y^3 - 3y^2 + 3y)> So although the constant term of two of the factors are divisible by> 7, and the third is co-prime to it (it is 22), in general *none* of the> factors are divisible by 7, as P(y) is only divisible by 7 in a limited> number of cases. So I am still wondering what JSH is trying to show> with his constant terms.> So am I. I'd be curious to figure out when a_1(x)/7 and a_2(x)/7> are algebraic integers, but admittedly it's not a priority.> You first are required to show what a_1, a_2 and a_3 are. Luckily that> is possible... (*)Yes, it is possible. However, I don't have to compute themexplicitly; I merely need show that the defining equationfor a_1(x)/7 etc. is not of the requisite form for most x.Then again, your way might be slightly cleaner, and you alsohave a constructive proof, whereas I merely have an existance proof.> It's> clear that generally speaking, they are not, although> a_1(0) = 0, a_2(0) = 0, a_3(0) = 3.> Your manipulation is interesting and simplifies the problem> considerably. :-) However, now one has to deal with y being> a multiple of 49, if x was originally an algebraic integer.> But it also shows that the factors are not divisible by 7 for *all* y,> so it shows that divisibility properties are erratic.Aye!> (I'm also curious as to the rest of his proof; this is> only a small snippet thereof. It's a bit like examining> a small area (sans the actual hole) of a flat tire to try> to figure out why the car won't move.)> The rest of his proof hinges on the fact that exactly two factors are> (FLT proof) or should be (definition error) divisible by 7.Not to mention that he assumes the factors are algebraic integers at all.After all, 1 = 1/49 * 49, which means one can divide 1 by 49. Butit wouldn't mean much. :-)> ----> (*) A very nice showing of that I found while looking around at the> solutions of cubic equations. All expositions I have seen miss a very> basic fact (I think), but the exposition at mathworld is closest (this> Let's calculate the roots of z^3 + a.z^2 + b.z + c = 0. Let us have> the following definitions:> Q = (12.b - 4.a^2)> R = 36.a.b - 108.c - 8.a^3> K1 = cbrt(R + sqrt(Q^3 + R^2))/2> K2 = cbrt(R - sqrt(Q^3 + R^2))/2> W = (-1 + i.sqrt(3))/2, W^2 = (-1 -i.sqrt(3))/2, W^3 = 1> then we have the following roots:> z_1 = (-a + W .K1 + W^2.K2)/3> z_2 = (-a + W^2.K1 + W .K2)/3> z_3 = (-a + K1 + K2)/3> filling in all stuff (and hoping I did not make a basic arithmetic> mistace), we find that the z's correspondend to James' a's in order.> The beauty of this presentation is (I think) how three cubic roots> of two different values are used.That's a nice distillery of the problem, admittedly.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === In sci.math, Virgil Hancher [...]> Jim, I'm going on a break, the discussion is over. You are > sitting at a comfy computer and you are free to do what you > want. I am at my computer because I am fighting for my life > while I am being tortured by a spy satellite for 2 years > continuously. it is the most hideous torture in all history, > I wish you would believe me, even give me a few hours benefit > of doubt because that is all it would take to confirm my > admitedly odd story. Don't watch your TV, its a lie.> You can get relief by wearing a hat of aluminium foil, or by blowing > your brains out.The first is probably easier on the brains...http://zapatopi.net/afdb.html:-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === ...> I have my program looking at n=10 now; it has run nearly 7 hours> already (in contrast to half-second, 3-second, and 4-minute timings> at n=7,8,9 on a 750MHz pentium) but hasn't found any solutions yet.That program eventually finished, after 71.6 hours cpu hours. It found the same solution set as the new program found in 15 minutes on my 450 MHz AMD-K6. Here is some output after an hour of processing at n=11 via program at http://pat7.com/jp/perp11b.c -- 52721211321 229611 29452737924 171618 74143477849 272293 25448863729 159527 12386799616 111296 27487318849 165793 13769144964 117342 17739842481 133191 39876894864 199692 22421468644 149738 14996941444 122462 === > ... Here is some output after an hour of processing> at n=11 via program at http://pat7.com/jp/perp11b.c --...The program finished in 6.8 hours and found 8 n=11 perplexes - 229611 171618 272293 159527 111296 165793 117342 133191 199692 149738 122462 290369 218896 198022 105629 212771 121426 234925 261263 135884 248686 128108 149119 156904 163332 178304 262157 219846 276441 255465 110665 247415 128334 271891 194623 314239 165758 220585 266791 106838 230093 284873 286815 138366 195396 284891 108746 276067 304178 242266 314619 247559 297568 134865 248384 105904 117285 167085 123039 236184 259535 240828 267865 149192 110568 255962 203276 132044 186966 153381 125121 127563 178427 165876 138264 271228 258093 287373 168632 233865 289816 190367 166839 212884 108785 119528 149192 307521For example, solution #6 squares out as: 11215657216 105904 13755771225 117285 27917397225 167085 15138595521 123039 55782881856 236184 67358416225 259535 57998125584 240828 71751658225 267865 22258252864 149192 12225282624 110568 65516545444 255962 === > ... Here is some output after an hour of processing> at n=11 via program at http://pat7.com/jp/perp11b.c --> ...> The program finished in 6.8 hours and found 8 n=11 perplexes -> 229611 171618 272293 159527 111296 165793 117342 133191 199692 149738 122462> ...[ I cancelled a reply I made, re N=12. The cancelled message that everyone should ignore === |In Grothendieck's EGA(Elements de Geometrie Algebrique) I (1960)|Proposition (9.5.1) p.176 states;|Let f:X --> Y be a morphism of schemes such that|f_*(O_X) is quasi-coherent sheaf of O_Y module.|Then there exists a closed subscheme Y' of Y with|the following property.|f splits into X --> Y' --> Y and Y' is the smallest|closed subscheme of Y with this property.||However, I think this proposition holds without the condition|on O_X. Am I missing here?Do you have an argument for the result without the condition,or are you just unable to come up with a counterexample?Perhaps it's to prevent situations where Y is not a separatedscheme?Keith Ramsay === > |In Grothendieck's EGA(Elements de Geometrie Algebrique) I (1960)> |Proposition (9.5.1) p.176 states;> |Let f:X --> Y be a morphism of schemes such that> |f_*(O_X) is quasi-coherent sheaf of O_Y module.> |Then there exists a closed subscheme Y' of Y with> |the following property.> |f splits into X --> Y' --> Y and Y' is the smallest> |closed subscheme of Y with this property.> |> |However, I think this proposition holds without the condition> |on O_X. Am I missing here?> Do you have an argument for the result without the condition,> or are you just unable to come up with a counterexample?I think I have a proof for the proposition without the condition. I posted the sketch of my proof. However, I'm not 100% sure of my proof, and EGA has authority... > Perhaps it's to prevent situations where Y is not a separated> scheme?It's unlikely since Grothendieck was trying hard to avoidunnecessary conditions on his results.Nobuo Saito === > In Grothendieck's EGA(Elements de Geometrie Algebrique) I (1960)> Proposition (9.5.1) p.176 states;> Let f:X --> Y be a morphism of schemes such that> f_*(O_X) is quasi-coherent sheaf of O_Y module.> Then there exists a closed subscheme Y' of Y with> the following property.> f splits into X --> Y' --> Y and Y' is the smallest> closed subscheme of Y with this property.> However, I think this proposition holds without the condition> on O_X. Am I missing here?My proof(sketch) of the proposition (9.5.1)If Y is an affine scheme Spec(A), then f: X --> Y isdetermined by homomorphism h: A --> O_X(X).Let I = Ker(h). Then Y' = Spec(A/I) satisfies the propertyof the proposition. If Y is not affine, then glue affine shemesobtained in the above method.N. Saito === everyone with this message:>asleep, the following problem popped in my mind:>Some additions:You can consider a linear variant of that problem. We should find a minimalstring which contains all binary strings of length n>1 (that one should be oflength 2^n+n-1). For example (n=3) the following string is solution:1110001011I can prove that that meta-string exists for any n>1. Also a circle (stringwith connected ends) of length 2^n exists too (in fact every meta-string could bereduced to circle by cutting off last n-1 bits). Proof afterspoiler spaceConsider a graph G which vertices are all strings of length n. If string b can beput after string a (the first n-1 bits of b are equal to last n-1 bits of a) thenan arrow from a to b is drawn. Thus every vertex has two ins and two outs.The graph is obviously connected. So there should be a cycle which goes throughevery vertex only once (ask Euler). That proves our theorem. Unfortunately this proof has nothing to do with 2D theorem.--|E1M1 :29 E1M5 :19 E2M1 :10 E3M2 :22 E4M5 :15, ============;|E1M2 :36 E1M6 :11 E2M3 :27 E4M1 :30 E4M6 :24|->grue3.tripod.com<--||E1M3_:43__E1M7_:14__E3M1_:43__E4M2_ :52__END__:37; =========[4*72]===> Well, it's probably hard to understand without an example.> Let n=1. So we have all 2*2 binary matrices:> OO OO XO OX OO XO OX XX> OO OX OO OO XO OX XO XX> XX XO OO OX XX XX OX XO> OO XO XX OX XO OX XX XX> That's torus> XXXO> XXOX> OXOO> XOOO> And that's matrix> XXXOX> XXOXX> OXOOO> XOOOX> XXXOX>The torus is kinda fascinating; all the possible combinations...couldpossibly be useful in artificial life applications using a cell grid. Haveyou come up with a use for such a compacting of information? It's almost aform of compression. === |Well, in fact I don't have the proof so you could try to find it. I just hope|elliptical curves have nothing to do with that problem...Elliptic curves. I'm sure they aren't. The one-dimensional version (asequence containing all subsequences of 0s and 1s of a given length)is pretty well known, so odds are good this one is too.Keith Ramsay === Worth to mention: Backup of all science related forums.Worth to mention:also good for scientists to find their topic right away.Website and forums order looks good to me as well. Anyways, I just though it's good to have it in our favourites. 7*---= 19 East/West-Coast Specialized Servers - Total Privacy via Encryption =--- === > ----------------------------- <^> <()> <^> -----------------------------> I have found several Chinese medicines that work much better> than the Western medicines for certain things.> When I young, and got colds,> I took those multi-purpose cold remedies,> that combined decongestants, antihistamines, antitussives,> expectorants, pain relievers, etc. in one pill.> As I wised up, I just took the specific ingredient> that I needed.> I also tried many grams of vitamin C with no results,> although I did find that zinc tablets sometimes helped.> I found the Chinese cold medicines worked better> than Western medicine medicines, and that ginger> was a common ingredient, so I bought some ginger,> and when I feel a cough or the sniffles coming on,> I snack on the ginger, and it seems to work very well.> You snack on RAW ginger? Brave, you are!It is a little hot, but otherwise it tastes pretty good, and it attacks cold symptoms immediately.Tom Potter === >I'll admit colds for me are extremely rare, although I will also>admit to washing my hands frequently after using the restroom>facilities, and using my elbows, feet, or arms instead of my hands>for opening the doors thereto and within.> Exactly. One of my tricks is to wear long-sleeved sweatshirts> and cover my hand before opening a door. While out, not touching> eyes and nose (that's when they always decide to get itchy) also> helps. Since women think it's cute to allow their sick kids to> touch everything in the grocery store, I wash everything I can> before I put them away. Always buy packaged produce. Make your> own hambuger after washing the meat.> /BAHI read about a study that found traces of 97 mens urine in the free peanuts at the bar.We should stop encouraging the 'working while sick' is pro company ethic, andencourage people to use their sick days at 1st sign.HercX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS016712394; === >I want to make a spreadsheet to calculate the number of feet of material on a>roll.>The user will enter the Outside Diameter, the Core Diameter (The tube that the>material is wrapped around) and the number of feet for a FULL roll first.>Then they will enter the Outside Diameter of the partial roll they want to>measure, and the Core Diameter. The spreadsheet will then calculate the number>of feet left on the roll.>I can't seem to come up with the algebraic expressions to calculate the>thickness of the material or the number of layers on the roll. I also want to>figure out the factor that will adjust the circumference for each successive>layer on the roll. It seems to me that the first section of entries (the OD,>the CORE and FEET) should be enough to extrapolate the number of layers on the>roll, and the thickness of the material. I know that if I had them enter the>material thickness to the equation it would be easier, but I don't want them to>have to measure the thickness.>any ideas?>MikeHere is an approximate way to attack the problem.If you look at the roll end-on, the cross-sectional area ofmaterial is approximately the material thickness times its length.A = pi * (r2 - r1)^2 = t * LSo if you measure r2 = outside diameter/2 and r1 = core diameter/2and you know L = material's length, you can get the thickness.Then apply same formula to another roll to find unknown length.Happy New Yearphil === >I want to make a spreadsheet to calculate the number of feet of material on a>roll.>The user will enter the Outside Diameter, the Core Diameter (The tube that the>material is wrapped around) and the number of feet for a FULL roll first.>Then they will enter the Outside Diameter of the partial roll they want to>measure, and the Core Diameter. The spreadsheet will then calculate the number>of feet left on the roll.>I can't seem to come up with the algebraic expressions to calculate the>thickness of the material or the number of layers on the roll. I also want to>figure out the factor that will adjust the circumference for each successive>layer on the roll. It seems to me that the first section of entries (the OD,>the CORE and FEET) should be enough to extrapolate the number of layers on the>roll, and the thickness of the material. I know that if I had them enter the>material thickness to the equation it would be easier, but I don't want them to>have to measure the thickness.>any ideas?>Mike>I lost the original post, but this reply is to Mike. I think theproblem is easier than you make it. If you call the radius of the tuber0 and the outer radius of a new roll r1 and you have the length of anew roll, call it L, then the length will be very close to linear as afunction of the radius r:length = L*(r - r0)/(r1-r0), for r0 <= r <= r1By basis for thinking this is that if the roll is modelled by a linearspiral r = k*theta + c, which I think it is, such a curve turns out tohave arc length a linear function of theta and hence a linear functionof the radius.Try it. I would be interested to hear how close it works out to thecorrect length in the shop.--Lynn === >By basis for thinking this is that if the roll is modelled by a linear>spiral r = k*theta + c, which I think it is, such a curve turns out to>have arc length a linear function of theta and hence a linear function>of the radius.Woops I should have checked this more carefully. It isn't linear.Please ignore my post.--LynnX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS016e12389; === Isn't 4 x 1 metre sidewalk too small for two men building it?Long time ago my dad asked me a question: If one man would dig a well in five hours, how much time would five workers need? I answered and then he said Imagine five big men digging one little well, huh?Smile!Mateusz === In sci.math, Mateusz Kwasnicki Isn't 4 x 1 metre sidewalk too small for two men building it?> Long time ago my dad asked me a question: If one man would dig> a well in five hours, how much time would five workers need?> I answered and then he said Imagine five big men digging one> little well, huh?> Smile!> Mateusz> Indeed. If a woman can have a baby in 8 1/2 months, how longwould it take 8 1/2 women (the 1/2 is working part time onanother project :-) ) ?-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > In sci.math, Mateusz Kwasnicki> Isn't 4 x 1 metre sidewalk too small for two men building it?> Long time ago my dad asked me a question: If one man would dig> a well in five hours, how much time would five workers need?> I answered and then he said Imagine five big men digging one> little well, huh?> Smile!> Mateusz> Indeed. If a woman can have a baby in 8 1/2 months, how long> would it take 8 1/2 women (the 1/2 is working part time on> another project :-) ) ?Be nice, people. I specifically stated in my original post that theproblem would not be modelled correctly unless the two workers workedwith perfect cooperation. Clearly women cannot cooperate on creatinga baby, althoughh these days that may no longer be as true as it oncewas. NO ATOMIC MEAURES! NO! NO! NO!. How would you then be able tocut the cake?As to the men working on the sidewalk. First of all this was theUnited States where most of us don't have a clue what a metre mightbe, and surely the cracks in side walks are not separated by a metreor even a meter as we say in the USA. Perhaps they were a yard apart. I don't remember. This was done in 1959 or thereabouts, and I amhaving trouble remembering the size of the sidewalk squares. I doremember that it was really, really difficult to make one step persquare for more than a couple of squares. I should also point outthat the problem never says that this isn't done with a GilbertSideWalk Kit that two boys could be using. I did say boys, not men,in the original post, although I did admit the somewaht egocentricnature of that assumption.At any rate, there were 5 men working digging a ditch, and theforemman was sitting on his ... watching the men work. One of thecame up and asked him why they had to do all the hard work and he gotto sit on his ... and watch them. He told the guy that it wasintelligence. Intelligence! What's that, said the worker. Theforeman held his hand up in front of a tree and said to the worker,might and the foreman pulled his hand away just in time so that theworker punched the tree with all of his might. He went back down intothe trench and was asked what the foreman had had to say. He said itwas intelligence, said the worker. Intelligence! What's that,replied his friend. Simple, he said, holding his hand in front ofAchavaX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS2gbb22101; === I recall having been taught not use operations such as +, -, /, x that will effect both sides(at the same time) of an identity to be proven. And so I told my students to follow this rule. One student respectfully and thoughtfully said why and went on to pose that the relationship between the two expressions we are trying to show are identical must be either >, >=, =, <= or < . So with the exception of multiplying or dividing by a negative value, the unknown relationship will stay the same(if we operate on both expressions at once).For example (trivial but it demonstrates the question) suppose you want to prove that 1 + sin(-x) = 1 - sin(x) .Normally we would apply the cofunction identity sin(-x) = - sin(x) the left hand expression and be done. But why not first subtract -1 from both expressions and then apply the cofunction identity?If the rule I remember being taught is valid there should exist some pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in fact identical but would appear to be identical if one operates on both expressions simulataneously. For example suppose one were trying to prove Expr1(x) = Expr2(x). In the process he/she divided both sides by cos(x)and found that the modified expresssions were in fact equal. I can't come up with such a pair of expressions that would validate the rule I remember. Can anyone help? === > I recall having been taught not use operations such as +, -, /, x >that will effect both sides(at the same time) of an identity to be >proven. And so I told my students to follow this rule. One student >respectfully and thoughtfully said why and went on to pose that the >relationship between the two expressions we are trying to show are >identical must be either >, >=, =, <= or < . So with the exception>of multiplying or dividing by a negative value, the unknown relationship>will stay the same(if we operate on both expressions at once).> For example (trivial but it demonstrates the question) suppose you >want to prove that 1 + sin(-x) = 1 - sin(x) .> Normally we would apply the cofunction identity sin(-x) = - sin(x) >the left hand expression and be done. But why not first subtract -1 from >both expressions and then apply the cofunction identity?> If the rule I remember being taught is valid there should exist some >pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in>fact.>identical but would appear to be identical if one operates on both> expressions simulataneously. For example suppose one were trying to prove >Expr1(x) = Expr2(x). In the process he/she divided both sides by>cos(x)and >found that the modified expresssions were in fact equal. I can't come up>with >such a pair of expressions that would validate the rule I remember. >Can anyone help? Firstly you must differentiate between an identity and an equality.Secondly, why must the alternatives to 'equals' be some inequality? Fineif you're dealing with real numbers, but what about complex numbers?In the example you've provided, all you;ve done is work backwards from thecorrect fact, that sin is an odd function (what's a cofunction identity bythe way?), to the statement you want. In this case all the steps you'vetaken are reversible, that is it is an if and only if deduction. Ingeneral in mathematics that doesn't happen, so it's better to teachstudents how to think 'properly'. An example where you shouldn't workbackwards (trivial, but important)showsin(x) '=' sin^3(x) + sin(x)cos^2(x)you can't factor out sin and divide because for some values of x you'redividing by zero. You may start from the identity 1 '=' sin^2 + cos^2 andmultiply through. The '=' bit means I really ought to write the three barsymbol. And that hidden dividing by zero thing needs to be watched. It is bad practice to assume the answer and work backwards. Don'tlet them start now and make maths teachers later tear their hair out. === >It is bad practice to assume the answer and work backwards. Don't>let them start now and make maths teachers later tear their hair out. Actually most of the time it's easier to work backwards, and perfectlyvalid _provided_ you know what you're doing: i.e. you know you're workingbackwards, and that the real formal proof will go forwards, so you make sure that the real proof will have valid inferences. To help in this,I like to write ?=, ?> etc. rather than = or > for relations I want to prove, but haven't yet.In your example: 1) sin(x) ?= sin^3(x) + sin(x)cos^2(x)is equivalent to2) sin(x) ?= sin(x) (sin^2(x) + cos^2(x))which is implied by3) 1 ?= sin^2(x) + cos^2(x)which is an identity we know.Note that it's OK to go from 2) to 3), precisely because we're reasoningbackwards: a b = a c does not imply b = c (because a might be 0), butit is implied by b = c. Now once you have the backwards proof, you canturn it around to make the real proof:1 = sin^2(x) + cos^2(x)sin(x) = sin(x)(sin^2(x) + cos^2(x))sin(x) = sin^3(x) + sin(x) cos^2(x)Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >It is bad practice to assume the answer and work backwards. Don't>let them start now and make maths teachers later tear their hair out. > Actually most of the time it's easier to work backwards, and perfectly> valid _provided_ you know what you're doing: i.e. you know you're working> backwards, and that the real formal proof will go forwards, so you make > sure that the real proof will have valid inferences. To help in this,> I like to write ?=, ?> etc. rather than = or > for relations I > want to prove, but haven't yet.> I think in some post that got lost by outlook (not usual choice) I agreed,and still do, that it is often practical to work backwards, though I tendto get students to rewrite it so it doesn't look as though they workedbackwards. And to make sure that all the if statements are only ifstatements too in this case. But the 'if you know what you're doing'proviso is key here. Surely it is better to go from what you know to betrue forwards. If the method you choose is deduced from thinking backwardsso be it. I agree the example below was pretty dire, and I certainly regret using it.> In your example: > 1) sin(x) ?= sin^3(x) + sin(x)cos^2(x)> is equivalent to> 2) sin(x) ?= sin(x) (sin^2(x) + cos^2(x))> which is implied by> 3) 1 ?= sin^2(x) + cos^2(x)> which is an identity we know.> Note that it's OK to go from 2) to 3), precisely because we're reasoning> backwards: a b = a c does not imply b = c (because a might be 0), but> it is implied by b = c. Now once you have the backwards proof, you can> turn it around to make the real proof:> 1 = sin^2(x) + cos^2(x)> sin(x) = sin(x)(sin^2(x) + cos^2(x))> sin(x) = sin^3(x) + sin(x) cos^2(x)> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 === [snip]> In the example you've provided, all you;ve done is work backwards from> the correct fact, that sin is an odd function (what's a cofunction> identity by the way?),I presume that the OP misused that term. Surely a cofunction identity issomething like cos(x) = sin(pi/2 - x).> to the statement you want. In this case all the> steps you've taken are reversible, that is it is an if and only if> deduction. In general in mathematics that doesn't happen, so it's better> to teach students how to think 'properly'. An example where you shouldn't> work backwards (trivial, but important)> show> sin(x) '=' sin^3(x) + sin(x)cos^2(x)> you can't factor out sin and divide because for some values of x you're> dividing by zero.One may factor out sin(x) on the right, and then notice that the otherfactor, sin^2(x) + cos^2(x), is always 1, thereby showing that the rightside simplifies to the left side.> You may start from the identity 1 '=' sin^2 + cos^2 and> multiply through. The '=' bit means I really ought to write the three bar> symbol. And that hidden dividing by zero thing needs to be watched.At least it needs to be watched when one is solving conditional equations,lest roots be lost.> It is bad practice to assume the answer and work backwards.In _general_, I wholeheartedly agree. But, in the specific context ofproving identities, I disagree. For example, very often, the easiest methodof proof is to simplify both sides of the supposed identity as much aspossible, in the hope that both sides will simplify to the same thing.David Cantrell === >I recall having been taught not use operations such as +, -, /, x>that will effect both sides(at the same time) of an identity to be>proven....>If the rule I remember being taught is valid there should exist>some pair of trigonometric expressions Expr1(x) and Expr2(x)that>are not in fact identical but would appear to be identical if one>operates on both expressions simulataneously.Something in what you have written here makes me think of thePerron Paradox. Perhaps that would be of help to you.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBS4GIB28000; === >I recall having been taught not use operations such as +, -, /, x that will effect both sides(at the same time) of an identity to be proven. And so I told my students to follow this rule. One student respectfully and thoughtfully said why and went on to pose that the relationship between the two expressions we are trying to show are identical must be either >, >=, =, <= or < . So with the exception of multiplying or dividing by a negative value, the unknown relationship will stay the same(if we operate on both expressions at once).>For example (trivial but it demonstrates the question) suppose you want to prove that 1 + sin(-x) = 1 - sin(x) .>Normally we would apply the cofunction identity sin(-x) = - sin(x) the left hand expression and be done. But why not first subtract -1 from both expressions and then apply the cofunction identity?>If the rule I remember being taught is valid there should exist some pair of trigonometric expressions Expr1(x) and Expr2(x)that are not in fact identical but would appear to be identical if one operates on both expressions simulataneously. For example suppose one were trying to prove Expr1(x) = Expr2(x). In the process he/she divided both sides by cos(x)and found that the modified expresssions were in fact equal. I can't come up with such a pair of expressions that would validate the rule I remember. Can anyone help? I have never heard of this rule. There is absolutely no reason why you can't perform the same operation on any equation, identity or not, and not arrive at a valid equation. One of the things you do have to be careful about is that each operation you use has to be invertible: that is you have to be careful about multiplying both sides of an equation by sin x since the inverse would be dividing both sides by sin x which is 0 for some x. The reason for this is that the real proof is working backwards. In order to prove, for example, that (tan x)(cos x)= sin x, we replace tan x with sin x/cos x so that the left side is (sin x/cos x)(cos x) which is, of course, sin x. But you can't prove a formula, or any statement, is true by starting from that statement! The real proof is working backwards.Looking at what we did, we know we an start from sin x= sin x, rewrite the left side as (sin x)(cos x/cos x)= (sin x/cos x)cos x= (tan x)(cos x). We don't have to write thatout because we know that everything we did originally is reversible. === > In order to prove, for example, that (tan x)(cos x)= sin x, we replace>tan x with sin x/cos x so that the left side is (sin x/cos x)(cos x)>which is, of course, sin x. That's not a proof. In fact there's nothing to prove since bydefinition tan x == sin x / cos x (this follows from elementarygeometry) so the result is simply a reformulation of the definition.Furthermore, it's not even an equality. What happens whenx = Pi/2? The right side is still well-defined but the left side mostcertainly not.> But you can't prove a formula, or any statement, is true by starting>from that statement! The real proof is working backwards.>Looking at what we did, we know we an start from sin x= sin x, rewrite>the left side as (sin x)(cos x/cos x)= (sin x/cos x)cos x= (tan x)(cos x).And how do you know that (sin x / cos x) = tan x when that is the veryresult you're trying to prove?>We don't have to write that out because we know that everything we did>originally is reversible. There is nothing that says you must work out your proofs backwards(what you call forwards) or that all proofs must be two-way. If westart from given truths and work our way using one-way implications(i.e. p => q) there is nothing that states the result is incorrect,just that the opposite (q => p) does not necessarily hold. === > In order to prove, for example, that (tan x)(cos x)= sin x, wereplace> tan x with sin x/cos x so that the left side is (sin x/cos x)(cosx)> which is, of course, sin x. > That's not a proof. In fact there's nothing to prove since by> definition tan x == sin x / cos x (this follows from elementary> geometry) so the result is simply a reformulation of the definition.There are several ways of defining tan, with or without any referenceto cos and sin. It is a matter of convention. So what is a proof andwhat is only a reformulation of the definition only depends on whatframe you're in.> Furthermore, it's not even an equality. What happens when> x = Pi/2? The right side is still well-defined but the left side most> certainly not.The right side is not defined either, at least in R, cos(Pi/2)=0.> But you can't prove a formula, or any statement, is true bystarting> from that statement! The real proof is working backwards.> We don't have to write that out because we know that everything wedid> originally is reversible. > There is nothing that says you must work out your proofs backwards> (what you call forwards) or that all proofs must be two-way. If we> start from given truths and work our way using one-way implications> (i.e. p => q) there is nothing that states the result is incorrect,> just that the opposite (q => p) does not necessarily hold.Both of you agree on that, I guess. I think that what you callbackwards is what he calls backwards too...But what you have just written is confusing, even if I guess you agreewith what I am going to say, I just want to clarify this point. There_is_ something that says that proof must go in one definite way: fromthe hypothesis to the result! This is the way that is meaningful.Sometimes, you can write the argument backwards because it isreversible. But what matters is the way (hyps+defs+axioms) => result.To come back to the original question, I would say that the rule Edgave is pointless with _purely_ symbolic proofs, and needs to besharpen when doing analysis. But I think it is interesting to explainto your students when such operations are allowed and when they arenot, and why. It depends on their level, but I think it is always abad idea to simply say NO!. Mathematics is not about calculus andits rules, it is about understanding. So you may allow them not torespect your rule, if they do it correctly.First of all, you want to consider functions, not expressions.You must be careful about the set of points on which what you aredoing is valid. It may not be R as a whole. You may need to excludesome points, like 0. Sometimes what you write is only valid on (0;Pi/2], and so on.And the simplest case which is not _purely_ symbolic is probably with<= and some multiplication on both sides of the inequality.Dividing by cos(x) on both sides is problematic when cos(x) did notappear in the numerators of each side of the previous expression.Because in such a case, you may have a problem when x = Pi/2. But insome cases, it may still be valid. If you have cotan(x) on both sides,or if both sides are equivalent to cos(x) when x tends to Pi/2, etc.The reason why you did not find any counter example (ie. some examplethat would validate the need for your rule) is that you are usingsimple expressions. If your expressions Expr1 and Expr2 containcos(x), sin(x), tan(x) and nothing else (but arithmetical operations),you can replace them by A, B, C and in practice, _any symboliccomputation_ on both sides of any equality written with these threeletters will remain valid with cos(x), sin(x) and tan(x). You can waituntil you're done to ask you where this is defined. Why?This is not elementary. These three functions are C-Infinityfunctions, but in some isolated points where they are not defined. Soeven if an expression, which is a step in your proof, written withcos, sin and tan is not defined in some point, you may easily considerits limit in this point. And if it is infinite, you know precisely thespeed of the convergence to the infinity. So that you may be able,in a latter step of your proof, to find a (finite) limit in the samepoint. And you will be able to state it rigorously.To say it simply, since they are explicit C-Infinity functions, cos,sin, and tan can be handled as polynomials, using Taylor's formula.I said previously that what you want to consider is functions and notexpressions. This is because expressions are only symbolic and tellsnothing about the values they can take when evaluated. And whatinterests you here is to identify expressions that take the samevalues in each point of some set. So that you identify functionsthrough their graph.So, what do you say to your students? Again, it depends on theirlevel, but I would recommend to abandon your rule, to concentrate onthe meaning of what they do. That doesn't mean you cannot say to them:be careful about that, in some cases it is false, thus I wouldrecommend you not to do this, unless you understand what you do andcan explain it, and you can require that they explain what they doeach time it is necessary. Try to tell them when it is safe: inparticular, for trigonometric formulas, tell them about definitiondomains (taking some example where you divised by 0 with cos(x) and x= Pi/2), the problem with limits and infinity if they are advancedargument if they start with the result.More generally, my personnal point of view is to always avoid blindrules in Mathematics. It is sometimes necessary to adopt such a pointof view: if you want to train them to compute derivatives quickly, forinstance. But this should always be accompanied with explanations, andif you can't, give them at least some clues to help answer the why?.My experience is that if a student can remember why, then you havewon. He/She may even begin to like Mathematics. /er. === > If uncountable S is a subset of R, then S contains> a subset T with the property that, for every distinct elements> x and y in T, there's a z in T between x and y.> This problem is equivalent to showing for uncountable S,> S is somewhere dense,> that is not nowhere dense; int cl S /= nulset>See retraction sci.math nowhere dense real subsets. <1072549678.37972@news.aic.at> === only one was willing to make the trip. Our small research group was thusfaced with the necessity of carrying n standard elephants themselves. Ina pilot project, it was established that our full staff was unable tolift, much less carry, a single standard elephant. It was also noted bythe project manager who was sprayed by the ill mannered standard elephantblowing it's nose upon him and again by the project director who had herbutt whooped by a standard trunk, that standard elephants may lackappropriate manners needed for cooperation on a long long trip. Anyfurther efforts on that field trip were immediately abandoned when n-1standard elephants taking notice of the ruckus, filled their standardtrunks with wet water and eagerly approached the experimental site.Upon recovery from this test project, the staff decided a better approachwould be to use the standard kangaroo model on the speculation that thestandardized equipment would hop itself into position. Before proceedingwith this project some facts need be established. As standard kangaroosare smaller that standard elephants, will n standard kangaroos suffice orwill 3n standard kangaroos be needed? Also once in place, how is astandard hopping kangaroo convinced to remain stationary? >By the way, thanks for the hint that standard elephants could maybe >decay into strange colorful left handed anti-quarks - we will >make some more calculations clarifying this point.Your welcome. Please keep us inform of your progress. Do be carefulhandling strange colorful left handed anti-quarks, those weirdo's havebeen noticed to change gender right in the middle of an.., experiment.Whence the name.---- === > An interesting approach. However upon asking 262,144 standard elephants,> only one was willing to make the trip. Our small research group was thus> faced with the necessity of carrying n standard elephants themselves. In> a pilot project, it was established that our full staff was unable to> lift, much less carry, a single standard elephant. It was also noted by> the project manager who was sprayed by the ill mannered standard elephant> blowing it's nose upon him and again by the project director who had her> butt whooped by a standard trunk, that standard elephants may lack> appropriate manners needed for cooperation on a long long trip. Any> further efforts on that field trip were immediately abandoned when n-1> standard elephants taking notice of the ruckus, filled their standard> trunks with wet water and eagerly approached the experimental site.>Oh ... we are very sorry that your research group suffered suchcomplications - somehow there is a very huge misunderstanding here :In order to measure the mass of the galactic black hole on should useVIRTUAL ELEPHANTS !!!The main advantage in such an approach would be, that thehandeling of virtual elephants is somehow more easy and convenient.Besides, the branching-fraction of a virtual standard elephant intoa left-handed kangaroo and an ensemble of photons is well known.> Upon recovery from this test project, the staff decided a better approach> would be to use the standard kangaroo model on the speculation that the> standardized equipment would hop itself into position. Before proceeding> with this project some facts need be established. As standard kangaroos> are smaller that standard elephants, will n standard kangaroos suffice or> will 3n standard kangaroos be needed? Also once in place, how is a> standard hopping kangaroo convinced to remain stationary?You are completely right that 3n standard kangaroos would beneeded for this task. (or more correctly 3*(n-1)*(1+epsilon),where epsilon is an empirical constant)We have made the observation, that standard kangaroos jump lessfrequently when we bind them together with an inelastic rope in pairs.Maybe this could help to make the measurement condition more stable ? <1072626573.232780@news.aic.at> === > elephant blowing it's nose upon him and again by the project > director who had her butt whooped by a standard trunk, that > standard elephants may lack appropriate manners needed for > cooperation on a long long trip. Any further efforts on that > field trip were immediately abandoned when n-1 standard elephants > taking notice of the ruckus, filled their standard trunks with > wet water and eagerly approached the experimental site.>Oh ... we are very sorry that your research group suffered such >complications - somehow there is a very huge misunderstanding here : >In order to measure the mass of the galactic black hole on should >use VIRTUAL ELEPHANTS !!!Again upon inquiring of 262,144 standard elephants from 256 herds we foundnone knew how to virtualize. We even introduced them to the Cheshire Cat,who if you recall would virtualize into thin air with a smile. At firstthey took note of him for smelling like a feline, but then when he smiledthey soon lost interest. We then enlisted the assistance of an elephanttrainer, but try as he may, even the most intelligent and well trainstandard elephants could or would not smile and smile from ear to ear likethe Cheshire Cat.One elephant however, the standard elephant who earlier expressed interestin travel to far and distance places, took interest in the Cheshire Cat.He was so bemused by the cat's smile that in response he widened andwidened his ears. This so please the Cheshire that they soon becamesteadfast friends. Not long thereafter they turned to staff and trainer;the Cheshire Cat smiling ever wider smiles at us, and the cooperativestandard elephant spreading his ears every wider and wider, graduallyvanished before us appearing as naught but a wide smile and wider earswhich eventually faded from sight.This second field trip into the standardized land of the elephants, thoexceedingly more pleasant, found no standard elephants who could or wouldvirtualize except for the one standard elephant friend of the Cheshire Catwho, with the Cat, virtualized into a virtual reality without telling uswhich of the virtually infinite virtual realities they went to.However our research has suddenly had to shift direction.You may will have noticed this astounding news story: anti-quark, was discovered at Los Alamos laboratories on having been observed changing gender right in the middle of an.., experiment, has Los Alamos scientists greatly perplex over chronom tunneling since it was detected a few days before it's discovery.your help to find the mass of the weirdo quickly by Monday before thescientists returned to the lab. That no longer appears possible, nor isit now of importance with the breaking of that story. Having reviewed andrechecked the events of the weekend we now know the weirdo appearedundetected at Livermore Research Center to be detected at Los Alamos. Westrongly suspect this is because associated with chronom tunneling isspatial distortion, displacement or tunneling.We attempted to model this possibility using the equations of generalrelativity on a super-computer. Preliminary results are perplexing. Wesuspect the super-computer adequately comprehended the phenomena and usedit. Indications are that, quickly running out of storage space to holdtemporary data, it resorted to storing it in the past and future. Thusthe perplexment, tho the program needs but a few more hours to run, we maynot know even when the results will be collated.One of our coworkers, fatigued by the pace of this weekend, fell asleepdreaming she had consulted with Chronos the Greek god of time and Kali theproblem, nay perhaps that of unified space/time field theory, wasunresolvable until the god and goddess of distance awoke. To awaken thesedeities one was to quietly and devoutly invoke their names three times ineach of the four directions on a mountain top inaccessible by helicopter.Can you assist our research efforts to find who thegod and goddess of distance are? >We have made the observation, that standard kangaroos jump less >frequently when we bind them together with an inelastic rope in >pairs.Being rushed by the pace of events, we released some standard kangaroosinto our back yard where our kids, who also like to jump up and down,could play with them. Being the offspring of scientists, they readilytook our suggestion to experiment binding standard kangaroos together inpairs. Results came soon: standard jumping kangaroos aren't intobondage; for best results use bungy cords; have fun.---- === > Can you assist our research efforts to find who the> god and goddess of distance are?>Indra & Prajapathi ...Despite of noting the obvious : In hinduistic mythologythere is only one single omni-potent, omni-present andomni-scient god.The multifarious 'deities' are nothing but different aspects ofthe same one.Ganesha is the aspect of 'everything that can be counted orcomprehended'.God is everything that exists and everything that does notexist - everything that can be thought of - and - everythingthat can not be thought of ...It is an illusion and delusion to hope to understand thiscompletely.All religions are one - we are all brothers and sisters on thisfragile planet, being less than dust or vanity in the universe. === Erratum> However for more ambitious measurements, such as the exact mass of> the Galatic black hole, does one use the Elephant in Musk Model?That's very easy in the framework of the Practical Elephant Model :>Put n standard-elephants into a stationary orbit around the galactic>black hole.>The measurement of the wobble-frequency and the radiation-spectra>of the elephants will provide you with the mass of your black hole>(in units of the standard-elephant)> An interesting approach. However upon asking 262,144 standard elephants,> only one was willing to make the trip. Our small research group was thus> faced with the necessity of carrying n standard elephants themselves. In'carrying n-1 standard elephants'> a pilot project, it was established that our full staff was unable to> lift, much less carry, a single standard elephant. It was also noted by> the project manager who was sprayed by the ill mannered standard elephant> blowing it's nose upon him and again by the project director who had her> butt whooped by a standard trunk, that standard elephants may lack> appropriate manners needed for cooperation on a long long trip. Any> further efforts on that field trip were immediately abandoned when n-1> standard elephants taking notice of the ruckus, filled their standard'when n-2 standard elephants'> trunks with wet water and eagerly approached the experimental site.> Upon recovery from this test project, the staff decided a better approach> would be to use the standard kangaroo model on the speculation that the> standardized equipment would hop itself into position. Before proceeding> with this project some facts need be established. As standard kangaroos> are smaller that standard elephants, will n standard kangaroos suffice or> will 3n standard kangaroos be needed? Also once in place, how is a'will 3(n-1) standard kangaroos and the willing standard elephant'> standard hopping kangaroo convinced to remain stationary?>By the way, thanks for the hint that standard elephants could maybe>decay into strange colorful left handed anti-quarks - we will>make some more calculations clarifying this point.> Your welcome. Please keep us inform of your progress. Do be careful> handling strange colorful left handed anti-quarks, those weirdo's have> been noticed to change gender right in the middle of an.., experiment.> Whence the name.> ---- === >If A is a nowhere dense subset of a > separable compact connected Hausdorff Baire space >is A countable? >The Cantor set is a nowhere dense set >with the cardinality of the reals.A most efficient proof. ;-)---- === > If A is a nowhere dense subset of the reals R,> then A is countable.That's impressive eggnog you're drinking. === > The Cantor set ... take out the trashHmmm... a nice way of defining it :-)Rainer Rosenthalr.rosenthal@web.de === Today, at the Barnes and Noble Bookstore, I saw the book The ColossalBook of Mathematics: Classic Puzzles, Paradoxes, and Problems byMartin Gardner. The book says thate^[pi*sqrt(163)] = 262,537,412,640,768,744which is an integer exactly.Specifically, the book says that Srinivasa Ramanujan(1887-1920)computed the number manually to 262,537,412,640,768,743.999999, butcould not go further; then someone in France computed two milliondigits after the decimal point which are all 9; then someoneingeniously used the Euler's Constant to prove that the result isexactly the integer as given above.A quick search in the Internet revealed that the above claim is nottrue and it first showed up as a April's Fool's joke by MartinGardner, and he subsequently admitted that it was a joke.Then how come it still got into the book? === > Today, at the Barnes and Noble Bookstore, I saw the book The Colossal> Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by> Martin Gardner. The book says that> e^[pi*sqrt(163)] = 262,537,412,640,768,744> which is an integer exactly.> Specifically, the book says that Srinivasa Ramanujan(1887-1920)> computed the number manually to 262,537,412,640,768,743.999999, but> could not go further; then someone in France computed two million> digits after the decimal point which are all 9; then someone> ingeniously used the Euler's Constant to prove that the result is> exactly the integer as given above.> A quick search in the Internet revealed that the above claim is not> true and it first showed up as a April's Fool's joke by Martin> Gardner, and he subsequently admitted that it was a joke.> Then how come it still got into the book?(because the author of The Colossal Book appears to be a disciple ofMartin Gardner ;-))According to Maple:> evalf(exp(Pi*sqrt(163)), 50);; 18 .26253741264076874399999999999925007259719818568865 10which is quite remarkable, though.I don't know if Ramanujan really did this computation, but thinkingabout it: how on earth can one come to think about it? Any clue? /er. === Eric Rannaud> e^[pi*sqrt(163)] = 262,537,412,640,768,744> ...> I don't know if Ramanujan really did this computation, but thinking> about it: how on earth can one come to think about it? Any clue?Please readThe Book of NumbersJohn H. Conway, Richard K. GuyCopernicus (Springer)ISBN 0-387-97993-Xand especially Chapter 8, part The Nine Magic Discriminants.Rainer Rosenthalr.rosenthal@web.de === > Today, at the Barnes and Noble Bookstore, I saw the book The Colossal> Book of Mathematics: Classic Puzzles, Paradoxes, and Problems by> Martin Gardner. The book says that> e^[pi*sqrt(163)] = 262,537,412,640,768,744> which is an integer exactly.> A quick search in the Internet revealed that the above claim is not> true and it first showed up as a April's Fool's joke by Martin> Gardner, and he subsequently admitted that it was a joke.> Then how come it still got into the book?>Who's the author of the book? ;-) === The author is Gardner himself. Well, today I checked the book again.At the end of the long chapter, there is actually an Addendum whichtalks about the fact that it was intended as a joke.> ...> Who's the author of the book? ;-) === > [snip]> The subject is typical representation of the OP's educational > inadequacies.You're obviously out of your depth here. Herc has strong and validopinions about logical and philosophical issues you probably aren'teven aware exist. He is also mentally ill. The two have none to dowith each other, and he still deserves respect.'cid === > Doing math is a series of small successes used to build a large house> of knowledge. I never understood how people could ever dislike it.> Presumably they never experienced one of those successes, and feelthat the large house of knowledge has long been built and they have totread carefully so as not to stir the ghosts in the dusty corners. === >I'm looking for criterion that ensure two groups are isomorphic.>[snip]> By group-theoretical invariants, I meant things like order and structure of> Z(G), order of [G,G], number of elements of order r for each r, etc.>Including cohomology groups of G over the ring of rational integers?Yes, I would include these as examples of group-theoretical invariants.I guess you are asking whether there exist non-isomorphic G and H suchthat H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sureabout that - it sounds an interesting question.Derek Holt. === >I guess you are asking whether there exist non-isomorphic G and H such>that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure>about that - it sounds an interesting question.How about G = Z/2 x Z/3 ( = Z/6 ) and H = Z/2 * Z/3 ( = PSL_2(Z) )?But yes, this is an interesting question for _finite_ groups, or atleast it was until Ian Leary bagged it. There are non-isomorphicfinite p-groups G and H for which H^n(G,Z) and H^n(H,Z) areisomorphic groups for every n.I think that in his example the _rings_ H^*(G,Z) and H^*(H,Z) arenot isomorphic, however. And of course even if the rings were isomorphic,one could ask about the structures of the rings as modules under theNote that cohomology is actually a functor of the corresponding group ring(Z[G] resp Z[H]) so that if it is possible to use cohomology todistinguish two groups, then it is because the group rings are alreadynon-isomorphic. So I suppose the greater challenge is to find two nonisomorphic (finite) groups whose integral group rings are isomorphic;this has been conjectured never to happen and last I heard, which wasquite a while ago, this conjecture was still open. Of course, I shouldpoint out that even if the integral group ring turns out to be aperfect invariant in this sense, this really only amounts tobegging the original question, since one would then have to decide howone can determine whether or not two rings (which happen to be finitely-generated Z-modules) are isomorphic. I don't see any reason to thinkthat's an easier question than the one we began with.[I should probably add that by ring, above, I mean ring with augmentationmap. I have this dim memory that there are examples where ZG and ZH areisomorphic as rings but the isomorphism cannot commute with augmentation.]Finally, let me respond to the person who mentioned taking cohomologywith coefficients in some other ring or field. That's a much weakerinvariant: from the Universal Coefficient Theorem one can pretty muchdetermine H^*(G,A) where A is any trivial G-module, as soon as oneknows H^*(G,Z). The reverse is far from true and one can easilyfind groups G, H with H^*(G,A) isomorphic to H^*(H,A); for example,if A is the field with 2 elements and G and H have odd order, or(less trivially) if G and H are dihedral groups of different orders.Ian's paper is MR1348713 (96j:20076) : $p$-groups are not determined by their integral cohomology groups. Bull. London Math. Soc. 27 (1995), no. 6, 585--589. 20J06 (20D15)dave === >I'm looking for criterion that ensure two groups are isomorphic.> [snip]> By group-theoretical invariants, I meant things like order and structure of> Z(G), order of [G,G], number of elements of order r for each r, etc.>Including cohomology groups of G over the ring of rational integers?> Yes, I would include these as examples of group-theoretical invariants.> I guess you are asking whether there exist non-isomorphic G and H such> that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure> about that - it sounds an interesting question. Derek Holt.Attach to the group G the sequence:EulerianFunction(G,1),EulerianFunction(G,2), EulerianFunction(G,3),...(this is implemented in GAP ) where the EulerianFunction(G,n) is thenumber of n-tuples of elements of G that generate G .Doesn't this sequence of numbers determine G up to isomorphism ?Tim === >I'm looking for criterion that ensure two groups are isomorphic.> [snip]> By group-theoretical invariants, I meant things like order and structure of> Z(G), order of [G,G], number of elements of order r for each r, etc.>Including cohomology groups of G over the ring of rational integers?> Yes, I would include these as examples of group-theoretical invariants.> I guess you are asking whether there exist non-isomorphic G and H such> that H^n(G,Z) and H^n(H,Z) are isomorphic for all n ? I am not sure> about that - it sounds an interesting question.> Derek Holt.>Attach to the group G the sequence:>EulerianFunction(G,1),EulerianFunction(G,2), EulerianFunction(G,3),...>(this is implemented in GAP ) where the EulerianFunction(G,n) is the>number of n-tuples of elements of G that generate G .>Doesn't this sequence of numbers determine G up to isomorphism ?>I don't think so.SmallGroup(32,13) and SmallGroup(32,14) give the same answer for n<=40.Derek Holt. === >In practice, groups are not normally given as multiplication tables - they are>usually defined as groups of permutations or matrices, or by means of>presentations. It is a much more difficult problem to decide whether or not>two subgroups G and H of S_n defined by generating permutations are>isomorphic. I am fairly sure that this is still in NP (assuming that the>putative isomorphism is given by means of images in H of generators of>G, you can compute in polynomial time a set of defining relators for G on>that generating set, and thereby check whether or not the map is a isomorphism)>but it may well be NP-complete.>If the elements are given as permutations or matrices, then you can create >the multiplication table in poly time (for a simple minded upper bound; As Keith Ramsay observed, for permutation groups input by generators, whichis the standard model for most complexity results and research concerningfinite permutation group algorithms, you cannot construct themultiplication table in polynomial time. There are many properties ofthe group which you can compute in polynomial time, however, such asits order, the order of its derived group, centre, ...Another standard model used in complexity theory of algorithms forfinite groups is the Black-Box group, introduced by L. Babai, andmatrix groups over finite fields fall into this category. ABlack-box group is one in which the elements are represented bybit-strings of bounded length, and the group-theoretical operations(composition and inversion) can be carried out in constant time.Again groups are defined by means of generators. But it is muchharder to prove results about Black-box groups - most of thealgorithms are probabilistic and have a small probability of givinga wrong answer.>er...I'm not so sure about a matrix representation because of the operations >needed to test equality of the matrix elements (with all those radicals >flying around)).>If the elements..er if the group is given as a finite presentation (set of >identities), then it's undecidable by ...uh ... a reduction from >undecidability of uh... well it should be from the word problem but I >can't see the obvious at the moment.Yes, but there are some particular types of presentation which are usefulcomputationally, such as the power-commutator presentation of finite solvable(or infinite polycyclic groups).Derek Holt. === >If the elements are given as permutations or matrices, then you can create >the multiplication table in poly time (for a simple minded upper bound; >As Keith Ramsay observed, for permutation groups input by generators, which>is the standard model for most complexity results and research concerning>finite permutation group algorithms, oh. my misunderstanding.>you cannot construct the>multiplication table in polynomial time. There are many properties of>the group which you can compute in polynomial time, however, such as>its order, the order of its derived group, centre, ...I suppose with just the permutation generators, the order could be exponential (e.g. 2 generators of degree n can generate S_n of size n!, whose number would be represented with O(n) bits). But how can you compute this order (in general) without actually constructing all elements?Mitch Harris === >If the elements are given as permutations or matrices, then you can create >the multiplication table in poly time (for a simple minded upper bound; >As Keith Ramsay observed, for permutation groups input by generators, which>is the standard model for most complexity results and research concerning>finite permutation group algorithms, >oh. my misunderstanding.>you cannot construct the>multiplication table in polynomial time. There are many properties of>the group which you can compute in polynomial time, however, such as>its order, the order of its derived group, centre, ...>I suppose with just the permutation generators, the order could be >exponential (e.g. 2 generators of degree n can generate S_n of size n!, >whose number would be represented with O(n) bits). But how can you compute >this order (in general) without actually constructing all elements?This is known as the Schreier-Sims algorithm. To summarize it very brieflyindeed, you calculate the orbit Orb(G,1) of 1 under G, then use a theoremof Schreier to compute generators of the stabilizer Stab(G,1) of 1 underG from the orbit and the generators of G. Then recursively compute|Stab(G,1)| and we have |G| = |Orb(G,1)||Stab(G,1)| by the orbit-stabilizertheorem.To get polynomial time, you have to be more careful, because thethe number of generators of Stab(G,1) coming from Schreier's Theorem isroughly |Orb(G,1)| times the number of original generators of G. To avoidan explosion in the number of generators, you maintain a subgroup H ofStab(G,1), initially trivial, for which you know the order. For each newSchreier generator of Stab(G,1) you first check if it already in H. Ifit is not, then you re-run the algorithm on H with the new generatoradjoined, and enlarge H.Derek Holt. === >There are non-isomorphic groups of order>32 (or maybe 64) which cannot be distinguished by means of any known>group-theoretical invariants. >Er...then how do you distinguish them? Non group-theoretical invariants? >Or is this a case of where there is a counting argument that is >nonconstructive?>Well, if you can't think of anything better, then you just try all>possible maps from the first group to the second and check that it is not>an isomorphism. In practice, you would just check maps from a generating>set of the first group to the second, because a homomorphism is>uniquely determined by its action on a generating set.>By group-theoretical invariants, I meant things like order and structure of>Z(G), order of [G,G], number of elements of order r for each r, etc.>I guess I am piqued by he use of the word cannot in your claim. Is there >a proof of that impossibility (given some arbitrary or not so arbitrary >set of group-theoretical invariants)?>I wasn't really trying to make a precise statement, and the idea ofa group-theoretical invariant is not really well-defined. But I would makethe following conjecture, just based on general experience, and without anyprospects of being able to prove anything precise. If you specify inadvance some collection of group-theoretical invariants, which may include things like Order and properties of Aut(G), (co)homology groups, numberof homomorphisms from G to standard groups like S_n, etc., then thereexist two non-isomorphic groups which agree in all of the invariants thatyou specified.Derek Holt. === >There are non-isomorphic groups of order>32 (or maybe 64) which cannot be distinguished by means of any known>group-theoretical invariants. ...>I guess I am piqued by he use of the word cannot in your claim. Is there >a proof of that impossibility (given some arbitrary or not so arbitrary >set of group-theoretical invariants)?>I wasn't really trying to make a precise statement, and the idea of>a group-theoretical invariant is not really well-defined. But I would make>the following conjecture, just based on general experience, and without any>prospects of being able to prove anything precise. If you specify in>advance some collection of group-theoretical invariants, which may include >things like Order and properties of Aut(G), (co)homology groups, number>of homomorphisms from G to standard groups like S_n, etc., then there>exist two non-isomorphic groups which agree in all of the invariants that>you specified.OK, I see. This sounds very similar to the idea (conjecture? theorem?) that there is no poolyomial size set of properties of two graphs todistinguish them.Mitch Harris === [snip]> I am not aware of any such criteria. There are non-isomorphic groups of order> 32 (or maybe 64) which cannot be distinguished by means of any known> group-theoretical invariants. [snip]> Derek Holt.Could you please give an example of a pair of such hard to distinguishgroups, in GAP notation?Alan === >[snip]> I am not aware of any such criteria. There are non-isomorphic groups of order> 32 (or maybe 64) which cannot be distinguished by means of any known> group-theoretical invariants. >[snip]> Derek Holt.>Could you please give an example of a pair of such hard to distinguish>groups, in GAP notation?SmallGroup(32,13) and SmallGroup(32,14) seem hard to distinguish using group-theoretical properties. Let me know if you succeed! Of course they can bedistinguished by the invariant Number of isomorphisms from given groupto SmallGroup(32,13). So I prefer to rephrase my original claim as:If you give me a collection of group-theoretical invariants, then I willfind two non-isomorphic groups which are not distinguished by them.Derek Holt. === >[snip]> I am not aware of any such criteria. There are non-isomorphic groups of order> 32 (or maybe 64) which cannot be distinguished by means of any known> group-theoretical invariants. > [snip]> Derek Holt.>Could you please give an example of a pair of such hard to distinguish>groups, in GAP notation?> SmallGroup(32,13) and SmallGroup(32,14) seem hard to distinguish using group-> theoretical properties. Let me know if you succeed! Of course they can be> distinguished by the invariant Number of isomorphisms from given group> to SmallGroup(32,13). So I prefer to rephrase my original claim as:> If you give me a collection of group-theoretical invariants, then I will> find two non-isomorphic groups which are not distinguished by them.> Derek Holt.Well, I don't know if this qualifies, but the number of elementshaving a square root or the maximal number of square roots of anysingle element seem to distinguish between those two. Of course,there will be non-isomorphic groups that cannot be told apart bythese, but such invariants can be generalized in many ways into aninfinite array of invariants, and I wouldn't count on there being nosuch an array that is capable of distinguishing any pair of finitenon-isomorphic groups.This, of course, says nothing about efficiency of computation - I'mreferring to sets of invariants that seem more natural than thenumber of isomorphisms to some fixed group G as G varies over allfinite groups.Alan === > If an abelian group G has two elements with order m and n respectively,> show G has an element whose order is the least common multiple of m and n.> a, b of order m and n respectively.> m is the smallest positive integer such that a^m = e.> n is the smallest positive integer such that b^n = e.One can show from the definitions of greatest common denominator (gcd) andleast common multiple (lcm) thatlcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say).(This is a handy thing to remember too.)so lg = mn ; m = qg ; n = rg since g divides both m and n.All we need to know about q and r is that they are integers. Thenl = mn/g = r*q*g = rm = qn ; so(ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = eand the order of ab is l = lcm(m,n). Note the if (m,n) = 1(i.e., are relatively prime), the order of the product ab is the product of the orders of a and b since [m,n] = mn in this case.Van === > One can show from the definitions of greatest common denominator (gcd) and> least common multiple (lcm) that> lcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say).> (This is a handy thing to remember too.)> so lg = mn ; m = qg ; n = rg since g divides both m and n.> All we need to know about q and r is that they are integers. Then> l = mn/g = r*q*g = rm = qn ; so> (ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = e> and the order of ab is l = lcm(m,n). Note the if (m,n) = 1> (i.e., are relatively prime), the order of the product ab is the> product of the orders of a and b since [m,n] = mn in this case.> VanThat's not correct.Two things:1) if l = lcm(m,n) (we know a and b commute) then it is obvious that (ab)^l= e2) that doesn't say l is the smallest positive integer such that (ab)^l =e,ie that doesn't say (ab) has order l.--Julien Santini === Your solution looks fine to me.>The problem is, what if a^r and b^s are multiplicative inverses ... then>this could equal 1. But there is no reason to assume that they are inverses, since a and b arearbitrary elements--you are creating an unnecessary problem for yourself.Yes, it is possible that in some particular case that a^r b^s = 1, but the problemis for the general case, for any m and n.Van === >Your solution looks fine to me.>The problem is, what if a^r and b^s are multiplicative inverses ... then>this could equal 1. >But there is no reason to assume that they are inverses, since a and b are>arbitrary elements--you are creating an unnecessary problem for yourself.>Yes, it is possible that in some particular case that a^r b^s = 1, but the problem>is for the general case, for any m and n.Sorry, but this is nonsense. While there is no reason to assume theyare inverses, there is in principle no way to discard that situationeither. As such, you ->must<- consider that possibility aswell. Saying that the elements are arbitrary does not make itimpossible for that situation to occur. If your argument is to workfor ANY two elements, then it must also work in that case.-- ============================================ ===It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ======================================= =========Arturo Magidinmagidin@math.berkeley.edu===> One can show from the definitions of greatest common denominator (gcd) and> least common multiple (lcm) that> lcm(m,n) = [m,n] = mn/(m,n) = l (say) ; (m,n) = gcd(m,n) = g (say).> (This is a handy thing to remember too.)> so lg = mn ; m = qg ; n = rg since g divides both m and n.> All we need to know about q and r is that they are integers. Then> l = mn/g = r*q*g = rm = qn ; so> (ab)^l = (ab)^(rqg) = a^(mr) b^(nq) = e^r e^q = e> and the order of ab is l = lcm(m,n). Note the if (m,n) = 1> (i.e., are relatively prime), the order of the product ab is the> product of the orders of a and b since [m,n] = mn in this case.I should not have used the letter l here, it looks too much like the number one (1).Also, I apologize for reading only the problem and not your solution;> That's not correct.> Two things:> 1) if l = lcm(m,n) (we know a and b commute) then it is obvious that (ab)^l> = eWell, even you took a few lines to show it.> 2) that doesn't say l is the smallest positive integer such that (ab)^l =e,> ie that doesn't say (ab) has order l.True, I wasn't thinking very clearly when I posted it.I'll try to come up with a proof of this.Van === -- set theory limitsIn set theory, for the powerset P(S) of a set Slimits of set sequences Aj are defined as lim Aj = liminf Aj = limsup Ajprovided liminf Aj = limsup Aj, liminf Aj = /{ /{ Aj | j >= n } | n in N } limsup Aj = /{ /{ Aj | j >= n } | n in N } / /; cap cup; intersection unionDoes this definition of limit produce a topology for P(S) ?-- powerset topologyThe powerset topology for a powerset P(S) is definedas the topology produced by subbase sets of the form { A in P(S) | a in A }, { A in P(S) | a not in A }A base for this topology are the sets of the form { X | A subset X subset SB }for A,B finite subset S.A local base for U in P(S) are the sets of the form { X | A subset X subset SB }A finite subset U, B finite subset SUWe have the following theorems:The powerset topology for P(S) is homeomorphic to the producttopology of {0,1}^P(S), the space of characteristic functions. {0,1} has discrete topology.Thus P(S) is zero-dimensional compact Hausdorff.P(S) is 1st countable iff S countable iff P(S) is 2nd countable-- limit equivalenceThe powerset topology has the desired property A = set-theory-lim Aj iff A = topology-lim AjHence when S is countable, the powerset topology can be describedby set theory limits. When S is uncountable, I would conjecture 1) the definition of set theory limits can be extended to nets 2) limit equivalence will be preserved 3) the powerset topology can be described by net/set theory limits-- QuestionsHas any use of the powerset topology been made?Have these notions been extended or refined?Does this topic show up in modern analysis?Who else has consider the powerset topology?Comments most welcome as well as additions, refutations or requests for proofs.-- latticed ordered topological spacesAs P(S) is the protype for complete complemented atomic distributivelattices or complete atomic Boolean algebras, the powerset topology can beextended to those spaces.Upon using 'subset' for the order <=, of P(S), the powerset topologypresents a base of convex sets and the T_2-ordered property, that <=is closed subset P(S)xP(S). Thus it is an appropriately well manneredtopology for a partially ordered set.Order and Topology, http://at.yorku.ca/t/a/i/c/05.htmAn ordered topological space is a topological space (X,T) equipped with apartial order <=. Usual compatibility conditions between the topology andorder include convexity (T has a basis of order-convex sets) and theT_2-ordered property: <=, ie { (x,y) | x <= y }, is closed in XxX.---- === > -- set theory limits> In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are defined as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj,> liminf Aj = /{ /{ Aj | j >= n } | n in N }> limsup Aj = /{ /{ Aj | j >= n } | n in N }> / /; cap cup; intersection union> Does this definition of limit produce a topology for P(S) ?Perhaps that I misunderstood the question, but I would say that nodefinition of limit produces a topology. Consider the set R of thereals and take the following definition of limit: the only sequencesa_1, a_2, a_3, ... that posess a limit are those such that, for somereal number l, a_n = l for every n large enough; in that case, thelimit of the sequence is l.The problem here is that you can define at least two topologies in Rsuch that the definition of limit made above is the notion of limitthat corresponds to those topologies: the discrete topology and thetopology such that the closed sets are R, the finite sets and thecountable sets.I hope that this helps.Jose Carlos Santos === > liminf Aj = /{ /{ Aj | j >= n } | n in N } > limsup Aj = /{ /{ Aj | j >= n } | n in N }> Does this definition of limit produce a topology for P(S) ? >Perhaps that I misunderstood the question,Perhaps I should have asked, When does this ... >but I would say that no definition of limit produces a topology.A definition of limit together with requirement first countable would.The issue of first countable P(S) is consider 2nd & 3rd sections. >Consider the set R of the reals and take the following definition of >limit: the only sequences a_1, a_2, a_3, ... that posess a limit are >those such that, for some real number l, a_n = l for every n large >enough; in that case, the limit of the sequence is l.A sequence converges iff it's eventually constant. >The problem here is that you can define at least two topologies in R >such that the definition of limit made above is the notion of limit >that corresponds to those topologies: the discrete topology and theWhich is first countable. Indeed cl A = A = limit points of A. >topology such that the closed sets are R, >the finite sets and the countable sets.An uncountable cocountable space isn't 1st countable.Here again A = limit pts of A, howevercl A = R when A uncountable, = nulset when A countable.---- === >but I would say that no definition of limit produces a topology.> A definition of limit together with requirement first countable would.> The issue of first countable P(S) is consider 2nd & 3rd sections.I have this curious habit of reading texts from the top to the bottom, notfrom the bottom to the top. And besides, what you mention below is that acertain specific topology about which you state that it is first countable.How was I supposed to deduce from that that, in the first section, you werelooking for a first countable topology?>Consider the set R of the reals and take the following definition oflimit: the only sequences a_1, a_2, a_3, ... that posess a limit are>those such that, for some real number l, a_n = l for every n large>enough; in that case, the limit of the sequence is l.> A sequence converges iff it's eventually constant.If that was all that I wanted to write, I would have done so. But whatconstant but, furthermore, that the limit of such a sequence is thatconstant (and yes, I know that it could not be otherwise).>The problem here is that you can define at least two topologies in R>such that the definition of limit made above is the notion of limit>that corresponds to those topologies: the discrete topology and the> Which is first countable. Indeed cl A = A = limit points of A.Don't you think that it is much more simple to see directly from the definitionthat the discrete topology is first countable? All that you have to do is tosee that, for any point p, {{p}} is a fundamental system of neighborhoods ofp.>topology such that the closed sets are R,>the finite sets and the countable sets.> An uncountable cocountable space isn't 1st countable.Indeed, but I never said otherwise.Jose Carlos Santos === > -- set theory limits> In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are defined as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj,> liminf Aj = /{ /{ Aj | j >= n } | n in N }> limsup Aj = /{ /{ Aj | j >= n } | n in N }> / /; cap cup; intersection union> Does this definition of limit produce a topology for P(S) ?> Perhaps that I misunderstood the question, but I would say that no> definition of limit produces a topology. Consider the set R of the> reals and take the following definition of limit: [...]> The problem here is that you can define at least two topologies in R> such that the definition of limit made above is the notion of limit> that corresponds to those topologies [...]Generally, one refers to the smallest topology where the stated limitsexist. Cf. the concept of induced topology (e.g., product topology). In particular, the topology whereto William refers is the naturalhomeomorphism of the product space {0,1}^A, where {0,1} is discrete. He states the result that the only topological limits of sequences inthis space are those which are also set-limits.William asked some interesting questions, some of which I posted inearlier threads. (Much of this was hidden in the thread on the set ofcountable ordinals.) In particular, is this same topology thesmallest where convergence of general set-nets are also topologicallimits? If so, are all topological limits also set-limits? Also, isthis the same as the topology induced by all f in [0,1]^A suchthat lim f a = f (lim a) for all set-convergent nets a in A?Stephen J. Herschkorn === > liminf Aj = /{ /{ Aj | j >= n } | n in N }> limsup Aj = /{ /{ Aj | j >= n } | n in N }Where are such limits used? Do they show up in modern analysis? >Generally, one refers to the smallest topology where the stated >limits exist. Cf. the concept of induced topology (e.g., product >topology). In particular, the topology whereto William refers is the >natural homeomorphism of the product space {0,1}^A, where {0,1} is >discrete. He states the result that the only topological limits of >sequences in this space are those which are also set-limits. >William asked some interesting questions, some of which I posted in >earlier threads. In particular, is this same topology the smallest >where convergence of general set-nets are also topological limits?I've looked closer into the details of this and I see no need basicneed to change the proofs of set limits equivalent topology limitsto a the proofs for net-set limits quivalent net-topology limitsThis is because the proofs when casted into the expressions of eventually inare identical; the only change to be made be the invisible change sequence-eventually to net-eventually.Again the key to the details is to use topology convergence to xhappens when a sequence/net is eventually in every subbase set of x. andlim Aj = A iff for all x in A, some n with for all j >= n, x in Aj for all x not in A, some n with for all j >= n, x not in Ajequivalently for all x in A, x eventually in (Aj)_j for all x not in A, x eventually in (SAj)_j >If so, are all topological limits also set-limits? Also, is >this the same as the topology induced by all f in [0,1]^A such >that lim f a = f (lim a) for all set-convergent nets a in A?Now that I don't see. I'll concur on {0,1}^P(A) for discrete {0,1}; but on [0,1]^A for [0,1] subspace reals?This you needs clarify.For {0,1}^P(S) we have, the projections are p_a:P(S) -> {0,1}, c_A -> c_A(a) = 'a in A'which are continuous and thus preserve limits, ie t-lim f(Aj) = f(t-lim Aj) = f(set-lim Aj)Would that suffice or do you want ? set-lim f(Aj) = f(set-lim Aj)But as the projections are into discrete {0,1} the notionof set-limits in the codomain of the projections experiences suchchaotic turbulance that such digression vanishes of its own accord.---- === > -- set theory limits> In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are defined as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj,> liminf Aj = /{ /{ Aj | j >= n } | n in N }> limsup Aj = /{ /{ Aj | j >= n } | n in N }> / /; cap cup; intersection union> Does this definition of limit produce a topology for P(S) ?> [...]> Generally, one refers to the smallest topology where the stated limits> exist. Cf. the concept of induced topology (e.g., product topology). > In particular, the topology whereto William refers is the natural> homeomorphism of the product space {0,1}^A, where {0,1} is discrete. > He states the result that the only topological limits of sequences in> this space are those which are also set-limits.>Correction: The *smallest* topology is the indiscrete one. Here wewant the *largest* topology whereunder the sequences converge to thedesignated limits. Let X = N U {infty} under the usual topology. The topology we seek is exactly that coinduced by all functions f inP(A)^X such that lim f = the set limit. This does not contradict thestatement about {0,1}^A. (At least I *think* there is no larger suchtopology.)Given a set Y and a collection F = {(X,f): X is a topologicalspace and f in Y^X}, the topology coinduced on Y by F is thelargest topology such that (X,f)in F implies f is continuous. Theprototypical example is the quotient topology, coinduced by thecanonical map to equivalence classes.I confess to not remembering right now the exact proof that such alargest topology exists. I.e., given a collection t of topologiessuch that each f is continuous with each topology in t, why is itthat each f is continuous in the topology generated by Ut? The prooffor induced topologies (which do refer to a smallest topology =intersection of topologies) is much more straightforward.Stephen J. Herschkorn === > provided liminf Aj = limsup Aj,> Generally, one refers to the smallest topology where the stated > limits exist. Cf. the concept of induced topology (e.g., product > topology). In particular, the topology whereto William refers is > the natural homeomorphism of the product space {0,1}^A, where {0,1} > is discrete. He states the result that the only topological limits > of sequences in this space are those which are also set-limits.>A needs be taken as P(S). The product topology is the intitial, I thinkthat's the induced, topology of the projections. By definition it's thesmallest topology making all the projections continuous. As the latticeof topologies is a complete lattice, (it's also complemented,non-distributive, atomic) such a smallest can be found and then proven tohave the continuity property. Easier is to construct a such topology andshow it's the smallest. For perspective, the largest topologycontinuingfying all the projections is the discrete topology. >Correction: The *smallest* topology is the indiscrete one. Here we >want the *largest* topology whereunder the sequences converge to the >designated limits. Let X = N U {infty} under the usual topology.Which is? If X is going to simulate a sequence or net, then it needs acountable set order isomorphic to N or an updirected domain of the net. >The topology we seek is exactly that coinduced by all functions f in >P(A)^X such that lim f = the set limit. This does not contradict >the statement about {0,1}^A. (At least I *think* there is no larger >such topology.)How are you using A? Do you actually mean P(P(S))^X or P(S)^X ? >Given a set Y and a collection F = {(X,f): X is a topological >space and f in Y^X}, the topology coinduced on Y by F is the >largest topology such that (X,f)in F implies f is continuous. >The prototypical example is the quotient topology, coinduced by the >canonical map to equivalence classes.f:X -> Y is embedding X into Y the disjoint sum of F copies of X.Each f in an embedding of an X into a unique portion of Y.Another application is the embeddings of subspaces into the mother space.The quotient space comes when the collection of functions is just one.Are you suggesting some sort of multiple quotients of a space? >I confess to not remembering right now the exact proof that such a >largest topology exists. I.e., given a collection t of >topologies such that each f is continuous with each topology in >t, why is it that each f is continuous in the topology generated by >Ut? The proof for induced topologies (which do refer to a smallest >topology = intersection of topologies) is much more straightforward.The smallest topology is the indiscrete topology. For the details of thefinal topology, I'd suggest a similar approach as for the initialtopology.---- === > In set theory, for the powerset P(S) of a set S> limits of set sequences Aj are defined as> lim Aj = liminf Aj = limsup Aj> provided liminf Aj = limsup Aj,> Generally, one refers to the smallest topology where the stated> limits exist. Cf. the concept of induced topology (e.g., product> topology). In particular, the topology whereto William refers is> the natural homeomorphism of the product space {0,1}^A, where {0,1}> is discrete. He states the result that the only topological limits> of sequences in this space are those which are also set-limits.> A needs be taken as P(S). The product topology is the intitial, I think> that's the induced, topology of the projections. By definition it's the> smallest topology making all the projections continuous. As the lattice> of topologies is a complete lattice, (it's also complemented,> non-distributive, atomic) such a smallest can be found and then proven to> have the continuity property. Easier is to construct a such topology and> show it's the smallest. For perspective, the largest topology> continuingfying all the projections is the discrete topology.Sorry, by A I meant S. And you are incorrect about A being P(S). The topology you describe makes P(S) homeomorphic to the product{0,1}^S under the product topology, where the homeomorphism is theusual identification of subsets with characteristic functions. Morebelow.>Correction: The *smallest* topology is the indiscrete one. Here we>want the *largest* topology whereunder the sequences converge to the>designated limits. Let X = N U {infty} under the usual topology.> Which is? If X is going to simulate a sequence or net, then it needs a> countable set order isomorphic to N or an updirected domain of the net.By the usual topology, I mean the one with the base {{n}, N[n,infty]: n in N}. Here, N[n,infty] denotes {n, n+1, n+2,... infty}, Anotherway to put it is X is the ordinal omega + 1 under the ordertopology.>The topology we seek is exactly that coinduced by all functions f in>P(A)^X such that lim f = the set limit. This does not contradict>the statement about {0,1}^A. (At least I *think* there is no larger>such topology.)> How are you using A? Do you actually mean P(P(S))^X or P(S)^X ?A = S, again. So a better way of describing the coinducing functions f are those in P(S)^X such that f(n) -> f(infty) (set-wise) as n-> infty.I am noting that this coinduced topology seems to be the same as thehomeomorph of the product topology on {0,1}^S (which by definition isan induced topology - by definition, the smallest topology satisfyingcertain conditions).BTW, given any set Y and a collection of sequences with designatedlimits in Y, the (largest) topology generated by thesesequence-limits is again the coinduced topology by the correspondingfunctions in Y^(omega + 1).>Given a set Y and a collection F = {(X,f): X is a topological>space and f in Y^X}, the topology coinduced on Y by F is the>largest topology such that (X,f)in F implies f is continuous.>The prototypical example is the quotient topology, coinduced by the>canonical map to equivalence classes.> f:X -> Y is embedding X into Y the disjoint sum of F copies of X.> Each f in an embedding of an X into a unique portion of Y.> Another application is the embeddings of subspaces into the mother space.> The quotient space comes when the collection of functions is just one.> Are you suggesting some sort of multiple quotients of a space?I am just stating the standard definition of the coinduced topology.>I confess to not remembering right now the exact proof that such a>largest topology exists. I.e., given a collection t of>topologies such that each f is continuous with each topology in>t, why is it that each f is continuous in the topology generated by>Ut? The proof for induced topologies (which do refer to a smallest>topology = intersection of topologies) is much more straightforward.> The smallest topology is the indiscrete topology. For the details of the> final topology, I'd suggest a similar approach as for the initial> topology.I think you misinterpret what I was saying. Being redundant here,Given a set X and a collection F of functions each of whose domainis X and whose range is some topological space, the topology on X induced by F is the smallest topology on X such that all thefunctions in F are continuous.- Note that the ranges of the functions in F need not be the samespaces.- Such a topology exists: It is the intersection of all topologies on X whereunder all the functions in F are continuous. Thecollection of such topologies is not empty, since it contains thediscrete topology.- Example 1. The product topology is the topology induced by theprojection maps.- Example 2. The relative topology is the topology induced by theinclusion map.Given a set Y and a collection F of functions each of whose domainis some topological space and whose range is Y, the topology on Y coinduced by F is the largest topology on Y such that all thefunctions in F are continuous.- Note that the domains of the functions in F need not be the samespaces.- Earlier, I wondered why we know such a topology exists. Havingslept on it, I realize the answer: Let t be the collection oftopologies on Y whereunder all functions in F are continuous. t is not empty, since it contains the indiscrete topology. Thecoinduced topology is the smallest topology containing Ut. It iseasy to check that if the inverse of a function maps sets in a subbaseto open sets, then it is continuous. (The last sentence is what I wasmissing earlier.)- Example. The quotient topology is the topology coinduced by thecanonical map to an equivalence class. In fact, I think this is themost general coinduced topology. Given F as in the definition, say x and y in Y are equivalent iff f(x) = f(y) for all f in F. I am not sure about this; at least this works if F is a singleton.One more example to compare the two concepts:Let X = [0,1] under the usual topology.Let Y = {(x,y) in R^2: x^2 + y^2 = 1}, the unit circle.Let Z = [-1,1] under the usual topology.Then the usual topology on Y can be generated in the following twoways:It is that induced by the projection maps from Y to Z^2.It is also that coinduced by f: X -> Y, x|->(cos 2pi x, sin 2pi x).Stephen J. HerschkornX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBSDc1w02521; === There are enough and coherent reasoning to prove that it is not possible to imagine the list of all real numbers without running against the nature of the natural numbers. For instance, if we admit the one-to-one correspondence between N and R, then we are confronting two different kinds of infinity, the potential infinite represented by the asymptotic approximation of the naturals to the infinite (never reaching it), and the actual infinite represented by the set of all real numbers; so that, a complete bijection between both sets is not possible. With another simple reasoning we arrive to the conclusion that the acceptance of the existence of that list would imply naturals with an infinite value (cardinal). On the other hand, there are constructive methods (at least I know one of them) that by simple induction they proof that such a list will require naturals with an infinite number of figures.Therefore, my question is this: If the list of all real numbers is not a valid concept, why must we accept it as premise in the Cantors proof of the diagonal? The theory of transfinite numbers would be able to be an air castle with such an argument in its foundations. Nicolas de la Foz === > For instance, if we admit the one-to-one correspondence between N and R,> then we are confronting two different kinds of infinity, the potential infinite> represented by the asymptotic approximation of the naturals to the infinite> (never reaching it), and the actual infinite represented by the set of all real> numbers;How do you handle the fact there are more than two grades of infinity? For example, the set F of all functions f: R->R is larger than theset of reals R, such that you cannot place these into one-to-onecorrespondence either. Yet you call the smaller set R an actualinfinity. What does that make F? An even more actual infinity?And how about the power set of F? That's bigger still.X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oftX-Terminate: SPA(GIS) === at 01:38 PM, nico80@jazzfree.com (Nicolas de la Foz) said:>There are enough and coherent reasoning to prove that it is not>possible to imagine the list of all real numbers without running>against the nature of the natural numbers.No. There is only posturing and the use of words in contexts wherethey have no meanings.>For instance, if we admit the one-to-one correspondence between N>and R, We can't, because there is no such 1-1 correspondence.>the potential infiniteWhat is a potential infinite and what does it have to do with eitherN or R?>asymptotic approximation of the naturals to the infiniteDo you believe that the above phrase has any meaning? If so, why?>the actual infiniteWhat is the actual infinite and what does it have to do with eitherN or R?>Therefore, my question is this: If the .99list of all real numbers.9a is>not a valid concept, why must we accept it as premise in the>Cantors proof of the diagonal? We don't. We show that such a premise would lead to a contradictionand is therefor false.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > There are enough and coherent reasoning to prove that it is not possibleto imagine the list of all real numbers without running against the natureof the natural numbers. For instance, if we admit the one-to-onecorrespondence between N and R, then we are confronting two different kindsof infinity, the potential infinite represented by the asymptoticapproximation of the naturals to the infinite (never reaching it), and theactual infinite represented by the set of all real numbers; so that, acomplete bijection between both sets is not possible. With another simplereasoning we arrive to the conclusion that the acceptance of the existenceof that list would imply naturals with an infinite value (cardinal). On theother hand, there are constructive methods (at least I know one of them)that by simple induction they proof that such a list will require naturalswith an infinite number of figures.> Therefore, my question is this: If the list of all real numbers is not avalid concept, why must we accept it as premise in the Cantor's proof of thediagonal? The theory of transfinite numbers would be able to be an aircastle with such an argument in its foundations.This is one of my own pet hobbyhorses. I am a firm believer in only oneinfinity, which is not even signed.Every presentation of the Cantor argument has turned upon a (partial) set ofreals, which is placed in one-to-one correspondence with a (partial) set ofintegers. Here is the example from G.9adel, Escher, Bach: 1 .1 4 1 5 9 2 6 5 3 2 .3 3 3 3 3 3 3 3 3 3 .7 1 8 2 8 1 8 2 8 4 .4 1 4 2 1 3 5 6 2 5 .5 0 0 0 0 0 0 0 0The argument then goes by constructing a new real by taking the elements ofthe diagonal and changing every digit. The argument then goes that (a) byconstruction, this new real is not in the list of reals (b) by construction,there is a real against every integer, (c) THEREFORE, there are more realsthan integers, even though the list of integers is infinite.My principal observation is that while the column containing the integers isquite obviously ordered, the column containing the reals is not. It is thisdisorder in the example list of reals that is the clincher. Once an orderingscheme is imposed on the reals, a true one-to-one correspondence,expressible as a simple formula, can be constructed. One example which hasbeen mooted here on a number of occasions is to reverse the order of thedecimal digits and disregard the decimal point [1], thus the real 0.5 mapsto the integer 5 as per the last line of the example, the real 0.890625 mapsto the integer 526098, etc.OK so all non-terminating decimals, recurring or not, map to infinity bythis method. No problem, there is enough room at infinity for all of them.Each successive Cantor diagonalisation of the list will give a newnonterminating [2] real and a new infinite integer to map it to.[1] This confines the reals to the range 0 to 1, but only a trivialmodification is required to remove this restriction. Simply pass theinfinite range of reals through a mathematical function that maps them to afinite range in 1:1 correspondence. The tanh function is a good one to usefor this purpose.[2] The diagonalisation must give a non-terminating decimal since allterminating ones are mapped to the finite integers by construction, so theyare all already in the list.--Paul V. S. TownsendInterchange the alphabetic elements to replyX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge: Micro$oft === at 03:46 PM, Prai Jei said:>My principal observation is that while the column containing the>integers is quite obviously ordered, the column containing the reals>is not.It clearly *IS* ordered, by the very act of arranging them into acolumn.>Once an ordering scheme is imposed on the reals, a true one-to-one >correspondence, expressible as a simple formula, can be constructed.You are confusing ordering with well ordering. Further, you areconfusing the statement to be disproved with a statement that has beenproven. In fact the diagonal argument proves that such a column isimpossible.>One example which has>been mooted here on a number of occasions is to reverse the order of>the decimal digits and disregard the decimal point [1],And what does 1/3 map to?>OK so all non-terminating decimals, recurring or not, map to>infinityWhat is infinity Does 1/3 map to the same infinity as 1/6?>there is enough room at infinity for all of them.What does that mean?>infinite integerWhat is an infinite integer?What you're doing isn't Mathematics; it isn't even Philosophy. Defineyour terms if you wish to be taken seriously, then use them only in afashion consistent with how you defined them.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === > This is one of my own pet hobbyhorses. I am a firm believer in only one> infinity, which is not even signed.> Every presentation of the Cantor argument has turned upon a (partial) set of> reals, which is placed in one-to-one correspondence with a (partial) set of> integers. Here is the example from G.9adel, Escher, Bach:> 1 .1 4 1 5 9 2 6 5 3> 2 .3 3 3 3 3 3 3 3 3> 3 .7 1 8 2 8 1 8 2 8> 4 .4 1 4 2 1 3 5 6 2> 5 .5 0 0 0 0 0 0 0 0> The argument then goes by constructing a new real by taking the elements of> the diagonal and changing every digit. The argument then goes that (a) by> construction, this new real is not in the list of reals (b) by construction,> there is a real against every integer, (c) THEREFORE, there are more reals> than integers, even though the list of integers is infinite. My principal observation is that while the column containing the integers is> quite obviously ordered, the column containing the reals is not. It is this> disorder in the example list of reals that is the clincher. Once an ordering> scheme is imposed on the reals, a true one-to-one correspondence,> expressible as a simple formula, can be constructed. One example which has> been mooted here on a number of occasions is to reverse the order of the> decimal digits and disregard the decimal point [1], thus the real 0.5 maps> to the integer 5 as per the last line of the example, the real 0.890625 maps> to the integer 526098, etc.> I'm having trouble understanding the flow of your response. This isthe simple formula, from [0,1] to Z, which you purport to fix below,since it has been mooted here before?> OK so all non-terminating decimals, recurring or not, map to infinity by> this method. No problem, there is enough room at infinity for all of them.It's at this point that I begin not to understand. What does this haveto do with the ordering you induce?Unfortunately, infinity isn't in the standard definitions of Z I'mfamiliar with, but this is not so much a hurdle. Besides, even if itwere, you said you were a firm believer in a single unsigned infinity,so every nonterminating real in [0,1] would be mapped to this singleelement; this mapping isn't a bijection and has little to do withcardinality.> Each successive Cantor diagonalisation of the list will give a new> nonterminating [2] real and a new infinite integer to map it to.> Says the firm believer in a single infinity? Are you suggesting thatwe define new elements and union them with Z to make this mappingwork? The problem is, we'll be adding an uncountable number ofinfinities to Z, so the set to which you're mapping won't be of thesame cardinality as Z. In other words, you've changed your mappingfrom [0,1]->Z to [0,1]->(Z union I), where I is the new set ofinfinities you define. Since there is no bijection between Z and (Zunion I), this shows us that they don't have the same cardinality.You're back where you began and haven't fixed this simple mapping.infinity. The argument against this fix still stands if you justlook at (Z union I), where we say that I is the set of infiniteintegers, which have never been part of Z and aren't countable. Allyou've really done is removed a decimal point from the analysis.> [1] This confines the reals to the range 0 to 1, but only a trivial> modification is required to remove this restriction. Simply pass the> infinite range of reals through a mathematical function that maps them to a> finite range in 1:1 correspondence. The tanh function is a good one to use> for this purpose.Let's look at f(x) = (tanh(x)+1)/2. What real number is mapped to 1and what is mapped to 0? I was under the impression that there were nosigned infinities in your system, in which case we're left with (0,1]and there is a quick detail to work out.> [2] The diagonalisation must give a non-terminating decimal since all> terminating ones are mapped to the finite integers by construction, so they> are all already in the list. === >[self-contradictory OP snipped]>This is one of my own pet hobbyhorses. I am a firm believer in only one>infinity, which is not even signed.I'm a firm believer in a flat earth. Doesn't make the earth flat.>Every presentation of the Cantor argument has turned upon a (partial) set of>reals, which is placed in one-to-one correspondence with a (partial) set of>integers. Here is the example from G.9adel, Escher, Bach:> 1 .1 4 1 5 9 2 6 5 3> 2 .3 3 3 3 3 3 3 3 3> 3 .7 1 8 2 8 1 8 2 8> 4 .4 1 4 2 1 3 5 6 2> 5 .5 0 0 0 0 0 0 0 0>The argument then goes by constructing a new real by taking the elements of>the diagonal and changing every digit. The argument then goes that (a) by>construction, this new real is not in the list of reals (b) by construction,>there is a real against every integer, (c) THEREFORE, there are more reals>than integers, even though the list of integers is infinite.>My principal observation is that while the column containing the integers is>quite obviously ordered, the column containing the reals is not. This has no relevance whatever. _By definition_, saying that there are more reals than integers means that if you have a mappingof integers to reals then there is at least one real not in the rangeof the mapping. Nothing there about columns being ordered.>It is this>disorder in the example list of reals that is the clincher. Once an ordering>scheme is imposed on the reals, a true one-to-one correspondence,>expressible as a simple formula, can be constructed. One example which has>been mooted here on a number of occasions is to reverse the order of the>decimal digits and disregard the decimal point [1], thus the real 0.5 maps>to the integer 5 as per the last line of the example, the real 0.890625 maps>to the integer 526098, etc.>OK so all non-terminating decimals, recurring or not, map to infinity by>this method. No problem, there is enough room at infinity for all of them.>Each successive Cantor diagonalisation of the list will give a new>nonterminating [2] real and a new infinite integer to map it to._By definition_ there is no such thing as an infinite integer. You'renot demonstrating anything here, all you're doing is giving wordsnew definitions.I can _prove_ that the Earth is flat. People point out that it's round, but that's no problem, because the desk in my officeis flat! Is that a proof that the Earth is flat? No, because thedesk in my office is not the Earth.If you think that there is a 1-1 correspondence betweenthe reals and the (finite) integers you're simply wrong. Butit doesn't look like you think that - it looks like you agreethere is no such 1-1 correspondence. If so then you _do_agree that there are more reals than integers, assumingwe give all the words their standard meanings - what youthink you accomplish by using words to mean thingsother than what they _do_ mean is not clear to me.>[1] This confines the reals to the range 0 to 1, but only a trivial>modification is required to remove this restriction. Simply pass the>infinite range of reals through a mathematical function that maps them to a>finite range in 1:1 correspondence. The tanh function is a good one to use>for this purpose.>[2] The diagonalisation must give a non-terminating decimal since all>terminating ones are mapped to the finite integers by construction, so they>are all already in the list.************************David C. Ullrich === >[self-contradictory OP snipped]>This is one of my own pet hobbyhorses. I am a firm believer in only one>infinity, which is not even signed.>I'm a firm believer in a flat earth. Doesn't make the earth flat.[snip argumentation why Cantor's diagonal is not wrong]I'd like to take a few moments to elaborate on a topic that irritatesme to no end:CANTOR'S PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVIL'S SPAWN!!!Well maybe not quite but it's certainly worse than L'Hospital's ruleon the scale of things that are so obvious no one can possiblymisunderstand them that are consequently misunderstood all the time.This is not aided by the fact that it's so deceptively simple thatit gets tacked on every single introductory freshman/high school mathtext and popularization from here to eternity.As a proof, it's... well, elegant. However it's so elegant that theunderlying finesse usually goes completely unnoticed, which means weget a steady stream of cranks^Walternative thinkers who are convincedthat the proof is wrong because . It's like mathematics is a house where one of the paintings hasa funny optical illusion that makes the painting look like it's tiltedwhile it's perfectly straight, and every joker who waltzes in pointsout that the painting looks tilted and thinks no one else has pointedthis out before.So please, to those in the position of influencing textbookpublishers, would it please be possible to remove this deceptivelysimple yet simply deceptive argument from texts where it is notrelevant and to replace it with a more formal two-page proof includingno more than five understandable English words and at least twohomeomorphisms? I think that would go a long way.And with these words we return to your scheduled gigantic Cantorthread... === >[self-contradictory OP snipped]>This is one of my own pet hobbyhorses. I am a firm believer in only one>infinity, which is not even signed.>I'm a firm believer in a flat earth. Doesn't make the earth flat.>[snip argumentation why Cantor's diagonal is not wrong]>I'd like to take a few moments to elaborate on a topic that irritates>me to no end:>CANTOR'S PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVIL'S SPAWN!!!>Well maybe not quite but it's certainly worse than L'Hospital's rule>on the scale of things that are so obvious no one can possibly>misunderstand them that are consequently misunderstood all the time.>This is not aided by the fact that it's so deceptively simple that>it gets tacked on every single introductory freshman/high school math>text and popularization from here to eternity.>As a proof, it's... well, elegant. However it's so elegant that the>underlying finesse usually goes completely unnoticed, which means we>get a steady stream of cranks^Walternative thinkers who are convinced>that the proof is wrong because here>. It's like mathematics is a house where one of the paintings has>a funny optical illusion that makes the painting look like it's tilted>while it's perfectly straight, and every joker who waltzes in points>out that the painting looks tilted and thinks no one else has pointed>this out before.>So please, to those in the position of influencing textbook>publishers, would it please be possible to remove this deceptively>simple yet simply deceptive argument from texts where it is not>relevant and to replace it with a more formal two-page proof including>no more than five understandable English words and at least two>homeomorphisms? I think that would go a long way.I can't decide whether you're serious. If you are then this is avery curious attitude... if you're not then you got me.>And with these words we return to your scheduled gigantic Cantor>thread...************************David C. Ullrich === [self-contradictory OP snipped]>This is one of my own pet hobbyhorses. I am a firm believer in only one>infinity, which is not even signed.>I'm a firm believer in a flat earth. Doesn't make the earth flat.> [snip argumentation why Cantor's diagonal is not wrong]> I'd like to take a few moments to elaborate on a topic that irritates> me to no end: CANTOR'S PROOF OF THE UNCOUNTABILITY OF REALS IS THE DEVIL'S SPAWN!!!> Well maybe not quite but it's certainly worse than L'Hospital's rule> on the scale of things that are so obvious no one can possibly> misunderstand them that are consequently misunderstood all the time.> This is not aided by the fact that it's so deceptively simple that> it gets tacked on every single introductory freshman/high school math> text and popularization from here to eternity.> Cantor's proof is the simplest, most accessible proof of something that's counterintuitive. That's why it stirs up the cranks so. === > There are enough and coherent reasoning to prove that it is not >possible to imagine the list of all real numbers without running >against the nature of the natural numbers. For instance, if we >admit the one-to-one correspondence between N and R, then we are >confronting two different kinds of infinity, the potential infinite >represented by the asymptotic approximation of the naturals to the >infinite (never reaching it), and the actual infinite represented by >the set of all real numbers; so that, a complete b ijection between >both sets is not possible. With another simple reasoning we arrive >to the conclusion that the acceptance of the existence of that list >would imply naturals with an infinite value (cardinal).So are you saying there are only finitely many natural numbers? > On the other >hand, there are constructive methods (at least I know one of them) >that by simple induction they proof that such a list will require >naturals with an infinite number of figures.> Therefore, my question is this: If the list of all real numbers >is not a valid concept, why must we accept it as premise in the Cantors >proof of the diagonal? The theory of transfinite numbers would be able >to be an air castle with such an argument in its foundations. > Nicolas de la FozI'm not sure I completely understand your problem with this. The diagonalargument shows that any countably infinite set of Real numbers cannotcontain all of them. And any countable set can be thought of as a 'list'. What's wrong with that? What asymptotic approximation of the infinite are you talking about? Itreally reads as though you are saying there are not infinitely manynatural numbers. Or that there isn't _an_ infinitely large natural number?If you don't like proof by contradiction, don't assume that the reals arecountable, but show that _any_ countable set of them is incomplete. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBSDc1102527; === hi im am a late nite science crack pot that is very interested in finding out what and how r universe was made and how it works i would like to get as mutch 411 on the subject as u can send me please send me as much as u can thank u === >would like to get as mutch 411 on the subject as u can send me In the movie Hercules, at some point an accident has happened and some one gives the throw away line Somebody call IXIIThat cracked me up. Mitch === >would like to get as mutch 411 on the subject as u can send me > In the movie Hercules, at some point an accident has happened and some > one gives the throw away