Mathematics
Click on a question to know the answer
Q. What is the current status of Fermat's last theorem?
(There are no positive integers x , y , z and n > 2 such
that x ^ n + y ^ n = z ^ n)
Ans. FLT was proven in 1993 working at Princeton.
Here is a brief history of attempts to solve FLT.
It is not intended to be 'deep',but rather is intended for non-specialists.
The theorem is broken into 2 cases. The first case assumes (abc,n) = 1.
The second case is the general case.
What has been proved
First Case:
It has been proven true up to 7.568 x 10 ^ 17 by the work of Wagstaff & Tanner,
Granville&Monagan, and Coppersmith.They all used extensions of the Wiefrich
criteria and improved upon work performed by Gunderson and Shanks&Williams.
The first case has been proven to be true for an infinite number of
exponents by Adelman, Frey, et. al. using a generalization of the
Sophie Germain criterion.
Second Case:
It has been proven true up to n = 150,000 by Tanner & Wagstaff. The work
used new techniques for computing Bernoulli numbers mod p and improved upon
work of Vandiver. The work involved computing the irregular primes up to
150,000. FLT is true for all regular primes by a theorem of Kummer.
In the case of irregular primes, some additional computations are needed.
Update:
Fermat's Last Theorem has been proved true up to exponent 2,000,000 in
the general case. The method used was that of Wagstaff: enumerating and
eliminating irregular primes by Bernoulli number computations. The
computations were performed on a set of NeXT computers
Since the genus of the curve a ^ n + b ^ n = 1, is greater than or equal to 2
for n > 3, it follows from Mordell's theorem [proved by Faltings], that for
any given n, there are at most a finite number of solutions.
Conjectures
There are many open conjectures that imply FLT. These conjectures come from
different directions, but can be basically broken into several classes:
(and there are interrelationships between the classes)
(a) conjectures arising from Diophantine approximation theory such as
The ABC conjecture, the Szpiro conjecture, the Hall conjecture, etc.
For an excellent survey article on these subjects see the article
in the Bulletin of the AMS, July 1990 entitled
"Old and new conjectured diophantine inequalities".
Masser and Osterle formulated the following known as the ABC conjecture:
Given epsilon > 0, there exists a number C(epsilon) such that for any
set of non-zero, relatively prime integers a,b,c such that a+b = c we
have
max( |a|, |b|, |c|) <= C (epsilon) N (abc) ^ ( 1 + epsilon )
where N ( x ) is the product of the distinct primes dividing x.
It is easy to see that it implies FLT asymptotically.
The conjecture was motivated by a theorem, due to Mason tha
essentially says the ABC conjecture is true for polynomials.
The ABC conjecture also implies Szpiro's conjecture [and vice-versa]
and Hall's conjecture. These results are all generally believed to be
true.
There is a generalization of the ABC conjecture [by Vojta] which is too
technical to discuss but involves heights of points on non-singular
algebraic varieties . Vojta's conjecture also imples Mordell's theorem.
[already known to be true]. There are also a number of inter-twined
conjectures involving heights on elliptic curves that are related to
much of this stuff. For a more complete discussion, see Lang 's article.
(b) conjectures arising from the study of elliptic curves and modular
forms. The Taniyama-Weil-Shmimura conjecture.
There is a very important and well known conjecture known as the
Taniyama-Weil-Shimura conjecture that concerns elliptic curves.
This conjecture has been shown by the work of Frey, Serre, Ribet, et. al.
to imply FLT uniformly, not just asymptotically as with the ABC conj.
The conjecture basically states that all elliptic curves can be
parameterized in terms of modular forms.
There is new work on the arithmetic of elliptic curves.
Sha, the Tate-Shafarevich group on elliptic curves of rank 0 or 1.
By the way. An interesting aspect of this work is that there is a close
connection between Sha, and some of the classical work on FLT. For
example, there is a classical proof that uses infinite descent to prove
FLT for n = 4. It can be shown that there is an elliptic curve associated
with FLT and that for n=4, Sha is trivial. It can also be shown that in
the cases where Sha is non-trivial, that infinite-descent arguments do
not work; that in some sense 'Sha blocks the descent'. Somewhat more
technically, Sha is an obstruction to the local-global principle [e.g.
the Hasse-Minkowski theorem].
(c) Conjectures arising from some conjectured inequalities involving
Chern classes and some other deep results/conjectures in arithmetic
algebraic gemoetry. [about which I know epsilon].
The diophantine and elliptic curve conjectures all involve deep properties
of integers. Until these conjecture were tied to FLT, FLT had been regarded
by most mathematicians as an isolated problem; a curiosity. Now it
can be seen that it follows from some deep and fundamental properties of
the integers. [not yet proven but generally believed].
This synopsis is quite brief. A full survey would run to many pages.
Q. Has the Four Colour Theorem been solved?
(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.)
Ans.
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 progam 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.
Q. What are the largest prime numbers?
Ans.
Largest known Mersenne prime
It is 2 ^ 756839 - 1. It was discovered by a Cray-2 in England in 1992.
It has 227,832 digits.
Largest known prime
The largest known prime was 391581 * 2 ^ 216193 - 1. See Brown, Noll,
Parady, Smith, Smith, and Zarantonello, Letter to the editor,
American Mathematical Monthly, vol. 97, 1990, p. 214.
Now the largest known prime is the Mersenne prime described above.
Largest known twin primes
The largest known twin primes are 1706595 * 2 ^ 11235 +- 1.
See B. K. Parady and J. F. Smith and S. E. Zarantonello,
Smith, Noll and Brown.
Largest known twin primes, Mathematics of Computation,
vol.55, 1990, pp. 381-382.
largest Fermat number with known factorization
F11 = ( 2 ^ ( 2 ^ 11 ) ) + 1 which was factored by Brent & Morain in
1988.
F9 = ( 2 ^ ( 2 ^ 9 ) ) + 1 = 2 ^ 512 + 1 was factored by
A.K. Lenstra, H.W. Lenstra Jr., M.S. Manasse & J.M. Pollard
in 1990.
The factorization for F10 is NOT known.
Current status on Mersenne primes
Mersenne primes are primes of the form 2 ^ p - 1. For 2 ^ p - 1 to be prime we
must have that p is prime. The following Mersenne primes are known.
| nr | p | year by |
| 1-5 | 2,3,5,7,13 | in or before the middle ages |
|
6-7 | 17,19 | 1588 Cataldi |
| 8 | 31 | 1750 Euler |
| 9 | 61 | 1883 Pervouchine |
| 10 | 89 | 1911 Powers |
| 11 | 107 | 1914 Powers |
| 12 | 127 | 1876 Lucas |
| 13-14 | 521,607 | 1952 Robinson |
| 15-17 | 1279,2203,2281 | 1952 Lehmer |
| 18 | 3217 | 1957 Riesel |
| 19-20 | 4253,4423 | 1961 Hurwitz & Selfridge |
| 21-23 | 9689,9941,11213 | 1963 Gillies |
| 24 | 19937 | 1971 Tuckerman |
| 25 | 21701 | 1978 Noll & Nickel |
| 26 | 23209 | 1979 Noll |
| 27 | 44497 | 1979 Slowinski & Nelson |
| 28 | 86243 | 1982 Slowinski |
| 29 | 110503 | 1988 Colquitt & Welsh jr. |
| 30 | 132049 | 1983 Slowinski |
| 31 | 216091 | 1985 Slowinski |
| 32? | 756839 | 1992 Slowinski & Gage |
The way to determine if 2^p-1 is prime is to use the Lucas-Lehmer test:
Lucas_Lehmer_Test(p):
u := 4
for i from 3 to p do
u := u ^ 2 - 2 mod 2 ^ p - 1
od
if u = = 0 then
2 ^ p - 1 is prime
else
2 ^ p - 1 is composite
fi
The following ranges have been checked completely:
2 - 355K and 430K - 520K
Q. I have this complicated symbolic problem (most likely
a symbolic integral or a DE system) that I can't solve.
What should I do?
Ans. 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.
Q. Where can I get Symbolic Computation Package?
Ans. This is not a comprehensive list. There are other
Computer Algebra packages available that may better
suit your needs.
Maple
Purpose: Symbolic and numeric computation, mathematical
programming, and mathematical visualization.
Contact: Waterloo Maple Software,
160 Columbia Street West,
Waterloo, Ontario, Canada N2L 3L3
Phone: (519) 747-2373
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
Pari
Purpose: Number-theoretic computations and simple numerical
analysis.
Available for Sun 3, Sun 4, generic 32-bit Unix, and
Macintosh II. This is a free package, available by ftp from
math.ucla.edu (128.97.64.16).
Contact: questions about pari can be sent to
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
Macsyma
Purpose: Symbolic and mathematical manipulations.
Contact: Symbolics, Inc.
8 New England Executive Park East
Burlington, Massachusetts 01803
United States of America
(617) 221-1250
Matlab
Purpose: `matrix laboratory' for tasks involving
matrices, graphics and general numerical computation.
Contact: The MathWorks, Inc.
21 Eliot Street
South Natick, MA 01760
508-653-1415
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
Q. What is 0 to the 0 power?
Ans. Note: Read 0 ** 0 as "0 to the 0 power."
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.
Q. Why is 0.9999... = 1?
Ans. 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.
Choose epsilon > 0. Suppose delta = 1 / - log_10 epsilon, thus
epsilon = 10 ^ ( -1 / delta ). For every m > 1 / delta we have that
So by the (epsilon-delta) definition of the limit we have
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.
Q. How can I find PI to extreme precision?
Ans. MAPLE or MATHEMATICA can give you 10,000 digits of Pi in a blink,
and they can compute another 20,000 - 500,000 overnight (range depends
on hardware platform).
It is possible to retrieve 1.25+ million digits of pi via anonymous
ftp from the site wuarchive.wustl.edu, in the files pi.doc.Z and
pi.dat.Z which reside in subdirectory doc/misc/pi.
References :
J. M. Borwein, P. B. Borwein, and D. H. Bailey,
"Ramanujan,
Modular Equations, and Approximations to Pi",
American Mathematical
Monthly, vol. 96, no. 3 (March 1989), p. 201 - 220.
P. Beckman
A history of pi
Golem Press, CO, 1971 (fourth edition 1977)
J.M. Borwein and P.B. Borwein
The arithmetic-geometric mean and fast computation of elementary
functions
SIAM Review, Vol. 26, 1984, pp. 351-366
J.M. Borwein and P.B. Borwein
More quadratically converging algorithms for pi
Mathematics of Computation, Vol. 46, 1986, pp. 247-253
J.M. Borwein and P.B. Borwein
Pi and the AGM - a study in analytic number theory and
computational complexity
Wiley, New York, 1987
Shlomo Breuer and Gideon Zwas
Mathematical-educational aspects of the computation of pi
Int. J. Math. Educ. Sci. Technol., Vol. 15, No. 2, 1984,
pp. 231-244
Y. Kanada and Y. Tamura
Calculation of pi to 10,013,395 decimal places based on the
Gauss-Legendre algorithm and Gauss arctangent relation
Computer Centre, University of Tokyo, 1983
Morris Newman and Daniel Shanks
On a sequence arising in series for pi
Mathematics of computation, Vol. 42, No. 165, Jan 1984,
pp. 199-217
E. Salamin
Computation of pi using arithmetic-geometric mean
Mathematics of Computation, Vol. 30, 1976, pp. 565-570
D. Shanks and J.W. Wrench, Jr.
Calculation of pi to 100,000 decimals
Mathematics of Computation, Vol. 16, 1962, pp. 76-99
Daniel Shanks
Dihedral quartic approximations and series for pi
J. Number Theory, Vol. 14, 1982, pp.397-423
David Singmaster
The legal values of pi
The Mathematical Intelligencer, Vol. 7, No. 2, 1985
Stan Wagon
Is pi normal?
The Mathematical Intelligencer, Vol. 7, No. 3, 1985
J.W. Wrench, Jr.
The evolution of extended decimal approximations to pi
The Mathematics Teacher, Vol. 53, 1960, pp. 644-650
Q. The existence of a projective plane of order 10 has long been
an outstanding problem in discrete mathematics and finite geometry.
Ans. More precisely, the question is: 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.
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. The n = 10 case has been settled as
not possible either See Am. Math. Monthly,
recent issue. 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.
Q. Is there a formula to determine the day of the week, given
the month, day and year?
Ans. Here is the Standard Method.
Take the last two digits of the year.
Divide by 4, discarding any fraction.
Add the day of the month.
Add the month's key value:
JFM AMJ JAS OND
144 025 036 146
Subtract 1 for January or February of a non-leap year.
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.
For a Julian date, add 1 for 1700's, and 1 for every additional
century you go back.
Add the year.
Now take the remainder when you divide by 7; 0 is Sunday, the first day
of the week, 1 is Monday, and so on.
Another formula is:
W = k + [ 2.6 m - 0.2 ] - 2 C + 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)
This formula is good for the Gregorian calendar
(introduced 1582 in parts of Europe, adopted in 1752 in Great Britain
and its colonies, and on various dates in other countries).
It handles century & 400 year corrections, but 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!
References:
Winning Ways by Conway, Guy, Berlekamp is supposed to have it.
Martin Gardner in "Mathematical Carnaval".
Michael Keith and Tom Craver, "The Ultimate Perpetual Calendar?",
Journal of Recreational Mathematics, 22:4, pp. 280-282, 1990.
K. Rosen, "Elementary Number Theory", p. 156.
Q. 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.
Ans. 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 ^( -x1 ^ 2 - x2 ^ 2 -...-xN ^ 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.
Q. 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?
Ans. 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
What a "piece" is, (polygon? Topological disk? Borel-set?
Lebesgue-measurable set? Arbitrary?)
how many pieces are permitted (finitely many? countably? uncountably?)
what motions are allowed in "reassembling" (translations?
rotations? orientation-reversing maps? isometries?
affine maps? homotheties? arbitrary continuous images? etc.)
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
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.
THM (Hadwiger-Glur, 1951) Two equal-area polygons P,Q are
equi-decomposable by translations only, if 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.
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.
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.
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", , 1985, Cambridge
University Press.
Also read in the Mathematical Intelligencier an article on
the Banach-Tarski Paradox.
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:
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
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 equidecompsable 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 unintuitive results?
Q.
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?
Ans. 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. 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, Amsterdam,
North-Holland, 1963.
A. Fraenkel, Y. Bar-Hillel, and A. Levy, Foundations of Set Theory,
Amsterdam, North-Holland, 1984 (2nd edition, 2nd printing), pp. 53-86.
Q. Is there a theory of quaternionic analytic functions, that is, a four-
dimensional analog to the theory of complex analytic functions?
Ans. Yes. This was developed in the 1930s by the mathematician
Fueter. It is based on a generalization of the Cauchy-Riemann
equations, since the possible alternatives of power series expansions
or quaternion differentiability do not produce useful theories.
A number of useful integral theorems follow from the theory.
Sudbery provides an excellent review. Deavours covers some of the same
material less thoroughly. Brackx discusses a further generalization
to arbitrary Clifford algebras.
Anthony Sudbery, Quaternionic Analysis, Proc. Camb. Phil. Soc.,
vol. 85, pp 199-225, 1979.
Cipher A. Deavours, The Quaternion Calculus, Am. Math. Monthly,
vol. 80, pp 995-1008, 1973.
F. Brackx and R. Delanghe and F. Sommen, Clifford analysis,
Pitman, 1983.
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