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 nball
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)7472373
FAX (519)7475284
email: info@maplesoft.on.ca
A: DOEMacsyma
Purpose: Symbolic and mathematical manipulations.
Contact: National Energy Software Center
Argonne National Laboratory 9700 South Cass Avenue
Argonne, Illinois 60439
Phone: (708) 9727250
A: Pari
Purpose: Numbertheoretic computations and simple numerical
analysis.
Available for most 32bit 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.ubordeaux.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 618207237
Phone: 1800441MATH
info@wri.com
A: Macsyma
Purpose: Symbolic numerical and graphical mathematics.
Contact: Macsyma Inc.
20 Academy Street
Arlington, MA 02174
tel: 6176464550
fax: 6176463161
email: infomacsyma@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
5086531415
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, JeanPierre 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 Yamaguchiken 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, ShingTung 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 18931986,
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), 167181.
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 nonzero
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.711712.
K. Appel and W. Haken, Every planar map is four colourable,
Illinois Journal of Mathematics, vol. 21, 1977, pp. 429567.
T. Saaty and Paul Kainen, The Four Colour Theorem: Assault and
Conquest, McGrawHill, 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 AddisonWesley, 2nd printing Dec, 1988.
References:
H. E. Vaughan, The expression '0^0', Mathematics Teacher 63 (1970),
pp.111112.
Louis M. Rotando & Henry Korn, "The Indeterminate Form 0^0",
Mathematics Magazine, Vol. 50, No. 1 (January 1977), pp. 4142.
L. J. Paige, A note on indeterminate forms, American Mathematical
Monthly, 61 (1954), 189190; reprinted in the Mathematical
Association of America's 1969 volume, Selected Papers on Calculus,
pp. 210211.
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 wellknown epsilondelta 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 (epsilondelta) 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,
SpringerVerlag, Berlin, 1965.
W. Rudin, Principles of Mathematical Analysis, McGrawHill, 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 archiverequest@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 (197677), 16.
13Q: What is the formula for the "Surface Area" of a sphere in
Euclidean NSpace. That is, of course, the volume of the N1
solid which comprises the boundary of an NSphere.
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 halfinteger. 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^(N1).
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 Nspace of
e^(x_1^2x_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^(N1)*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 "builtin" 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 Cray1) it can be
called the most timeintensive 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 8893.
C. Lam, Amer.Math.Monthly 98 (1991), 305318.
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 fouryear
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 [2.6m0.2] term relates to the repetitive pattern that the 30day
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. 280282, 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 HahnBanach theorem are examples. Indeed,
Tychonoff's theorem is equivalent to AC. Similarly, AC is equivalent
to the thesis that every set can be wellordered. 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 BanachTarski 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 constructionnote 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 twoperson 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 counterexample 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 ZermeloFraenkel (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. 490500, and "Believing the Axioms II" in v.53, no. 3.
Gregory H. Moore, Zermelo's Axiom of Choice, New York, SpringerVerlag,
1982.
H. Rubin and J. E. Rubin, Equivalents of the Axiom of Choice II,
NorthHolland/Elsevier Science, 1985.
A. Fraenkel, Y. BarHillel, and A. Levy, Foundations of Set Theory,
Amsterdam, NorthHolland, 1984 (2nd edition, 2nd printing), pp. 5386.
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? Borelset?
Lebesguemeasurable set? Arbitrary?)
(B) how many pieces are permitted (finitely many? countably? uncountably?)
(C) what motions are allowed in "reassembling" (translations?
rotations? orientationreversing 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 equalarea polygons are equidecomposable.
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 HadwigerGlur theorem implies that any two equalarea polygons are
equidecomposable using only TRANSLATIONS and ROTATIONS BY 180
DEGREES.
2) THM (HadwigerGlur, 1951) Two equalarea polygons P,Q are
equidecomposable 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
equidecomposable by means of translations and rotations, by
cutting into finitely many subpolyhedra, 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, 345354). 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 noncountable 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 "BanachTarski" paradox or the
"BanachTarskiHausdorff" paradox (Hausdorff did an early version of
it). The "pieces" here are nonmeasurable 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 BanachTarski is:
"The BanachTarski Paradox", by Stan Wagon, 1985, Cambridge
University Press.
Robert M. French, The BanachTarski theorem, The Mathematical
Intelligencer 10 (1988) 2128.
The pieces are not (Lebesgue) measurable, since measure is preserved
by rigid motion. Since the pieces are nonmeasurable, they do not
have reasonable boundaries. For example, it is likely that each piece's
topologicalboundary is the entire ball.
The full BanachTarski paradox is stronger than just doubling the
ball. It states:
5) Any two bounded subsets (of 3space) with nonempty interior, are
equidecomposable 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
nonmeasurable 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. LopezOrtiz
This FAQ is Copyright (C) 1994 by Alex LopezOrtiz. This text,
in whole or in part, may not be sold in any medium, including,
but not limited to electronic, CDROM, or published in print,
without the explicit, written permission of Alex LopezOrtiz.

Questions and Answers Edited and Compiled by:
Alex LopezOrtiz alopezo@maytag.UWaterloo.ca
Department of Computer Science University of Waterloo
Waterloo, Ontario Canada

Alex LopezOrtiz alopezo@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
© 19992011 Gamedev.net. All rights reserved. Terms of Use Privacy Policy
Comments? Questions? Feedback? Click here!
