SciMath FAQ Table of Contents ----------------- 1Q.- Fermat's Last Theorem, status of .. 2Q.- Values of Record Numbers 3Q.- Formula for prime numbers... 4Q.- Digits of Pi, computation and references 5Q.- Odd Perfect Number 6Q.- Computer Algebra Systems, application of .. 7Q.- Computer Algebra Systems, references to .. 8Q.- Fields Medal, general info .. 9Q.- Four Colour Theorem, proof of .. 10Q.- 0^0=1. A comprehensive approach 11Q.- 0.999... = 1. Properties of the real numbers .. 12Q.- There are three doors, The Monty Hall problem, Master Mind and other games .. 13Q.- Surface and Volume of the n-ball 14Q.- f(x)^f(x)=x, name of the function .. 15Q.- Projective plane of order 10 .. 16Q.- How to compute day of week of a given date 17Q.- Axiom of Choice and/or Continuum Hypothesis? 18Q.- Cutting a sphere into pieces of larger volume 19Q.- Pointers to Quaternions 20Q.- Erdos Number 21Q.- Why is there no Nobel in mathematics? 22Q.- General References and textbooks... 23Q.- Interest Rate... 24Q.- Euler's formula e^(i Pi) = - 1 ... 6Q: I have this complicated symbolic problem (most likely a symbolic integral or a DE system) that I can't solve. What should I do? A: Find a friend with access to a computer algebra system like MAPLE, MACSYMA or MATHEMATICA and ask her/him to solve it. If packages cannot solve it, then (and only then) ask the net. 7Q: Where can I get ? THIS IS NOT A COMPREHENSIVE LIST. There are other Computer Algebra packages available that may better suit your needs. There is also a FAQ list in the group sci.math.symbolic. It includes a much larger list of vendors and developers. (The FAQ list can be obtained from math.berkeley.edu via anonymous ftp). A: Maple Purpose: Symbolic and numeric computation, mathematical programming, and mathematical visualization. Contact: Waterloo Maple Software, 450 Phillip Street Waterloo, Ontario N2L 5J2 Phone (519)747-2373 FAX (519)747-5284 email: info@maplesoft.on.ca A: DOE-Macsyma Purpose: Symbolic and mathematical manipulations. Contact: National Energy Software Center Argonne National Laboratory 9700 South Cass Avenue Argonne, Illinois 60439 Phone: (708) 972-7250 A: Pari Purpose: Number-theoretic computations and simple numerical analysis. Available for most 32-bit machines, including 386+387 and 486. This is a copyrighted but free package, available by ftp from math.ucla.edu (128.97.4.254) and ftp.inria.fr (128.93.1.26). Contact: questions about pari can be sent to pari@ceremab.u-bordeaux.fr and for the Macintosh versions to bernardi@mathp7.jussieu.fr A: Mathematica Purpose: Mathematical computation and visualization, symbolic programming. Contact: Wolfram Research, Inc. 100 Trade Center Drive Champaign, IL 61820-7237 Phone: 1-800-441-MATH info@wri.com A: Macsyma Purpose: Symbolic numerical and graphical mathematics. Contact: Macsyma Inc. 20 Academy Street Arlington, MA 02174 tel: 617-646-4550 fax: 617-646-3161 email: info-macsyma@macsyma.com A: Matlab Purpose: matrix laboratory' for tasks involving matrices, graphics and general numerical computation. Contact: The MathWorks, Inc. 21 Prime Park Way Natick, MA 01760 508-653-1415 info@mathworks.com A: Cayley Purpose: Computation in algebraic and combinatorial structures such as groups, rings, fields, modules and graphs. Available for: SUN 3, SUN 4, IBM running AIX or VM, DEC VMS, others Contact: Computational Algebra Group University of Sydney NSW 2006 Australia Phone: (61) (02) 692 3338 Fax: (61) (02) 692 4534 cayley@maths.su.oz.au 8Q: Let P be a property about the Fields Medal. Is P(x) true? A: Institution is meant to be the Institution to which the researcher in question was associated to at the time the medal was awarded. Year Name Birthplace Age Institution ---- ---- ---------- --- ----------- 1936 Ahlfors, Lars Helsinki Finland 29 Harvard U USA 1936 Douglas, Jesse New York NY USA 39 MIT USA 1950 Schwartz, Laurent Paris France 35 U of Nancy France 1950 Selberg, Atle Langesund Norway 33 Adv.Std.Princeton USA 1954 Kodaira, Kunihiko Tokyo Japan 39 Princeton U USA 1954 Serre, Jean-Pierre Bages France 27 College de France France 1958 Roth, Klaus Breslau Germany 32 U of London UK 1958 Thom, Rene Montbeliard France 35 U of Strasbourg France 1962 Hormander, Lars Mjallby Sweden 31 U of Stockholm Sweden 1962 Milnor, John Orange NJ USA 31 Princeton U USA 1966 Atiyah, Michael London UK 37 Oxford U UK 1966 Cohen, Paul Long Branch NJ USA 32 Stanford U USA 1966 Grothendieck, Alexander Berlin Germany 38 U of Paris France 1966 Smale, Stephen Flint MI USA 36 UC Berkeley USA 1970 Baker, Alan London UK 31 Cambridge U UK 1970 Hironaka, Heisuke Yamaguchi-ken Japan 39 Harvard U USA 1970 Novikov, Serge Gorki USSR 32 Moscow U USSR 1970 Thompson, John Ottawa KA USA 37 U of Chicago USA 1974 Bombieri, Enrico Milan Italy 33 U of Pisa Italy 1974 Mumford, David Worth, Sussex UK 37 Harvard U USA 1978 Deligne, Pierre Brussels Belgium 33 IHES France 1978 Fefferman, Charles Washington DC USA 29 Princeton U USA 1978 Margulis, Gregori Moscow USSR 32 InstPrblmInfTrans USSR 1978 Quillen, Daniel Orange NJ USA 38 MIT USA 1982 Connes, Alain Draguignan France 35 IHES France 1982 Thurston, William Washington DC USA 35 Princeton U USA 1982 Yau, Shing-Tung Kwuntung China 33 IAS USA 1986 Donaldson, Simon Cambridge UK 27 Oxford U UK 1986 Faltings, Gerd 1954 Germany 32 Princeton U USA 1986 Freedman, Michael Los Angeles CA USA 35 UC San Diego USA 1990 Drinfeld, Vladimir Kharkov USSR 36 Phys.Inst.Kharkov USSR 1990 Jones, Vaughan Gisborne N Zealand 38 UC Berkeley USA 1990 Mori, Shigefumi Nagoya Japan 39 U of Kyoto? Japan 1990 Witten, Edward Baltimore USA 38 Princeton U/IAS USA References : International Mathematical Congresses, An Illustrated History 1893-1986, Revised Edition, Including 1986, by Donald J.Alberts, G. L. Alexanderson and Constance Reid, Springer Verlag, 1987. Tropp, Henry S., The origins and history of the Fields Medal,'' Historia Mathematica, 3(1976), 167-181. 9Q: Has the Four Colour Theorem been proved? Four Color Theorem: Every planar map with regions of simple borders can be coloured with 4 colours in such a way that no two regions sharing a non-zero length border have the same colour. A: This theorem was proved with the aid of a computer in 1976. The proof shows that if aprox. 1,936 basic forms of maps can be coloured with four colours, then any given map can be coloured with four colours. A computer program coloured this basic forms. So far nobody has been able to prove it without using a computer. In principle it is possible to emulate the computer proof by hand computations. References: K. Appel and W. Haken, Every planar map is four colourable, Bulletin of the American Mathematical Society, vol. 82, 1976 pp.711-712. K. Appel and W. Haken, Every planar map is four colourable, Illinois Journal of Mathematics, vol. 21, 1977, pp. 429-567. T. Saaty and Paul Kainen, The Four Colour Theorem: Assault and Conquest, McGraw-Hill, 1977. Reprinted by Dover Publications 1986. K. Appel and W. Haken, Every Planar Map is Four Colourable, Contemporary Mathematics, vol. 98, American Mathematical Society, 1989, pp.741. F. Bernhart, Math Reviews. 91m:05007, Dec. 1991. (Review of Appel and Haken's book). 10Q: What is 0^0 ? A: According to some Calculus textbooks, 0^0 is an "indeterminate form". When evaluating a limit of the form 0^0, then you need to know that limits of that form are called "indeterminate forms", and that you need to use a special technique such as L'Hopital's rule to evaluate them. Otherwise, 0^0=1 seems to be the most useful choice for 0^0. This convention allows us to extend definitions in different areas of mathematics that otherwise would require treating 0 as a special case. Notice that 0^0 is a discontinuity of the function x^y. Rotando & Korn show that if f and g are real functions that vanish at the origin and are _analytic_ at 0 (infinitely differentiable is not sufficient), then f(x)^g(x) approaches 1 as x approaches 0 from the right. From Concrete Mathematics p.162 (R. Graham, D. Knuth, O. Patashnik): "Some textbooks leave the quantity 0^0 undefined, because the functions x^0 and 0^x have different limiting values when x decreases to 0. But this is a mistake. We must define x^0 = 1 for all x, if the binomial theorem is to be valid when x=0, y=0, and/or x=-y. The theorem is too important to be arbitrarily restricted! By contrast, the function 0^x is quite unimportant." Published by Addison-Wesley, 2nd printing Dec, 1988. References: H. E. Vaughan, The expression '0^0', Mathematics Teacher 63 (1970), pp.111-112. Louis M. Rotando & Henry Korn, "The Indeterminate Form 0^0", Mathematics Magazine, Vol. 50, No. 1 (January 1977), pp. 41-42. L. J. Paige, A note on indeterminate forms, American Mathematical Monthly, 61 (1954), 189-190; reprinted in the Mathematical Association of America's 1969 volume, Selected Papers on Calculus, pp. 210-211. 11Q: Why is 0.9999... = 1? A: In modern mathematics, the string of symbols "0.9999..." is understood to be a shorthand for "the infinite sum 9/10 + 9/100 + 9/1000 + ...." This in turn is shorthand for "the limit of the sequence of real numbers 9/10, 9/10 + 9/100, 9/10 + 9/100 + 9/1000, ..." Using the well-known epsilon-delta definition of limit, one can easily show that this limit is 1. The statement that 0.9999... = 1 is simply an abbreviation of this fact. oo m --- 9 --- 9 0.999... = > ---- = lim > ---- --- 10^n m->oo --- 10^n n=1 n=1 Choose epsilon > 0. Suppose delta = 1/-log_10 epsilon, thus epsilon = 10^(-1/delta). For every m>1/delta we have that | m | | --- 9 | 1 1 | > ---- - 1 | = ---- < ------------ = epsilon | --- 10^n | 10^m 10^(1/delta) | n=1 | So by the (epsilon-delta) definition of the limit we have m --- 9 lim > ---- = 1 m->oo --- 10^n n=1 An *informal* argument could be given by noticing that the following sequence of "natural" operations has as a consequence 1 = 0.9999.... Therefore it's "natural" to assume 1 = 0.9999..... x = 0.99999.... 10x = 9.99999.... 10x - x = 9 9x = 9 x = 1 Thus 1 = 0.99999.... References: E. Hewitt & K. Stromberg, Real and Abstract Analysis, Springer-Verlag, Berlin, 1965. W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, 1976. 12Q: There are three doors, and there is a car hidden behind one of them, Master Mind and other games .. A: Read frequently asked questions from rec.puzzles, as well as their archive file'' where the problem is solved and carefully explained. (The Monty Hall problem). MANY OTHER MATHEMATICAL GAMES ARE EXPLAINED IN THE REC.PUZZLES FAQ AND ARCHIVES. READ IT BEFORE ASKING IN SCI.MATH. Your chance of winning is 2/3 if you switch and 1/3 if you don't. For a full explanation from the rec.puzzles' archive, send to the address archive-request@questrel.com an email message consisting of the text send monty.hall Also any other FAQ list can be obtained through anonymous ftp from rtfm.mit.edu. References American Mathematical Monthly, January 1992. For the game of Master Mind it has been proven that no more than five moves are required in the worst case. For references look at One such algorithm was published in the Journal of Recreational Mathematics; in '70 or '71 (I think), which always solved the 4 peg problem in 5 moves. Knuth later published an algorithm which solves the problem in a shorter # of moves - on average - but can take six guesses on certain combinations. Donald E. Knuth, The Computer as Master Mind, J. Recreational Mathematics 9 (1976-77), 1-6. 13Q: What is the formula for the "Surface Area" of a sphere in Euclidean N-Space. That is, of course, the volume of the N-1 solid which comprises the boundary of an N-Sphere. A: The volume of a ball is the easiest formula to remember: It's r^N times pi^(N/2)/(N/2)!. The only hard part is taking the factorial of a half-integer. The real definition is that x! = Gamma(x+1), but if you want a formula, it's: (1/2+n)! = sqrt(pi)*(2n+2)!/(n+1)!/4^(n+1) To get the surface area, you just differentiate to get N*pi^(N/2)/(N/2)!*r^(N-1). There is a clever way to obtain this formula using Gaussian integrals. First, we note that the integral over the line of e^(-x^2) is sqrt(pi). Therefore the integral over N-space of e^(-x_1^2-x_2^2-...-x_N^2) is sqrt(pi)^n. Now we change to spherical coordinates. We get the integral from 0 to infinity of V*r^(N-1)*e^(-r^2), where V is the surface volume of a sphere. Integrate by parts repeatedly to get the desired formula. It is possible to derive the volume of the sphere from first principles''. 14Q: Does anyone know a name (or a closed form) for f(x)^f(x)=x Solving for f one finds a "continued fraction"-like answer f(x) = log x ----- log (log x ------ ........... A: This question has been repeated here from time to time over the years, and no one seems to have heard of any published work on it, nor a published name for it (D. Merrit proposes "lx" due to its (very) faint resemblance to log). It's not an analytic function. The "continued fraction" form for its numeric solution is highly unstable in the region of its minimum at 1/e (because the graph is quite flat there yet logarithmic approximation oscillates wildly), although it converges fairly quickly elsewhere. To compute its value near 1/e, use the bisection method which gives good results. Bisection in other regions converges much more slowly than the "logarithmic continued fraction" form, so a hybrid of the two seems suitable. Note that it's dual valued for the reals (and many valued complex for negative reals). A similar function is a "built-in" function in MAPLE called W(x). MAPLE considers a solution in terms of W(x) as a closed form (like the erf function). W is defined as W(x)*exp(W(x))=x. An extensive treatise on the known facts of Lambert's W function is available for anonymous ftp at daisy.uwaterloo.ca in the /pub/maple/5.2/share/LambertW.ps. 15Q: Does there exist a projective plane of order 10? More precisely: Is it possible to define 111 sets (lines) of 11 points each such that: For any pair of points there is precisely one line containing them both and for any pair of lines there is only one point common to them both? A: Analogous questions with n^2 + n + 1 and n + 1 instead of 111 and 11 have been positively answered only in case n is a prime power. For n=6 it is not possible, more generally if n is congruent to 1 or 2 mod 4 and can not be written as a sum of two squares, then an FPP of order n does not exist. The n=10 case has been settled as not possible either by Clement Lam. As the "proof" took several years of computer search (the equivalent of 2000 hours on a Cray-1) it can be called the most time-intensive computer assisted single proof. The final steps were ready in January 1989. References R. H. Bruck and H. J. Ryser, "The nonexistence of certain finite projective planes," Canadian Journal of Mathematics, vol. 1 (1949), pp 88-93. C. Lam, Amer.Math.Monthly 98 (1991), 305-318. 16Q: Is there a formula to determine the day of the week, given the month, day and year? A: First a brief explanation: In the Gregorian Calendar, over a period of four hundred years, there are 97 leap years and 303 normal years. Each normal year, the day of January 1 advances by one; for each leap year it advances by two. 303 + 97 + 97 = 497 = 7 * 71 As a result, January 1 year N occurs on the same day of the week as January 1 year N + 400. Because the leap year pattern also recurs with a four hundred year cycle, a simple table of four hundred elements, and single modulus, suffices to determine the day of the week (in the Gregorian Calendar), and does it much faster than all the other algorithms proposed. Also, each element takes (in principle) only three bits; the entire table thus takes only 1200 bits, or 300 bytes; on many computers this will be less than the instructions to do all the complicated calculations proposed for the other algorithms. Incidental note: Because 7 does not divide 400, January 1 occurs more frequently on some days than others! Trick your friends! In a cycle of 400 years, January 1 and March 1 occur on the following days with the following frequencies: Sun Mon Tue Wed Thu Fri Sat Jan 1: 58 56 58 57 57 58 56 Mar 1: 58 56 58 56 58 57 57 Of interest is that (contrary to most initial guesses) the occurrence is not maximally flat. The Gregorian calendar was introduced in 1582 in parts of Europe; it was adopted in 1752 in Great Britain and its colonies, and on various dates in other countries. It replaced the Julian Calendar which has a four-year cycle of leap years; after four years January 1 has advanced by five days. Since 5 is relatively prime to 7, a table of 4 * 7 = 28 elements is necessary for the Julian Calendar. There is still a 3 day / 10,000 year error which the Gregorian calendar does not take. into account. At some time such a correction will have to be done but your software will probably not last that long :-) ! Here is a standard method suitable for mental computation: A. Take the last two digits of the year. B. Divide by 4, discarding any fraction. C. Add the day of the month. D. Add the month's key value: JFM AMJ JAS OND 144 025 036 146 E. Subtract 1 for January or February of a leap year. F. For a Gregorian date, add 0 for 1900's, 6 for 2000's, 4 for 1700's, 2 for 1800's; for other years, add or subtract multiples of 400. G. For a Julian date, add 1 for 1700's, and 1 for every additional century you go back. H. Add the last two digits of the year. I. Divide by 7 and take the remainder. Now 1 is Sunday, the first day of the week, 2 is Monday, and so on. The following formula, which is for the Gregorian calendar only, may be more convenient for computer programming. Note that in some programming languages the remainder operation can yield a negative result if given a negative operand, so "mod 7" may not translate to a simple remainder. W == (k + [2.6m - 0.2] - 2C + Y + [Y/4] + [C/4]) mod 7 where [] denotes the integer floor function (round down), k is day (1 to 31) m is month (1 = March, ..., 10 = December, 11 = Jan, 12 = Feb) Treat Jan & Feb as months of the preceding year C is century (1987 has C = 19) Y is year (1987 has Y = 87 except Y = 86 for Jan & Feb) W is week day (0 = Sunday, ..., 6 = Saturday) Here the century & 400 year corrections are built into the formula. The [2.6m-0.2] term relates to the repetitive pattern that the 30-day months show when March is taken as the first month. References: Winning Ways by Conway, Guy, Berlekamp is supposed to have it. Martin Gardner in "Mathematical Carnival". Michael Keith and Tom Craver, "The Ultimate Perpetual Calendar?", Journal of Recreational Mathematics, 22:4, pp. 280-282, 19 K. Rosen, "Elementary Number Theory", p. 156. 17Q: What is the Axiom of Choice? Why is it important? Why some articles say "such and such is provable, if you accept the axiom of choice."? What are the arguments for and against the axiom of choice? A: There are several equivalent formulations: -The Cartesian product of nonempty sets is nonempty, even if the product is of an infinite family of sets. -Given any set S of mutually disjoint nonempty sets, there is a set C containing a single member from each element of S. C can thus be thought of as the result of "choosing" a representative from each set in S. Hence the name. >Why is it important? All kinds of important theorems in analysis require it. Tychonoff's theorem and the Hahn-Banach theorem are examples. Indeed, Tychonoff's theorem is equivalent to AC. Similarly, AC is equivalent to the thesis that every set can be well-ordered. Zermelo's first proof of this in 1904 I believe was the first proof in which AC was made explicit. AC is especially handy for doing infinite cardinal arithmetic, as without it the most you get is a *partial* ordering on the cardinal numbers. It also enables you to prove such interesting general facts as that n^2 = n for all infinite cardinal numbers. > What are the arguments for and against the axiom of choice? The axiom of choice is independent of the other axioms of set theory and can be assumed or not as one chooses. (For) All ordinary mathematics uses it. There are a number of arguments for AC, ranging from a priori to pragmatic. The pragmatic argument (Zermelo's original approach) is that it allows you to do a lot of interesting mathematics. The more conceptual argument derives from the "iterative" conception of set according to which sets are "built up" in layers, each layer consisting of all possible sets that can be constructed out of elements in the previous layers. (The building up is of course metaphorical, and is suggested only by the idea of sets in some sense consisting of their members; you can't have a set of things without the things it's a set of). If then we consider the first layer containing a given set S of pairwise disjoint nonempty sets, the argument runs, all the elements of all the sets in S must exist at previous levels "below" the level of S. But then since each new level contains *all* the sets that can be formed from stuff in previous levels, it must be that at least by S's level all possible choice sets have already been *formed*. This is more in the spirit of Zermelo's later views (c. 1930). (Against) It has some supposedly counterintuitive consequences, such as the Banach-Tarski paradox. (See next question) Arguments against AC typically target its nonconstructive character: it is a cheat because it conjures up a set without providing any sort of *procedure* for its construction--note that no *method* is assumed for picking out the members of a choice set. It is thus the platonic axiom par excellence, boldly asserting that a given set will always exist under certain circumstances in utter disregard of our ability to conceive or construct it. The axiom thus can be seen as marking a divide between two opposing camps in the philosophy of mathematics: those for whom mathematics is essentially tied to our conceptual capacities, and hence is something we in some sense *create*, and those for whom mathematics is independent of any such capacities and hence is something we *discover*. AC is thus of philosophical as well as mathematical significance. It should be noted that some interesting mathematics has come out of an incompatible axiom, the Axiom of Determinacy (AD). AD asserts that any two-person game without ties has a winning strategy for the first or second player. For finite games, this is an easy theorem; for infinite games with duration less than \omega and move chosen from a countable set, you can prove the existence of a counter-example using AC. Jech's book "The Axiom of Choice" has a discussion. An example of such a game goes as follows. Choose in advance a set of infinite sequences of integers; call it A. Then I pick an integer, then you do, then I do, and so on forever (i.e. length \omega). When we're done, if the sequence of integers we've chosen is in A, I win; otherwise you win. AD says that one of us must have a winning strategy. Of course the strategy, and which of us has it, will depend upon A. From a philosophical/intuitive/pedagogical standpoint, I think Bertrand Russell's shoe/sock analogy has a lot to recommend it. Suppose you have an infinite collection of pairs of shoes. You want to form a set with one shoe from each pair. AC is not necessary, since you can define the set as "the set of all left shoes". (Technically, we're using the axiom of replacement, one of the basic axioms of Zermelo-Fraenkel (ZF) set theory.) If instead you want to form a set containing one sock from each pair of an infinite collection of pairs of socks, you now need AC. References: Maddy, "Believing the Axioms, I", J. Symb. Logic, v. 53, no. 2, June 1988, pp. 490-500, and "Believing the Axioms II" in v.53, no. 3. Gregory H. Moore, Zermelo's Axiom of Choice, New York, Springer-Verlag, 1982. H. Rubin and J. E. Rubin, Equivalents of the Axiom of Choice II, North-Holland/Elsevier Science, 1985. A. Fraenkel, Y. Bar-Hillel, and A. Levy, Foundations of Set Theory, Amsterdam, North-Holland, 1984 (2nd edition, 2nd printing), pp. 53-86. 18Q: Cutting a sphere into pieces of larger volume. Is it possible to cut a sphere into a finite number of pieces and reassemble into a solid of twice the volume? A: This question has many variants and it is best answered explicitly. Given two polygons of the same area, is it always possible to dissect one into a finite number of pieces which can be reassembled into a replica of the other? Dissection theory is extensive. In such questions one needs to specify (A) what a "piece" is, (polygon? Topological disk? Borel-set? Lebesgue-measurable set? Arbitrary?) (B) how many pieces are permitted (finitely many? countably? uncountably?) (C) what motions are allowed in "reassembling" (translations? rotations? orientation-reversing maps? isometries? affine maps? homotheties? arbitrary continuous images? etc.) (D) how the pieces are permitted to be glued together. The simplest notion is that they must be disjoint. If the pieces are polygons [or any piece with a nice boundary] you can permit them to be glued along their boundaries, ie the interiors of the pieces disjoint, and their union is the desired figure. Some dissection results 1) We are permitted to cut into FINITELY MANY polygons, to TRANSLATE and ROTATE the pieces, and to glue ALONG BOUNDARIES; then Yes, any two equal-area polygons are equi-decomposable. This theorem was proven by Bolyai and Gerwien independently, and has undoubtedly been independently rediscovered many times. I would not be surprised if the Greeks knew this. The Hadwiger-Glur theorem implies that any two equal-area polygons are equi-decomposable using only TRANSLATIONS and ROTATIONS BY 180 DEGREES. 2) THM (Hadwiger-Glur, 1951) Two equal-area polygons P,Q are equi-decomposable by TRANSLATIONS only, iff we have equality of these two functions: PHI_P() = PHI_Q() Here, for each direction v (ie, each vector on the unit circle in the plane), let PHI_P(v) be the sum of the lengths of the edges of P which are perpendicular to v, where for such an edge, its length is positive if v is an outward normal to the edge and is negative if v is an inward normal to the edge. 3) In dimension 3, the famous "Hilbert's third problem" is: "If P and Q are two polyhedra of equal volume, are they equi-decomposable by means of translations and rotations, by cutting into finitely many sub-polyhedra, and gluing along boundaries?" The answer is "NO" and was proven by Dehn in 1900, just a few months after the problem was posed. (Ueber raumgleiche polyeder, Goettinger Nachrichten 1900, 345-354). It was the first of Hilbert's problems to be solved. The proof is nontrivial but does *not* use the axiom of choice. "Hilbert's Third Problem", by V.G.Boltianskii, Wiley 1978. 4) Using the axiom of choice on non-countable sets, you can prove that a solid sphere can be dissected into a finite number of pieces that can be reassembled to two solid spheres, each of same volume of the original. No more than nine pieces are needed. The minimum possible number of pieces is FIVE. (It's quite easy to show that four will not suffice). There is a particular dissection in which one of the five pieces is the single center point of the original sphere, and the other four pieces A, A', B, B' are such that A is congruent to A' and B is congruent to B'. [See Wagon's book]. This construction is known as the "Banach-Tarski" paradox or the "Banach-Tarski-Hausdorff" paradox (Hausdorff did an early version of it). The "pieces" here are non-measurable sets, and they are assembled *disjointly* (they are not glued together along a boundary, unlike the situation in Bolyai's thm.) An excellent book on Banach-Tarski is: "The Banach-Tarski Paradox", by Stan Wagon, 1985, Cambridge University Press. Robert M. French, The Banach-Tarski theorem, The Mathematical Intelligencer 10 (1988) 21-28. The pieces are not (Lebesgue) measurable, since measure is preserved by rigid motion. Since the pieces are non-measurable, they do not have reasonable boundaries. For example, it is likely that each piece's topological-boundary is the entire ball. The full Banach-Tarski paradox is stronger than just doubling the ball. It states: 5) Any two bounded subsets (of 3-space) with non-empty interior, are equi-decomposable by translations and rotations. This is usually illustrated by observing that a pea can be cut up into finitely pieces and reassembled into the Earth. The easiest decomposition "paradox" was observed first by Hausdorff: 6) The unit interval can be cut up into COUNTABLY many pieces which, by *translation* only, can be reassembled into the interval of length 2. This result is, nowadays, trivial, and is the standard example of a non-measurable set, taught in a beginning graduate class on measure theory. References: In addition to Wagon's book above, Boltyanskii has written at least two works on this subject. An elementary one is: "Equivalent and equidecomposable figures" in Topics in Mathematics published by D.C. HEATH AND CO., Boston. It is a translation from the 1956 work in Russian. Also, the article "Scissor Congruence" by Dubins, Hirsch and ?, which appeared about 20 years ago in the Math Monthly, has a pretty theorem on decomposition by Jordan arcs. Banach and Tarski had hoped that the physical absurdity of this theorem would encourage mathematicians to discard AC. They were dismayed when the response of the math community was Isn't AC great? How else could we get such counterintuitive results?' '' Copyright Notice Copyright (c) 1993 A. Lopez-Ortiz This FAQ is Copyright (C) 1994 by Alex Lopez-Ortiz. This text, in whole or in part, may not be sold in any medium, including, but not limited to electronic, CD-ROM, or published in print, without the explicit, written permission of Alex Lopez-Ortiz. -------------------------------------------------------------------------- Questions and Answers Edited and Compiled by: Alex Lopez-Ortiz alopez-o@maytag.UWaterloo.ca Department of Computer Science University of Waterloo Waterloo, Ontario Canada -- Alex Lopez-Ortiz alopez-o@neumann.UWaterloo.ca Department of Computer Science University of Waterloo Waterloo, Ontario Canada  Discuss this article in the forums Date this article was posted to GameDev.net: 7/16/1999 (Note that this date does not necessarily correspond to the date the article was written) See Also: General Math © 1999-2011 Gamedev.net. All rights reserved. Terms of Use Privacy Policy Comments? Questions? Feedback? Click here!