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RE: [math-fun] AB = B.AA seems to imply commutativity?
by Torgerson, Mark D 14 Feb '06

14 Feb '06

After David's nice little proof below, he writes >so, while not everything has to be idempotent, the multiplication action of every element is the same as that of its square. Of course, without associativity, this does not mean that the multiplication action is idempotent! Maybe a bit more can be said, or at least said differently. aa=a(aa)=(aa)(aa) So squares of elements are idempotent. Mark -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov(a)mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of David P. Moulton Sent: Tuesday, February 14, 2006 7:38 AM To: math-fun Subject: Re: [math-fun] AB = B.AA seems to imply commutativity? On Tue, 14 Feb 2006, Schroeppel, Richard wrote: > I've been exploring the binary operator *, which satisifies the rule > A*B = B*(A*A). I like to write it in dot notation as AB = B.AA . > > ... > > Empirically, and surprisingly (to me at least), all 85196 tables are > commutative. Ditto for smaller tables. And also, apparently there's > always at least one idempotent element. In fact, this axiom implies commutativity, since we have ab = b(aa) = (aa)(bb) = (bb)[(aa)(aa)] = (bb)[a(aa)] = (bb)(aa) = a(bb) = ba. Therefore, we also have ab = (aa)b, so, while not everything has to be idempotent, the multiplication action of every element is the same as that of its square. Of course, without associativity, this does not mean that the multiplication action is idempotent! David _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] AB = B.AA seems to imply commutativity?
by dasimov＠earthlink.net 14 Feb '06

14 Feb '06

David Moulton writes: << ab = b(aa) = (aa)(bb) = (bb)[(aa)(aa)] = (bb)[a(aa)] = (bb)(aa) = a(bb) = ba. >> Nice one! --Dan

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[math-fun] AB = B.AA seems to imply commutativity?
by Schroeppel, Richard 14 Feb '06

14 Feb '06

I've been exploring the binary operator *, which satisifies the rule A*B = B*(A*A). I like to write it in dot notation as AB = B.AA . There are 85196 non-isomorphic tables of 5 elements that satisfy the rule. (Roughly 5^10/5!). It's clear enough that an idempotent table (satisfying the rule AA=A) will mean that AB = B.AA can be morphed to AB = BA . And it's also clear that any table which is both idempotent and commutative will also satisfy AB = B.AA . However, there are lots of tables that are not fully idempotent. Empirically, and surprisingly (to me at least), all 85196 tables are commutative. Ditto for smaller tables. And also, apparently there's always at least one idempotent element. Anyway, if it can be proven that all *-tables are commutative, and that the algebraic relation AB=B.AA does not algebraically imply AB=BA, this would provide a fine example of a truth about the system that can't be proven within the system. (Maybe because it's only true in the finite case?) Rich

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Re: [math-fun] AB = B.AA seems to imply commutativity?
by dasimov＠earthlink.net 14 Feb '06

14 Feb '06

Rich writes: << I've been exploring the binary operator *, which satisifies the rule A*B = B*(A*A). I like to write it in dot notation as AB = B.AA . >> (Groups won't help since any group satisfying AB = B.AA is clearly trivial.) << There are 85196 non-isomorphic tables of 5 elements that satisfy the rule. (Roughly 5^10/5!). Empirically, and surprisingly (to me at least), all 85196 tables are commutative. Ditto for smaller tables. . . . . . . (Maybe because it's only true in the finite case?) >> I just learned the word "magma" for a set S together with a binary operation *. Call this magma (S,*). QUESTION for Rich: Or are you actually assuming |S| <= 5 Or, as seems more likely, are you considering all magmas (S,*) that are *generated* by <= 5 elements of S ? --> If the latter, any idea how to tell which tables yield a finite S vs. an infinite S ? ---------------------------------------------------------------------------------- In particular there is a "free magma" FM(S) on any set of generators S. (Cf. Wikipedia, "Magma".) This is of course all finite strings of elements of S with all possible parenthesizations (or in Rich's notation, all periodizations), and A*B = A.B is just concatenation. Then there should exist a "quotient magma" which is just FM(S) with specified identifications, such as M = FM({x,y}) / AB ~ B.AA If this M = FM({x,y}) / AB ~ B.AA turns out to be commutative, then it should be true that any magma with the axiom AB = B.AA is commutative. QUESTION: Is this M commutative? --Dan

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Re: [math-fun] parity of floor(2^n / n)
by Michael Reid 13 Feb '06

13 Feb '06

> mreid>> (This may be asking way too much: I don't even know > > how to prove that the floor of 2^n / n is odd for > > infinitely many values of n.) > > Note that n = A090659 plus its conjectured continuation, > 25, 91, 703, 1891, 12403, 38503,79003 ,88831 ,146611,188191, 218791, 269011, 286903, 385003, 497503, 597871, 736291, 765703, 954271, 1056331, 1314631, 1869211, 2741311, 3270403, 3913003, 4255903 ,4686391 ,5292631 > > gives all odds. > --rwg i didn't quite follow this. apart from the first term (25), the others are all of the form pq where p is a prime = 3 mod 4 , and q = 2 p - 1 . there is a comment in the online encyclopedia that later terms in the sequence are conjectured to be of this form (although it does not mention that p = 3 mod 4 ). for such numbers, we have 2^(pq) = 0 mod 2 , 2^(pq) = 2^q = 2 mod p , and 2^(pq) = 2^p = -2 mod q (using the fact that 2 is a quadratic non-residue mod q ). put these together to get 2^(pq) = (2p - 4) q - 2 mod 2pq . if p > 3 , then the remainder, (2p - 4) q - 2 , is in the range [pq, 2pq) , so indeed floor(2^(pq) / pq) is odd. (this also explains why 15 doesn't work.) conjecturely, there are infinitely many primes p = 3 mod 4 , for which 2p - 1 is also prime, but i don't believe it's been proven. nevertheless, these terms should grow slower (i.e. there are more of them) than those in the provably infinite "sequence" i gave, whose plateaus grow doubly exponentially. none of this speaks to jim's original question, which asked if, as N goes to infinity, floor(2^n / n) is even for asymptotically half the positive integers less than N . mike

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[math-fun] Half-pipes & cycloids
by Henry Baker 13 Feb '06

13 Feb '06

While watching Shaun White win the gold medal in the "half-pipe" snowboarding event at the Olympics yesterday (on TV), it occurred to me that the best cross section for the half-pipe would be the cycloid. The brachistochrone property of the cycloid should allow the snowboarder/skateboarder to do stunts, each of which take approximately the same time & thereby set up a better rhythm. I did some searching on the web, and discovered that the current halfpipe curve has evolved from a true semicircle to a flat area with two quarter circle "corners". So it appears that the "half-pipe" may eventually evolve into a cycloid. Regardless of the suggestions in the references below about the bottom of the half-pipe being flat, the actual contour of the snow in the Torino half-pipe appears to me to be much more like a cycloid. I guess a question for the DIY crowd would be how to describe to your son/nephew/grandson the easiest way to construct a proper cycloid without building a huge wheel. Is there a way to construct a large cycloid using strings & pegs, etc., in a manner analogous to constructing an ellipse? (I guess this discussion provides one answer to high school kids who ask why anyone would want to study math.) http://en.wikipedia.org/wiki/Half-pipe http://www.abc-of-snowboarding.com/snowboardinghalfpipe.asp http://mathworld.wolfram.com/Cycloid.html

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[math-fun] If you can't paint good,
by R. William Gosper 11 Feb '06

11 Feb '06

paint big. www.tweedledum.com/rwg/hyperb.GIF is a large (4800x3500, 2.27MB, too big for IE?) captioned demonstration that the hyperbola as conic section is the locus of constant distance difference. (I deleted the monster bmp.) --rwg

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RE: [math-fun] how many 4x4x1 blocks
by Pacher Christoph 10 Feb '06

10 Feb '06

Regarding the attachment, no need to worry! this is the not-so-dangerous content of the attachment: _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun Just save it and open it with an ascii editor to convince yourself. I love linux ;-) regards Christoph -- DI Christoph Pacher Optoelectronic Engineering - Quantum Technologies ARC Seibersdorf research GmbH Tech Gate Tower, 3rd Floor A-1220 Wien, Donau-City-Str. 1 fon: +43(0)50550 ext. 4165, fax: ext. 4190 mobile: +43(0)664/8251077 mailto:christoph.pacher@arcs.ac.at http://www.smart-systems.at -----Original Message----- From: math-fun-bounces+christoph.pacher=arcs.ac.at(a)mailman.xmission.com on behalf of dasimov(a)earthlink.net Sent: Fri 2/10/2006 5:54 PM To: math-fun Subject: RE: [math-fun] how many 4x4x1 blocks Rich, did you know your math-fun e-mail as quoted came with an attachment? It's listed by my mailer as "Unknown 166-B". Also my Earthlink mail program shows a red flag next to it, which I take is a warning. --Dan << I've got a proof. I'll hold it back for a couple of days to give others a chance to play with the puzzle. Rich >> _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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RE: [math-fun] how many 4x4x1 blocks
by dasimov＠earthlink.net 10 Feb '06

10 Feb '06

Rich, did you know your math-fun e-mail as quoted came with an attachment? It's listed by my mailer as "Unknown 166-B". Also my Earthlink mail program shows a red flag next to it, which I take is a warning. --Dan << I've got a proof. I'll hold it back for a couple of days to give others a chance to play with the puzzle. Rich >>

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[math-fun] Re: Packing circles into a finite rectangle
by James Propp 09 Feb '06

09 Feb '06

>> If you have circles of radii 1/1, 1/2, 1/3, ..., 1/n for all n >= 1, >> then their total area is finite, although the sum of their radii is >> infinite. >> >> Can one fit these circles into a finite-sized rectangle? > > yes, they will fit inside a 1 x (11/6) rectangle for example. put each > circle inside a square, and then pack the squares like this: > http://www.stetson.edu/~efriedma/harmonic/ It is apparently still unknown whether one can pack squares of sides-lengths 1/1, 1/2, 1/3, etc. into a 1-by-(pi squared over six) rectangle. See Clive Tooth's web-page www.pisquaredoversix.force9.co.uk/Tiling.htm for an example of how one might get started. I asked Clive for background on this tiling, and he kindly forwarded to me copies of some relevant sci.math postings from last year (one by Clive himself and one by David Cantrell): ************************************************************************** From: "The Last Danish Pastry" <TheLastDanishPas...(a)yahoo.com> Newsgroups: sci.math Subject: Packing squares into a 1 x Zeta[2] rectangle Date: Sun, 14 Mar 2004 13:41:34 -0000 Since the sum of the reciprocals of the squares of the positive integers is pi^2/6, the question arises as to whether squares with sides 1, 1/2, 1/3, etc can be packed into a rectangle of size 1 by pi^2/6. A picture of such a packing appears at http://www.pisquaredoversix.force9.co.uk/Tiling.htm I know of no proof that such a packing is possible or impossible. A program which I wrote has recently packed the first one million such squares into such a rectangle. I wrote the program in Mathematica and it uses exact arithmetic throughout. The algorithm is simple: The program maintains a list of rectangles. Each member of this list is itself a list containing exactly two members: the short side of a rectangle followed by the long side of the rectangle. The program does not store the position or orientation of any rectangle. The list is stored in ascending short side order. Initially the list contains one entry: {1, pi^2/6}. The program inserts the squares in decreasing size order, starting with the 1x1 square. At each stage, to insert the square with side 1/n, the program finds the rectangle with the shortest short side which will accommodate the square. [In the case that there are two or more such rectangles with equal short sides, the program will pick the youngest one.] Suppose that this rectangle is a x b, where 1/n<=a<b. The program deletes this rectangle from the list and inserts two new rectangles (b-1/n) x a and (a-1/n) x 1/n into the list. - If any side of one of these new rectangles is zero, the rectangle is not inserted. - Each new rectangle is stored in {short side, long side} form. Thus, the list of rectangles grows by one per inserted square, except in the rare cases in which a rectangle of exactly the right width is found. A trivial optimisation, which took me a long time to notice, is that if the number of squares to be inserted is decided at the beginning, as N, say, then any rectangles narrower than 1/N can be discarded as soon as they appear. This has a dramatic effect on the memory requirements and running time of the program. In the case of N=10^6 the maximum length of the rectangle list is only 7,493 which occurs just after inserting the square with side 1/55,205. Thereafter the length of the list slowly decreases. After inserting the last square, with side 10^-6, there are just 7 rectangles in the list. Their dimensions are, roughly: 0.000001000264 x 0.000371508565 [1,000,000] 0.000001000289 x 0.000006561855 [ 999,736] 0.000001000332 x 0.000011950546 [ 999,710] 0.000001000424 x 0.000099911008 [ 999,668] 0.000001001705 x 0.000013366392 [ 999,575] 0.000001001771 x 0.000555285230 [ 998,297] 0.000621399972 x 0.000621402936 [ 998,232] The number in square brackets is the square insertion step on which the rectangle was created. Note that each rectangle, except the last, has a short side just exceeding 10^-6. The last one is huge in comparison to the others. Also, it is almost square, having an aspect ratio of about 1.00000477 . The exact size of that last rectangle may be expressed as: (pi^2/6 - ShortNumerator/ShortDenominator) by (LongNumerator/LongDenominator) Where ShortNumerator = 88148577597184641314023698687804198322622003067950 90277185502203375757868045576236778623499141581835 91095283100445795053089253938670874859361703421694 93589311823604254188777249571387340885945805010588 59158565482204748239750452173829861295626045336259 37611823762149119593932203343599960000593198902456 62151237628335467123582313067030461952701063806541 09667253352708582423851542412389977650553557102776 48867766805593783542952432276691372915179943637567 30921614687225494242879103415386639772227105415075 60643644416702161056703911777435451470868967261015 03285333687384626370723064421360623232628962182637 33785897530162588640506103091960735379749803817168 27337105343328553249474140253583305664128829433056 13496001065570446619534699711595669897241621797339 93388113729020710706312912651983440135427698863738 63455691221381048718834049233446561869585869886892 11066449759482338440176987851999633382542943944743 83052271691048660471857179765488027759900589960305 37378583137068870187987573969424873832345344118539 33368635075465276924658419905001027235726511505469 90353609271920928926254763071708700239940565565014 69168650106503612867362486635886531701170797485510 99375741498187773039269448405524049571692907463933 52292512770766034352574830725344567781634199947344 39128864912738319438009565992988317045941602472518 84296172734535520954249457165155850357846464984168 55299222999724423825677947429849212042460805178788 23214473791347512926915739195451188103089711423995 69891250713418835019305494967439261535807066257765 86799633983386818438909781178910531233498247023179 22489232680161587870238609995903489290522620579734 27913072409961919211050524749783993630732833119534 51221464170732847284504244076797981093088600479209 053071348915420741889 ShortDenominator = 53608160645434703140761211994317971700783074368336 23325975705041657962199994733358382994458318582660 77598233503372093023408990096331259105144227503324 99931830157705570560424971647310855286927378535483 51519419895212616165032657829920140448566211296935 08925318960134998789890442610040864611429495599467 55671385275409484022080990210675967224987014208598 03335236373663242927780013944648864149878730486125 39041613975393755631755935261748017063478349180906 68435262002209072815068783227098957825899390123056 87116154057047656878878907753207321889684985791605 16008019019744195226005265347943729214648741668979 90732236259531674523664568538104488968273268732774 74645971065348455713992347183400863276778739619943 89625588842652615699820699583835957017913827580111 39619239553717364061586421301662144411944880262864 13395969932087668263688142360819354641427777701996 11575083489100468980371536759296335301136691198138 39559100156995546068054947710363960181181289364408 47219039882502818687162480694856587042340603392188 93192097190331493671006154660882637398429894947140 66724834302759980279336075917538263518035481261929 09959571761983813160895335069812335369906345916646 57307263315174319702690110108769119880417800559541 00846379467033278426493445187132240090150811316237 11870955392306082427560405688265993364931620972062 42335013731582665984021398103261291575096750373611 78793366106368150944101270721191063173365978409196 32419016511795819355629216137822167151147383280072 17692953976640188298210375720290857531988737370969 01207985676688854219832216404101906724620701738590 65777947161752685898547108814129100143613045915161 23374032620084188029394058765229087587568694264890 44233796349785411756140522085279648064146173069741 478636801148014080000 LongNumerator = 86047691660315035868968439680618750650385496662337 69341618423224868557176707904056350747739644607152 19341301092244142985694427567208621603825424715808 50112490209730250303812592410205852634552809617406 54471220118615832220663814127702644186714156802514 89272173167125109946591599082250813475432136844437 68434150604827750301142594110541508119955806775639 48463821703879298880841441517861092382325647927555 17115984413353561811152746794319549286695056542231 75370181749913246697604501180515750468588947850814 96706860842334628047016110265861987131951269186913 55908484624794372227266755207694315503681976948427 59732973255837108405693520922709400140190847636749 60091554508265275883766083172672994896381234279724 35485615485280999872894822261731570552709443192476 07685741096374165452217373319152585412622236248087 75856424918727148070254116048744362672852782317829 99207549345144244124571110563211160343448412194752 79110755652853486307738886342254777707077572587992 29405091182393418319601369763056120887765700618802 27139753549245993550447831136020976532760818138519 98216552674127086312240961717818956038535226187300 43026597993187644795910836481706823332909020130603 09729716420449811214331443400182398534278484763571 27904525989212662382850630094162630962359551947273 97311922973034069310043168935737831265241199089047 46559558404697050977201086017144902884602248144108 71005578441670661613107026657746377160408235952122 70087369832576028781964223879066392055590320740847 64872057694761838719127407266757016871496087930551 58594303661605591550655979493954945142285417155556 97061447617516531128930865552526142619566789459050 16802730874016347300046435875463493515247956092289 80788952041557673741641666771939979618817508420910 11050430185421809507580791626321726210797393934779 33362836600904869583181 LongDenominator = 13847326202197605511604348541676281864027795818043 31494335574789145450920249049910013265993162156505 60831474662540178414423726822350848233301461067293 21368995360010230519749558542277057071473577106811 23548536664652068929533780687547853336157369555086 86435599910456774645119837370840414802536199558214 07527909448749766750406409444125057733978666254122 19353807926168688488715863115960118447262678159252 66974533884810419996115934738184899912380787401311 77856867483197701324817440576351765023155329523060 68640715686158929369832278778842403496825357945180 51999476544062632945516973292520225337991834072383 82846228280041100692245330583478338241014202245803 30323143236417275386997321382053499979484896390576 45426647084624716067636294896758072966317096343334 28169063530634898773783098553798071542939121086910 08006721058240228461124891602875033925435245224172 90909661066024272174911721070106393261965722571136 18608587173321485889447514408382277645342459166650 66976600733716756193252940889415454560334136529577 70915746535414564177067690689656587269581203051464 54295311030944948219808305598301111195229732416082 42817235468158091436079088616794858652804473035517 82817817155337872004025893330437425894039012377193 76447539452009335637045807403996380650616246802471 05691575600545667349590304669247493320414626833379 52612693846867728391114410877249463229225980473996 60123182871902607011087292407939606651288648107916 20645836880002779839862309998364888537462475216562 83515669632032306317521726424095159019732485843596 26389117354057755490411586373128011034072899097872 57238596906896860163242623358201011919041838989031 36072177994374094882330699298066695784371692394667 11940033790850742395135563324255474227834715815498 61014376098413906103604397447388393432391565219123 065068758213935671227536000 -- Clive Tooth http://www.clivetooth.dk ************************************************************************** Subject: Re: Packing squares into a 1 x Zeta[2] rectangle From: David W. Cantrell <DWCantr...(a)sigmaxi.org> Date: 18 Mar 2004 16:31:52 GMT Newsgroups: sci.math "The Last Danish Pastry" <TheLastDanishPas...(a)yahoo.com> wrote: > Since the sum of the reciprocals of the squares of the positive integers > is pi^2/6, the question arises as to whether squares with sides 1, 1/2, > 1/3, etc can be packed into a rectangle of size 1 by pi^2/6. > > A picture of such a packing appears at > http://www.pisquaredoversix.force9.co.uk/Tiling.htm > > I know of no proof that such a packing is possible or impossible. [snip] It would be good if people interested in this thread were aware of some of the history of this problem. To that end: _Unsolved Problems in Geometry_ by Croft, Falconer and Guy (Springer-Verlag, 1991) "D5. Packing unequal rectangles and squares in a square." (pp. 112-113) mentions a question similar to the one considered in this thread. D5 caused me, in 1996, to consider several packing problems of this sort. I found simple algorithms, which I implemented in Mathematica, which seemed to pack rectangles of decreasing size "perfectly". The algorithms worked for classic problems, such as packing the rectangles 1/n by 1/(n+1) into the unit square, as well as for interesting new packing problems. I did not publish my results then, hoping first to prove that the algorithms were adequate for packing all of the rectangles. In November 1999, spurred by Clive Tooth, I posted to sci.math a brief description of what I had done. See both parts of "Packing Rectangles of Decreasing Size" at <http://mathforum.org/discuss/sci.math/a/t/241167>. (Readers may also be interested in looking at the packing algorithm which Clive had used back then.) Unbeknownst to me in 1999, an algorithm for such packings had appeared previously in the literature: An Algorithm for Packing Squares, Marc M. Paulhus, _J. Combin. Theory Ser. A_ 82 (1997) 147-157. His algorithm is somewhat different from any of the algorithms which I had used (but I wonder if perhaps it is the same as Clive's current algorithm). For packing squares of side lengths 1/2, 1/3, 1/4,... into a rectangle 1/2 by 2(pi^2/6 - 1), Paulhus gives two rules: "Rule 1. Always place the next square in the corner of the smallest width rectangle into which it will fit." "Rule 2. After placing a square into the corner of a rectangle, always cut the remaining area into two rectangular pieces by cutting from the free corner of the square tot the longer side of the original rectangle." For older references, see those cited by Paulhus. The most recent articles known to me concerning such packings -- but again, I may not be up-to-date with the literature -- are the following: Perfect Square Packings, Adam Chalcraft, _J. Combin. Theory Ser. A_ 92 (2000) 158-172. Compactness Theorems for Geometric Packings, Greg Martin, <http://arxiv.org/PS_cache/math/pdf/0005/0005054.pdf>. Perfect Packings of Squares Using the Stack-Pack Strategy, Johan Wastlund, <www.mai.liu.se/~jowas/dcg.ps <http://www.mai.liu.se/~jowas/dcg.ps>>. Thoughtful comments are welcome. Regards, David W. Cantrell ************************************************************************** (Clive tells me that there's more discussion of this problem in sci.math, though I haven't checked.) Jim Propp

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