FREQUENTLY ASKED QUESTIONS IN MATHEMATICS
The Sci.Math FAQ Team.
Editor: Alex Lspez-Ortiz
e-mail: alopez-o@neumann.UWaterloo.ca
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archive-name: sci-math-faq/tableofcontents
Last-modified: Dec 8, 1994
Version: 6.1
The NEW FAQ is out, and will be posted by instalments. Here is the table
of contents.
Contents
* Contents
* Introduction
+ Why a list of Frequently Asked Questions?
+ Frequently Asked Questions in Mathematics?
* Fermat's Last Theorem
+ History of Fermat's Last Theorem
+ What is the current status of FLT?
+ Wiles' line of attack
+ If not, then what?
o What has been proved
o Conjectures
+ Did Fermat prove this theorem?
* Prime Numbers
+ Largest known Mersenne prime
+ Largest known prime
+ Largest known twin primes
+ Largest Fermat number with known factorization
+ Algorithms to factor integer numbers
+ List of record numbers
+ What is the current status on Mersenne primes?
+ Formulae to compute prime numbers
* What are numbers?
+ Introduction
+ Construction of the Number System
+ Construction of N
+ Construction of Z
+ Construction of Q
+ Construction of R
+ Construction of C
+ Rounding things up
+ What's next?
* Special Numbers and Functions
+ How to compute digits of pi ?
+ Indiana bill sets the value of pi to 3
+ Euler's formula: e^(i pi) = -1
+ What is 0^0
+ Why is 0.9999... = 1 ?
+ Name for f(x)^(f(x)) = x
* Unsolved Problems
+ Does there exist a number that is perfect and odd?
+ Collatz Problem
* Symbolic Computation Packages
* Fields Medal
+ Historical Introduction
+ Table of Awardees
* Famous Problems in Mathematics
+ The Four Colour Theorem
+ The Trisection of an Angle
* Mathematical Games
+ The Monty Hall problem
+ Master Mind
* Projective plane of order 10
* The Axiom of Choice
+ Relevance of the Axiom of Choice
+ Cutting a sphere into pieces of larger volume
* The Continuum Hypothesis
* Theory of quaternionic analytic functions
* Mathematical Oddities
+ Erdos Number
+ Why is there no Nobel in mathematics?
+ Who is N. Bourbaki?
* Which are the 23 Hilbert Problems?
* Formulas of General Interest
+ How to determine the day of the week, given the month, day
and year
+ Formula for the Surface Area of a sphere in Euclidean N
-Space
+ Formula to compute compound interest.
* General Bibliography
* The Sci.Math FAQ Team
* Copyright Notice
* About this document ...
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Why is 0.9999... = 1 ?
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 the limit (you can find it in
any of the given references on analysis), one can easily show that
this limit is 1 . The statement that 0.9999... = 1 is simply an
abbreviation of this fact.
0.9999... = sum_(n = 1)^(oo) (9)/(10^n) = lim_(m --> oo) sum_(n = 1)^m
(9)/(10^n)
Choose varepsilon > 0 . Suppose delta = 1/- log_(10) varepsilon , thus
varepsilon = 10^(-1/delta) . For every m > 1/delta we have that
\left| sum_(n = 1)^m (9)/(10^n) - 1 \right| = (1)/(10^m) <
(1)/(10^(1/delta)) = varepsilon
So by the varepsilon - delta definition of the limit we have
lim_(m --> oo) sum_(n = 1)^m (9)/(10^n) = 1
Not formal enough? In that case you need to go back to the
construction of the number system. After you have constructed the
reals (Cauchy sequences are well suited for this case, see
[Shapiro75]), you can indeed verify that the preceding proof correctly
shows 0.9999... = 1 .
An informal argument could be given by noticing that the following
sequence of ``natural'' operations has as a consequence 0.9999... = 1
. Therefore it's ``natural'' to assume 0.9999... = 1 .
x = 0.9999... ; 10x = 10 o 0.9999... ; 10x = 9.9999... ; 10x - x =
9.9999... - 0.9999... ; 9x = 9 ; x = 1 ;
Thus 0.9999... = 1 .
An even easier argument multiplies both sides of 0.3333... = 1/3 by 3
. The result is 0.9999... = 3/3 = 1 .
Another informal argument is to notice that all periodic numbers such
as 0.46464646... are equal to the period divided over the same number
of 9 s. Thus 0.46464646... = 46/99 . Applying the same argument to
0.9999... = 9/9 = 1 .
Although the three informal arguments might convince you that
0.9999... = 1 , they are not complete proofs. Basically, you need to
prove that each step on the way is allowed and is correct. They are
also ``clumsy'' ways to prove the equality since they go around the
bush: proving 0.9999... = 1 directly is much easier.
You can even have that while you are proving it the ``clumsy'' way,
you get proof of the result in another way. For instance, in the first
argument the first step is showing that 0.9999... is real indeed. You
can do this by giving the formal proof stated in the beginning of this
FAQ question. But then you have 0.9999... = 1 as corollary. So the
rest of the argument is irrelevant: you already proved what you wanted
to prove.
References
R.V. Churchill and J.W. Brown. Complex Variables and Applications.
5^(th) ed., McGraw-Hill, 1990.
E. Hewitt and K. Stromberg. Real and Abstract Analysis.
Springer-Verlag, Berlin, 1965.
W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, 1976.
L. Shapiro. Introduction to Abstract Algebra. McGraw-Hill, 1975.
This subsection of the FAQ is Copyright (c) 1994 Hans de Vreught. Send
comments and or corrections relating to this part to
hdev@cp.tn.tudelft.nl.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Thu Dec 08 20:49:28 EST 1994
Archive-Name: sci-math-faq/specialnumbers/0to0
Last-modified: December 8, 1994
Version: 6.1
What is 0^0
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.
See also Knuth "Two notes on notation" (AMM 99 no. 5 (May 1992),
403-422).
On a discussion of the use of the function 0^(0^x) by an Italian
mathematician named Guglielmo Libri.
[T]he paper [33] did produce several ripples in mathematical waters
when it originally appeared, because it stirred up a controversy
about whether 0^0 is defined. Most mathematicians agreed that 0^0 =
1 , but Cauchy [5, page 70] had listed 0^0 together with other
expressions like 0/0 and oo - oo in a table of undefined forms.
Libri's justification for the equation 0^0 = 1 was far from
convincing, and a commentator who signed his name simply ``S'' rose
to the attack [45]. August Mvbius [36] defended Libri, by presenting
his former professor's reason for believing that 0^0 = 1 (basically
a proof that lim_(x --> 0+) x^x = 1 ). Mvbius also went further and
presented a supposed proof that lim_(x --> 0+) f(x)^(g(x)) whenever
lim_(x --> 0+) f(x) = lim_(x --> 0+) g(x) = 0 . Of course ``S'' then
asked [3] whether Mvbius knew about functions such as f(x) =
e^(-1/x) and g(x) = x . (And paper [36] was quietly omitted from the
historical record when the collected words of Mvbius were ultimately
published.) The debate stopped there, apparently with the conclusion
that 0^0 should be undefined.
But no, no, ten thousand times no! Anybody who wants the binomial
theorem (x + y)^n = sum_(k = 0)^n (n\choose k) x^k y^(n - k) to hold
for at least one nonnegative integer n must believe that 0^0 = 1 ,
for we can plug in x = 0 and y = 1 to get 1 on the left and 0^0 on
the right.
The number of mappings from the empty set to the empty set is 0^0 .
It has to be 1.
On the other hand, Cauchy had good reason to consider 0^0 as an
undefined limiting form, in the sense that the limiting value of
f(x)^(g(x)) is not known a priori when f(x) and g(x) approach 0
independently. In this much stronger sense, the value of 0^0 is less
defined than, say, the value of 0 + 0 . Both Cauchy and Libri were
right, but Libri and his defenders did not understand why truth was
on their side.
[3] Anonymous and S... Bemerkungen zu den Aufsatze |berschrieben,
`Beweis der Gleichung 0^0 = 1 , nach J. F. Pfaff', im zweiten Hefte
dieses Bandes, S. 134, Journal f|r die reine und angewandte
Mathematik, 12 (1834), 292-294.
[5] OE uvres Complhtes. Augustin-Louis Cauchy. Cours d'Analyse de
l'Ecole Royale Polytechnique (1821). Series 2, volume 3.
[33] Guillaume Libri. Mimoire sur les fonctions discontinues,
Journal f|r die reine und angewandte Mathematik, 10 (1833),
303-316.
[36] A. F. Mvbius. Beweis der Gleichung 0^0 = 1 , nach J. F. Pfaff.
Journal f|r die reine und angewandte Mathematik,
12 (1834), 134-136.
[45] S... Sur la valeur de 0^0 . Journal f|r die reine und
angewandte Mathematik 11, (1834), 272-273.
References
Knuth. Two notes on notation. (AMM 99 no. 5 (May 1992), 403-422).
H. E. Vaughan. The expression ' 0^0 '. Mathematics Teacher 63 (1970),
pp.111-112.
Louis M. Rotando and 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.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/trisection
Last-modified: December 8, 1994
Version: 6.1
The Trisection of an Angle
Theorem. It is not possible to split an arbitrary angle in three equal
parts using a compass and an unmarked ruler.
This problem, together with Doubling the Cube, Constructing the
regular Heptagon and Squaring the Circle were posed by the Greeks in
antiquity, and remained open until modern times.
The solution all of them is rather inelegant from a geometric
perspective. No geometric proof has been offered [check?], however, a
very clever solution was found using fairly basic results from
extension field and modern algebra.
It turns out that trisecting the angle is equivalent to solving a
cubic equation. Constructions with ruler and compass may only compute
the solution of a limited set of such equations, even when restricted
to integer coefficients. In particular, the equation for theta = 60
degrees cannot be solved by ruler and compass and thus the trisection
of the angle is not possible.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/fourcolour
Last-modified: December 8, 1994
Version: 6.1
The Four Colour Theorem
Theorem [Four Colour 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.
An equivalent combinatorial interpretation is
Theorem [Four Colour Theorem] Every loopless planar graph admits a
vertex-colouring with at most four different colours.
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 these 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.
The number of basic forms, or configurations has been reduced to 633.
The basic trust of the known proofs is to show that none of those
basic configurations may appear in a minimal counterexample to the
Four Colour Theorem because if one appeared, it could be replaced by
something smaller, contradicting minimality.
Then it is shown that all minimal counterexamples must contain one of
the basic configurations.
A recent simplification of the Four Colour Theorem proof, by
Robertson, Sanders, Seymour and Thomas, greatly simplifies the
original proof by Appel and Haken.
Further, the set of computer programs involved, will be made available
via anonymous ftp for all to check.
If correct, this new proof should dispel all criticisms of the
computer assisted proof.
References
K. Appel and W. Haken. Every planar map is four colorable. Bulletin of
the American Mathematical Society, vol. 82, 1976 pp.711-712.
K. Appel and W. Haken. Every planar map is four colorable. Illinois
Journal of Mathematics, vol. 21, 1977, pp. 429-567.
N. Robertson, D. Sanders, P. Seymour, R. Thomas The Four Colour
Theorem Preprint, February 1994.
The Four Color Theorem: Assault and Conquest T. Saaty and Paul Kainen.
McGraw-Hill, 1977. Reprinted by Dover Publications 1986.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/AC/relevance
Last-modified: December 8, 1994
Version: 6.1
Relevance of the Axiom of Choice
THE AXIOM OF CHOICE
There are many equivalent statements of the Axiom of Choice. The
following version gave rise to its name:
For any set X there is a function f , with domain X ; (0) , so that
f(x) is a member of x for every nonempty x in X .
Such an f is called a "choice function" on X . [Note that X ; (0)
means X with the empty set removed. Also note that in Zermelo-Fraenkel
set theory all mathematical objects are sets so each member of X is
itself a set.]
The Axiom of Choice (AC) is one of the most discussed axioms of
mathematics, perhaps second only to Euclid's parallel postulate. The
axioms of set theory provide a foundation for modern mathematics in
the same way that Euclid's five postulates provided a foundation for
Euclidean geometry, and the questions surrounding AC are the same as
the questions that surrounded Euclid's Parallel Postulate:
1. Can it be derived from the other axioms?
2. Is it consistent with the other axioms?
3. Should we accept it as an axiom?
For many sets, including any finite set, the first six axioms of set
theory (abbreviated ZF) are enough to guarantee the existence of a
choice function but there do exist sets for which AC is required to
show the existence of a choice function. The existence of such sets
was proved in 1963 by Paul Cohen. This means that AC cannot be derived
from the other six axioms; in other words "AC is independent of ZF."
This answers question [1] posed above.
The question of whether AC is consistent with the other axioms
(question [2] above) was answered by Goedel in 1938. Goedel showed
that if the other axioms are consistent then AC is consistent with
them. This is a "relative consistency" proof which is the best we can
hope for because of Goedel's Second Incompleteness Theorem.
The third question, "Should we accept it as an axiom?", moves us into
the realm of philosophy. Today there are three major schools of
thought concerning the use of AC:
1. Accept it as an axiom and use it without hesitation.
2. Accept it as an axiom but use it only when you cannot find a proof
without it.
3. AC is unacceptable.
Most mathematicians today belong to school A. Mathematicians who are
in school B are usually there because of a belief in Occam's Razor
(use as few assumptions as possible when explaining something) or an
interest in metamathematics. There are a growing number of people
moving to school C, especially computer scientists who work on
automated reasoning using constructive type theories.
Underlying the schools of thought about the use of AC are views about
truth and the nature of mathematical objects. Three major views are
platonism, constructivism, and formalism.
Platonism
A platonist believes that mathematical objects exist independent of
the human mind, and a mathematical statement, such as AC, is
objectively either true or false. A platonist accepts AC only if it is
objectively true, and probably falls into school A or C depending on
her belief. If she isn't sure about AC's truth then she may be in
school B so that once she finds out the truth about AC she will know
which theorems are true.
Constructivism
A constructivist believes that the only acceptable mathematical
objects are ones that can be constructed by the human mind, and the
only acceptable proofs are constructive proofs. Since AC gives no
method for constructing a choice set constructivists belong to school
C.
Formalism
A formalist believes that mathematics is strictly symbol manipulation
and any consistent theory is reasonable to study. For a formalist the
notion of truth is confined to the context of mathematical models,
e.g., a formalist would say "The parallel postulate is false in
Riemannian geometry." but she wouldn't say "The parallel postulate is
false." A formalist will probably not allign herself with any school.
She will comfortably switch between A, B, and C depending on her
current interests.
So: Should you accept the Axiom of Choice? Here are some arguments for
and against it.
Against
* It's not as simple, aesthetically pleasing, and intuitive as the
other axioms.
* It is equivalent to many statements which are not intuitive such
as "Every set can be well ordered." How, for example, would you
well order the reals?
* With it you can derive non-intuitive results, such as the
existence of a discontinuous additive function, the existence of a
non-measurable set of reals, and the Banach-Tarski Paradox (see
the next section of the sci.math FAQ).
* It is nonconstructive - it conjures up a set without providing any
sort of procedure for its construction.
For
The acceptance of AC is based on the belief that our intuition about
finite sets can be extended to infinite sets. The main argument for
accepting it is that it is useful. Many important, intuitively
plausible theorems are equivalent to it or depend on it. For example
these statements are equivalent to AC:
* Every vector space has a basis.
* Trichotomy of Cardinals: For any cardinals k and l , either k < l
or k = l or k > l .
* Tychonoff's Theorem: The product of compact spaces is compact in
the product topology.
* Zorn's Lemma: Every nonempty partially ordered set P in which each
chain has an upper bound in P has a maximal element.
And these statements depend on AC (i.e., they cannot be proved in ZF
without AC):
* The union of countably many countable sets is countable.
* Every infinite set has a denumerable subset.
* The Loewenheim-Skolem Theorem: Any first-order theory which has a
model has a denumerable model.
* The Baire Category Theorem: The reals are not the union of
countably many nowhere dense sets (i.e., the reals are not
meager).
* The Ultrafilter Theorem: Every Boolean algebra has an ultrafilter
on it.
Alternatives to AC
* Accept only a weak form of AC such as the Denumerable Axiom of
Choice (every denumerable set has a choice function) or the Axiom
of Dependent Choice.
* Accept an axiom that implies AC such as the Axiom of
Constructibility ( V = L ) or the Generalized Continuum Hypothesis
(GCH).
* Adopt AC as a logical axiom (Hilbert suggested this with his
epsilon axiom). If set theory is done in such a logical formal
system the Axiom of Choice will be a theorem.
* Accept a contradictory axiom such as the Axiom of Determinacy.
* Use a completely different framework for mathematics such as
Category Theory. Note that within the framework of Category Theory
Tychonoff's Theorem can be proved without AC (Johnstone, 1981).
Test Yourself: When is AC necessary?
If you are working in Zermelo-Fraenkel set theory without the Axiom of
Choice, can you choose an element from...
1. a finite set?
2. an infinite set?
3. each member of an infinite set of singletons (i.e., one-element
sets)?
4. each member of an infinite set of pairs of shoes?
5. each member of inifinite set of pairs of socks?
6. each member of a finite set of sets if each of the members is
infinite?
7. each member of an infinite set of sets if each of the members is
infinite?
8. each member of a denumerable set of sets if each of the members is
infinite?
9. each member of an infinite set of sets of rationals?
10. each member of a denumerable set of sets if each of the members is
denumberable?
11. each member of an infinite set of sets if each of the members is
finite?
12. each member of an infinite set of finite sets of reals?
13. each member of an infinite set of sets of reals?
14. each member of an infinite set of two-element sets whose members
are sets of reals?
The answers to these questions with explanations are accessible
through http://www.jazzie.com/ii/math/index.html
References
Benacerraf, Paul and Putnam, Hilary. "Philosophy of Mathematics:
Selected Readings, 2nd edition." Cambridge University Press, 1983.
Dauben, Joseph Warren. "Georg Cantor: His Mathematics and Philosophy
of the Infinite." Princeton University Press, 1979.
A. Fraenkel, Y. Bar-Hillel, and A. Levy with van Dalen, Dirk.
"Foundations of Set Theory, Second Revised Edition." North-Holland,
1973.
Johnstone, Peter T. "Tychonoff's Theorem without the Axiom of Choice."
Fundamenta Mathematica 113: 21-35, 1981.
Leisenring, Albert C. "Mathematical Logic and Hilbert's
Epsilon-Symbol." Gordon and Breach, 1969.
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.
Moore, Gregory H. "Zermelo's Axiom of Choice: Its Origins,
Development, and Influence." Springer-Verlag, 1982.
Rubin, Herman and Rubin, Jean E. "Equivalents of the Axiom of Choice
II." North-Holland, 1985.
This section of the FAQ is Copyright (c) 1994 Nancy McGough. Send
comments and or corrections relating to this part to nancym@ii.com.
The most up to date version of this section of the sci.math FAQ is
accesible through http://www.jazzie.com/ii/math/index.html
_________________________________________________________________
This section of the FAQ is Copyright (c) 1994 Nancy McGough. Send
comments and or corrections relating to this part to nancym@ii.com.
The most up to date version of this section of the sci.math FAQ is
accesible through http://www.jazzie.com/ii/math/index.html
Archive-Name: sci-math-faq/bibliography
Last-modified: December 8, 1994
Version: 6.1
GENERAL BIBLIOGRAPHY
The books listed in this section are both good introductory books to
mathematics, and classic references for the advanced mathematician.
* Algebra
Lang, Serge. Algebra
Birkhoff, McLane, Algebra
* Analysis
Rudin
Hewitt, Stromberg
* Geometry
David Hilbert. Foundations of Geometry 2nd English edition, tr. by
Leo Unger, publ. by Open Court, 1971.
* Number Theory
Hardy, Littlewood.
* Morris Kline Mathematical Thought from Ancient to Modern Times
*
Guillemin, Victor and Alan Pollack: Differential Topology. Spivak,
Michael: A Comprehensive Introduction to Differential Geometry,
Vol. I
* Morgan, Frank: Riemannian Geometry: A Beginner's Guide
* Greenberg, Martin and (?) Harper: Algebraic Topology: An
Introduction.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/AC/cuttingSphere
Last-modified: December 8, 1994
Version: 6.1
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?
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 is a ``piece"? (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.
*
Theorem [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.
* 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.
References
Hilbert's Third Problem. 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.
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. Stan Wagon. Cambridge University Press,
985
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:
* 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.
*
Theorem. There is a finite collection of disjoint open sets in the
unit cube in R^3 which can be moved by isometries to a finite
collection of disjoint open sets whose union is dense in the cube
of size 2 in R^3.
This result is by Foreman and Dougherty.
References
Boltyanskii. Equivalent and equidecomposable figures. in Topics in
Mathematics published by D.C. HEATH AND CO., Boston.
Dubins, Hirsch and ? Scissor Congruence American Mathematical Monthly.
``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?' ''
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/bourbaki
Last-modified: December 8, 1994
Version: 6.1
_________________________________________________________________
Next: Which are the Up: Mathematical Oddities Previous: Why is there
_________________________________________________________________
Who is N. Bourbaki?
A group of mostly French mathematicians which began meeting in the
1930s, aiming to write a thorough unified account of all mathematics.
They had tremendous influence on the way math is done since. For a
very accessible sampler see Dieudonne MATHEMATICS: THE MUSIC OF REASON
(orig. POUR l'HONNEUR DE l'ESPRIT HUMAIN).
The founding is described in Andre Weil's autobiography, titled
something like "memoir of an apprenticeship" (orig. SOUVENIRS
D'APPRENTISSAGE). There is a usable book BOURBAKI by J. Fang. Liliane
Beaulieu has a book forthcoming, which you can sample in "A Parisian
Cafe and Ten Proto-Bourbaki Meetings 1934-1935" in the MATHEMATICAL
INTELLIGENCER 15 no.1 (1993) 27-35.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/compoundInterest
Last-modified: December 8, 1994
Version: 6.1
Formula to compute compound interest.
Here's a formula which can be used in 123, Excel, Wings and Dynaplan:
------- Input this data -------------------------------
principal amount = E9 ( in dollars )
Amortization Period = d10 ( in years ie 6 mon = .5 )
Payments / year = D11 ( 12 = monthly, 52 = weekly )
Published Interest rate = D12 ( ie 9 % = 0.09 )
Times per year Int calculated = d13 ( CDN mortgage use 2
US mortgage use 12
all other loans use 12 )
----- Calculate the proper rate of interest -----------
e14 = Effective annual rate = EXP(D13*LN(1+(D12/D13)))-1
e15 = Interest rate per payment = (EXP(LN(E14+1)/(D10*D11))-1)*D10*D11
e17 = Payments = APMT(E9,E15/D11,D10*D11) ( both these functions are
= PMT (E9,E15/D11,D10*D11) ( identical,diff spreadsheet)
APMT( principal amount,interest rate per period,# periods )
( this is a standard function on any true commercial spreadsheet)
OR use the following if done using a calculator
= Payments = P*I/[1-(I+1)^-T]
= E9*(E15/D11)/(1-((E15/D11) +1)**(-1*D10*D11))
Total interest cost = E17*D10*D11-E9
-- Use these formulas if you wish to generate an amortization table --
always add up to 'Payments (e17)'
Interest per payment = current balance * ( E15 / D11 )
Principal per payment = current balance - Interest per payment
new current balance = current balance - Principal per payment -
(extra payment)
keep repeating until 'new current balance' = 0
Derivation of Compound Interest Rate Formula
Suppose you deposited a fixed payment into an interest bearing account
at regular intervals, say monthly, at the end of each month. How much
money would there be in the account at the end of the nth month (at
which point you've made i payments)?
Let x be the monthly interest rate as a fraction of principle.
Let n be the amount deposited each month.
Let p[k] be the total number of months.
Let k be the principle after p[n] = x + ((1 + i) p[n - 1]) eq 1
months.
So the recursive formula is:
p[n] = sum_(k = 0)^(n - 1)x (1 + i)^k
This yields the summation:
(1 + i)
The way to solve this is to multiply through by pi p[n] = x ((1 + i)^n
- 1) and subtract the original equation from the resulting equation.
Observe that all terms in the summation cancel except the last term of
the multiplied equation and the first term of the original equation:
p[n] = x ((1 + i)^n - 1)/i
or
p
Now suppose you borrow i at constant interest rate x . You make
monthly payments of p . It turns out that this problem is identical to
taking out a balloon loan of x (that is it's all due at the end of
some term) and putting payments of n into a savings account. At the
end of the term you use the principle in the savings account to pay
off the balance of the loan. The loan and the savings account, of
course, must be at the same interest rate. So what we want to know is:
what monthly payment is needed so that the balance of the savings
account will be identical to the balance of the balloon loan after n
payments?
The formula for the principal of the balloon loan at the end of the
p[n] = p[0] (1 + i)^n th month is:
p[0] (1 + i)^n = x ((1 + i)^n - 1)/i
So we set this expression equal to the expression for the the savings
account, and we get:
x = p[0] (1 + i)^n i/((1 + i)^n - 1)
or solving for x:
(1 + i)^n
If n is large enough (say greater than 5), here is an approximation
for determining x from p , i , and n ~= -ln(ln(x/(ip)))/ln(1 + i) :
ln(y - 1) ~= ln y - 1/y
The above approximation is based upon the following approximation:
y >= 5
Which is within 2 For example, a $100000loan at 1%monthly, paying
$1028.61 per month should be paid in 360 months. The approximation
yields 358.9 payments.
If this were your 30 year mortgage and you were paying $1028.61 per
month and you wanted to see the effect of paying $1050per month, the
approximation tells you that it would be paid off in 303.5 months (25
years and 3.5 months). If you stick 304 months into the equation for x
, you get $1051.04, so it is fairly close. This approximation does not
work, though, for very small interest rates or for a small number of
payments. The rule is to get a rough idea first of what (1 + i)^n is.
If that is greater than 5, the approximation works pretty well. In the
examples given, (1 + i)^n is about 36.
Finding i given n , x , and p is not as easy. If i is less than 5%per
payment period, the following equation approximately holds for i:
i = -(1/n) ln(1 - ip/x)
There is no direct solution to this, but you can do it by
Newton-Raphson approximation. Begin with a guess, i[0]. Then apply:
i[k + 1] = i[k] - (x(1 - i[k]p/x) (ni[k] + ln(1 - i[k]p/x)))/(xn(1 -
i[k]p/x) - p)
You must start with i too big, because the equation for i has a
solution at i = 0 , and that's not the one you want to end up with.
Example: Let the loan be for p = $10000 , x = $50 per week for 5 years
( n = 260 ). Let i[0] = 20 %per annum or 0.3846%per week. Since i must
be a fraction rather than a percent, i[0] = 0.003846 . Then, applying
eq 11:
i[1] = 0.003077 ; i[2] = 0.002479 ; i[3] = 0.002185 ; i[4] = 0.002118
; i[5] = 0.002115 ;
The series is clearly beginning to converge here.
To get i[5] as an annual percentage rate, multiply by 52 weeks in a
year and then by 100%, so i[5] = 10.997 %per annum. Substituting i[5]
back into eq 7, we get x = $50.04 , so it works pretty well.
References
The theory of interest. Stephen G. Kellison. Homewood, Ill., R. D.
Irwin, 1970.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/specialnumbers/computePi
Last-modified: December 8, 1994
Version: 6.1
How to compute digits of pi ?
Symbolic Computation software such as Maple or Mathematica can compute
10,000 digits of pi in a blink, and another 20,000-1,000,000 digits
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. New York's
Chudnovsky brothers have computed 2 billion digits of pi on a homebrew
computer.
There are essentially 3 different methods to calculate pi to many
decimals.
1. One of the oldest is to use the power series expansion of atan(x)
= x - x^3/3 + x^5/5 - ... together with formulas like pi =
16*atan(1/5) - 4*atan(1/239) . This gives about 1.4 decimals per
term.
2. A second is to use formulas coming from Arithmetic-Geometric mean
computations. A beautiful compendium of such formulas is given in
the book pi and the AGM, (see references). They have the advantage
of converging quadratically, i.e. you double the number of
decimals per iteration. For instance, to obtain 1 000 000
decimals, around 20 iterations are sufficient. The disadvantage is
that you need FFT type multiplication to get a reasonable speed,
and this is not so easy to program.
3. A third one comes from the theory of complex multiplication of
elliptic curves, and was discovered by S. Ramanujan. This gives a
number of beautiful formulas, but the most useful was missed by
Ramanujan and discovered by the Chudnovsky's. It is the following
(slightly modified for ease of programming):
Set k_1 = 545140134; i k_2 = 13591409; k_3 = 640320; k_4 =
100100025; k_5 = 327843840; k_6 = 53360;
Then pi = (k_6 sqrt(k_3))/(S) , where
S = sum_(n = 0)^oo (-1)^n ((6n)!(k_2 +
nk_1))/(n!^3(3n)!(8k_4k_5)^n)
The great advantages of this formula are that
1) It converges linearly, but very fast (more than 14 decimal
digits per term).
2) The way it is written, all operations to compute S can be
programmed very simply since it only involves
multiplication/division by single precision numbers. This is why
the constant 8k_4k_5 appearing in the denominator has been written
this way instead of 262537412640768000. This is how the
Chudnovsky's have computed several billion decimals.
The following 160 character C program, written by Dik T. Winter at
CWI, computes pi to 800 decimal digits.
int a=10000,b,c=2800,d,e,f[2801],g;main(){for(;b-c;)f[b++]=a/5;
for(;d=0,g=c*2;c-=14,printf("%.4d",e+d/a),e=d%a)for(b=c;d+=f[b]*a,
f[b]=d%--g,d/=g--,--b;d*=b);}
References
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.
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.
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.
David Chudnovsky and Gregory Chudnovsky. The computation of classical
constants. Columbia University, Proc. Natl. Acad. Sci. USA, Vol. 86,
1989.
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
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.
A history of pi . P. Beckman. Golem Press, CO, 1971 (fourth edition
1977)
pi and the AGM - a study in analytic number theory and computational
complexity. J.M. Borwein and P.B. Borwein. Wiley, New York, 1987.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/AC/ContinuumHyp
Last-modified: December 8, 1994
Version: 6.1
THE CONTINUUM HYPOTHESIS
A basic reference is Godel's "What is Cantor's Continuum Problem?",
from 1947 with a 1963 supplement, reprinted in Benacerraf and Putnam's
collection Philosophy of Mathematics. This outlines Godel's generally
anti-CH views, giving some "implausible" consequences of CH.
"I believe that adding up all that has been said one has good reason
to suspect that the role of the continuum problem in set theory will
be to lead to the discovery of new axioms which will make it possible
to disprove Cantor's conjecture."
At one stage he believed he had a proof that C = aleph_2 from some new
axioms, but this turned out to be fallacious. (See Ellentuck, "Godel's
Square Axioms for the Continuum", Mathematische Annalen 1975.)
Two people pointed me to a recent paper by Maddy: "Believing the
Axioms", Journal of Symbolic Logic 1988 (in 2 parts). This is an
extremely interesting paper and a lot of fun to read. A bonus is that
it gives a non-set-theorist who knows the basics a good feeling for a
lot of issues in contemporary set theory.
Most of the first part is devoted to "plausible arguments" for or
against CH: how it stands relative to both other possible axioms and
to various set-theoretic "rules of thumb". One gets the feeling that
the weight of the arguments is against CH, although Maddy says that
many "younger members" of the set-theoretic community are becoming
more sympathetic to CH than their elders. There's far too much here
for me to be able to go into it in much detail.
Some highlights from Maddy's discussion, also incorporating a few
things that other people sent me:
1. Cantor's reasons for believing CH aren't all that persuasive
today.
2. Godel's proof of the consistency of CH shows that CH follows from
ZFC plus the Axiom of Constructibility ( V = L , roughly that the
set-theoretic universe = the constructible universe). However,
most set-theorists seem to find Constructiblity implausible and
much too restrictive. It's an example of a "minimizing" principle,
which tends to cut down on the number of sets admitted to one's
universe. Apparently "maximizing" principles meet with much more
sympathy from set theorists. Such principles are more compatible
with not CH than with CH.
3. If GCH is true, this implies that aleph_0 has certain unique
properties: e.g. that it's that cardinal before which GCH is false
and after which it is true. Some would like to believe that the
set-theoretic universe is more "uniform" (homogeneous) than that,
without this kind of singular occurrence. Such a "uniformity"
principle tends to imply not GCH.
4. Most of those who disbelieve CH think that the continuum is likely
to have very large cardinality, rather than aleph_2 (as Godel
seems to have suggested). Even Cohen, a professed formalist,
argues that the power set operation is a strong operation that
should yield sets much larger than those reached quickly by
stepping forward through the ordinals:
"This point of view regards C as an incredibly rich set given to us
by a bold new axiom, which can never be approached by any piecemeal
process of construction."
5. There are also a few arguments in favour of CH, e.g. there's an
argument that not CH is restrictive (in the sense of (2) above).
Also, CH is much easier to force (Cohen's method) than not -CH.
And CH is much more likely to settle various outstanding results
than is not CH, which tends to be neutral on these results.
6. Most large cardinal axioms (asserting the existence of cardinals
with various properties of hugeness: these are usually derived
either from considering the hugeness of aleph_0 compared to the
finite cardinals and applying uniformity, or from considering the
hugeness of V (the set-theoretic universe) relative to all sets
and applying "reflection") don't seem to settle CH one way or the
other.
7. Various other axioms have some bearing. Axioms of determinacy
restrict the class of sets of reals that might be counterexamples
to CH. Various forcing axioms (e.g. Martin's axiom), which are
"maximality" principles (in the sense of (2) above), imply not CH.
The strongest (Martin's maximum) implies that C = aleph_2 . Of
course the "truth" or otherwise of all these axioms is
controversial.
8. Freiling's principle about "throwing darts at the real line" is a
seemingly very plausible principle, not involving large cardinals
at all, from which not CH immediately follows. Freiling's paper
(JSL 1986) is a good read. More on this at the end of this
message.
Of course I've conspicuously avoided saying anything about whether
it's even reasonable to suppose that CH has a determinate truth-value.
Formalists will argue that we may choose to make it come out whichever
way we want, depending on the system we work in. On the other hand,
the mere fact of its independence from ZFC shouldn't immediately lead
us to this conclusion - this would be assigning ZFC a privileged
status which it hasn't necessarily earned. Indeed, Maddy points out
that various axioms within ZFC (notably the Axiom of Choice, and also
Replacement) were adopted for extrinsic reasons (e.g. "usefulness") as
well as for "intrinsic" reasons (e.g. "intuitiveness"). Further
axioms, from which CH might be settled, might well be adopted for such
reasons.
One set-theorist correspondent said that set-theorists themselves are
very loathe to talk about "truth" or "falsity" of such claims.
(They're prepared to concede that 2 + 2 = 4 is true, but as soon as
you move beyond the integers trouble starts. e.g. most were wary even
of suggesting that the Riemann Hypothesis necessarily has a
determinate truth-value.) On the other hand, Maddy's contemporaries
discussed in her paper seemed quite happy to speculate about the
"truth" or "falsity" of CH.
Personally, not only do I see the integers as bedrock, but I'm also
prepared to take any finite number of power sets before I have any
problems. At least that far, I'm a diehard Platonist. I'm considerably
less sure whether to be a diehard Platonist about the paradise of
ridiculously big cardinals. But my intuition is happy with reals and
sets of reals, so as a naive non-set-theorist I'm sticking to the
belief that CH is determinate one way or the other. As one
correspondent suggested, the question of the determinateness of CH is
perhaps the single best way to separate the Platonists from the
formalists.
And is it true or false? Well, I'd always found CH somewhat
intuitively plausible. But after reading all this, it does seem that
the weight of evidence tend to point the other way. On the other hand,
that's only a second-order approximation based on limited knowledge -
more subtle and sophisticated arguments (third-order approximations)
might begin to push things the other way. Still, if I had to lay money
on it (God, of course, knows the right answer and could settle the
bet), I'd feel safer with the money on not CH.
I've enclosed (with permission) a brief but enlightening message from
Bill Allen on Freiling's Axiom of Symmetry. This is a good one to run
your intuitions by.
Many thanks to Bill Allen, James Cummings, David Feldman, Torkel
Franzen, Calvin Ostrum, Keith Ramsay, and Peter Suber. I'd be very
interested to hear more about this subject. Let A be the set of
functions mapping Real Numbers into countable sets of Real Numbers.
Given a function f in A , and some arbitrary real numbers x and y , we
see that x is in f(y) with probability 0, i.e. x is not in f(y) with
probability 1. Similarly, y is not in f(x) with probability 1. Let AX
be the axiom which states
"for every f in A , there exist x and y such that x is not in f(y) and
y is not in f(x) "
The intuitive justification for AX is that we can find the x and y by
choosing them at random.
In ZFC, AX = not CH. proof: If CH holds, then well-order R as r_0,
r_1, .... , r_x, ... with x < aleph_1 . Define f(r_x) as { r_y : y >=
x } . Then f is a function which witnesses the falsity of AX.
If CH fails, then let f be some member of A . Let Y be a subset of R
of cardinality aleph_1 . Then Y is a proper subset. Let X be the union
of all the sets f(y) with y in Y , together with Y . Then, as X is an
aleph_1 union of countable sets, together with a single aleph_1 size
set Y , the cardinality of X is also aleph_1 , so X is not all of R .
Let a be in R X , so that a is not in f(y) for any y in Y . Since f(a)
is countable, there has to be some b in Y such that b is not in f(a) .
Thus we have shown that there must exist a and b such that a is not in
f(b) and b is not in f(a) . So AX holds.
I like Freiling's proof, since it does not invoke large cardinals or
intense infinitary combinatorics to make the point that CH implies
counter-intuitive propositions. Freiling has also pointed out that the
natural extension of AX is AXL (notation mine), where AXL is AX with
the notion of countable replaced by Lebesgue Measure zero. Freiling
has established some interesting Fubini-type theorems using AXL.
See "Axioms of Symmetry: Throwing Darts at the Real Line", by
Freiling, Journal of Symbolic Logic, 51, pages 190-200. An extension
of this work appears in "Some properties of large filters", by
Freiling and Payne, in the JSL, LIII, pages 1027-1035, but Chris tells
me he's not as fond of the latter paper as he is the former.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Thu Dec 08 20:49:28 EST 1994
Archive-Name: sci-math-faq/dayWeek
Last-modified: December 8, 1994
Version: 6.1
How to determine the day of the week, given the month, day and year
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 + 4004 . 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 over 10,000 years 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:
1. Take the last two digits of the year.
2. Divide by 4, discarding any fraction.
3. Add the day of the month.
4. Add the month's key value: JFM AMJ JAS OND 144 025 036 146
5. Subtract 1 for January or February of a leap year.
6. 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.
7. For a Julian date, add 1 for 1700's, and 1 for every additional
century you go back.
8. Add the last two digits of the year.
9. 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 + floor(2.6m - 0.2) - 2C + Y + floor(Y/4) +
floor(C/4)) mod 7 where floor() denotes the integer floor function,
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 and 400 year corrections are built into the formula.
The floor(2.6m - 0.2) term relates to the repetitive pattern that the
30-day months show when March is taken as the first month.
The following short C program works for a restricted range
dow(m,d,y){y-=m<3;return(y+y/4-y/100+y/400+"-bed=pen+mad."[m]+d)%7;}
The program appeared was posted by sakamoto@sm.sony.co.jp (Tomohiko
Sakamoto) on comp.lang.c on March 10th, 1993.
A good mnemonic rule to help on the computation of the day of the week
is as follows. In any given year the following days come on the same
day of the week:
4/4
6/6
8/8
10/10
12/12
to remember the next for remember that I work from 9-5 at a 7-11 so
9/5
5/9
7/11
11/7
and the last day of Feb.
This year they come on a Monday. Every year this advances one (so next
year they come on a Tuesday) other than leap-years which advance 2.
Thus every 4 years it advances 5 days (there is one more minor
correction at the century mark).
Even ignoring the pattern over for a period of years this is still
useful since you can generally figure out what day of the week a given
date is on faster than someone else can look it up with a calender if
the calender is not right there. (A useful skill that.)
References
Winning Ways for your mathematical plays. Guy Conway and Elwyn
Berlekamp London ; Toronto : Academic Press, 1982.
Mathematical Carnival. Martin Gardner. New York : Knopf, c1975.
Elementary Number Theory and its applications. Kenneth Rosen. Reading,
Mass. ; Don Mills, Ont. : Addison-Wesley Pub. Co., c1993. p. 156.
Michael Keith and Tom Craver. The Ultimate Perpetual Calendar? Journal
of Recreational Mathematics, 22:4, pp. 280-282, 19
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/erdos
Last-modified: December 8, 1994
Version: 6.1
Erdos Number
Form an undirected graph where the vertices are academics, and an edge
connects academic X to academic Y if X has written a paper with Y .
The Erdos number of X is the length of the shortest path in this graph
connecting X with Erdos.
Erdos has Erdos number 0. Co-authors of Erdos have Erdos number 1.
Einstein has Erdos number 2, since he wrote a paper with Ernst Straus,
and Straus wrote many papers with Erdos.
The Extended Erdos Number applies to co-authors of Erdos. For People
who have authored more than one paper with Erdos, their Erdos number
is defined to be 1/# papers-co-authored. Ron Graham has the smallest,
non-zero, Erdos number.
Why people care about it?
Nobody seems to have a reasonable answer...
Who is Paul Erdos?
Paul Erdos is an Hungarian mathematician. He obtained his PhD from the
University of Manchester and has spent most of his efforts tackling
"small" problems and conjectures related to graph theory,
combinatorics, geometry and number theory.
He is one of the most prolific publishers of papers; and is also and
indefatigable traveller.
References
Caspar Goffman. And what is your Erdos number? American Mathematical
Monthly, v. 76 (1969), p. 791.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/specialnumbers/eulerFormula
Last-modified: December 8, 1994
Version: 6.1
Euler's formula: e^(i pi) = -1 , What does it mean and why is it equal to -1
The definition and domain of exponentiation has been changed several
times. The original operation x^y was only defined when y was a
positive integer. The domain of the operation of exponentation has
been extended, not so much because the original definition made sense
in the extended domain, but because there were (almost) unique ways to
extend exponentation which preserved many of what seemed to be the
"important" properties of the original operation. So in part, these
definitions are only convention, motivated by reasons of aesthetics
and utility.
The original definition of exponentiation is, of course, that x^y = x
*x * ... * x, where x is multiplied by itself y times. This is only a
reasonable definition for y = 1, 2, 3, ... (It could be argued that it
is reasonable when y = 0 , but that issue is taken up in a different
part of the FAQ). This operation has a number of properties, including
1. x^1 = x
2. For any x , n , m , x^n x^m = x^(n + m) .
3. If x is positive, then x^n is positive.
Now, we can try to see how far we can extend the domain of
exponentiation so that the above properties (and others) still
hold. This naturally leads to defining the operation x^y on the
domain x positive real; y rational, by setting x^(p/q) = the
q^(th) root of x^p . This operation agrees with the original
definition of exponentiation on their common domain, and also
satisfies (1), (2) and (3). In fact, it is the unique operation on
this domain that does so. This operation also has some other
properties:
4. If x > 1 , then x^y is an increasing function of y .
5. If 0 < x < 1 , then x^y is a decreasing function of y .
Again, we can again see how far we can extend the domain of
exponentiation while still preserving properties (1)-(5). This
leads naturally to the following definition of x^y on the domain x
positive real; y real:
If x > 1 , x^y is defined to be sup_q x^q , where q runs over all
rationals less than or equal to y .
If x < 1 , x^y is defined to be in f_q x^q , where q runs over all
rationals less than or equal to y .
If x = 1 , x^y is defined to be 1 .
Again, this operation satisfies (1)-(5), and is in fact the only
operation on this domain to do so.
The next extension is somewhat more complicated. As can be proved
using the methods of calculus, if we define e to be the number
e = 1 + 1/1! + 1/2! + 1/3! + ... = 2.71828...
it turns out that for every real number x ,
6. e^x = 1 + x/1! + x^2/2! + x^3/3! + ...
e^x is also denoted exp(x) . (This series always converges
regardless of the value of x ).
One can also define an operation ln(x) on the positive reals,
which is the inverse of the operation of exponentiation by e. In
other words, exp(ln(x)) = x for all positive x . Moreover,
7. If x is positive, then x^y = exp(y ln(x)) . Because of this, the
natural extension of exponentiation to complex exponents, seems to
be to define
exp(z) = 1 + z/1! + z^2/2! + z^3/3! + ...
for all complex z (not just the reals, as before), and to define
x^z = exp(z ln(x))
when x is a positive real and z is complex.
This is the only operation x^y on the domain x positive real, y
complex which satisfies all of (1)-(7). Because of this and other
reasons, it is accepted as the modern definition of
exponentiation.
From the identities
sin x = x - x^3/3! + x^5/5! - x^7/7! + ...
cos x = 1 - x^2/2! + x^4/4! - x^6/6! + ...
which are also obtainable from calculus, one sees that, for any
real x,
8. exp(ix) = cos x + i sin x.
Thus, we get Euler's famous formula
e^(pi i) = -1
and
e^(2pi i) = e^0 = 1.
One can also obtain the classical addition formulae for sine and
cosine from (8) and (1).
All of the above extensions have been restricted to a positive real
for the base. When the base x is not a positive real, it is not as
clear-cut how to extend the definition of exponentiation. For example,
(-1)^(1/2) could well be i or -i, (-1)^(1/3) could be -1 , 1/2 +
sqrt(3)i/2 , or 1/2 - sqrt(3)i/2 , and so on. Some values of x and y
give infinitely many candidates for x^y , all equally plausible. And
of course x = 0 has its own special problems. These problems can all
be traced to the fact that the exp function is not injective on the
complex plane, so that ln is not well defined outside the real line.
There are ways around these difficulties (defining branches of the
logarithm, for example), but we shall not go into this here.
The operation of exponentiation has also been extended to other
systems like matrices and operators. The key is to define an
exponential function by (6) and work from there. [Some reference on
operator calculus and/or advanced linear algebra?]
References
Complex Analysis. Ahlfors, Lars V. McGraw-Hill, 1953.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/fields
Last-modified: December 8, 1994
Version: 6.1
FIELDS MEDAL
_________________________________________________________________
* Historical Introduction
* Table of Awardees
_________________________________________________________________
Historical Introduction
This is the original letter by Fields creating the endowment for the
medals that bear his name. It is thought to have been written during
the few onths before his death. Notice that no mention is made about
the age of the recipients (currently there is a 40 year-old limit),
and that the medal should not be attached to any person, private or
public, meaning that it shouldn't bear anybody's name.
It is proposed to found two gold medals to be awarded at successive
International Mathematical Congress for outstanding achievements in
mathematics. Because of the multiplicity of the branches of
mathematics and taking into account the fact that the interval
between such congresses is four years it is felt that at least two
medals should be available. The awards would be open to the whole
world and would be made by an International Comitee.
The fund for the founding of the medals is constituted by balance
left over after financing the Toronto congress held in 1924. This
must be held in trust by the Government or by some body authorized
by government to hold and invest such funds. It would seem that a
dignified method for handling the matter and one which in this
changing world should most nearly secure permanency would be for the
Canadian Government to take over the fund and appoint as his
custodian say the Prime Minister of the Dominion or the Prime
Minister in association with the Minister of Finance. The medals
would be struck at the Mint in Ottawa ans the duty of the custodian
would be simply to hand over the medals at the proper time to the
accredited International Committee.
As things are at present a practical course of procedure would seem
to be for the Executive Committee of a Congress to appoint a small
international committee authorized to add to its number and call
into consultation other mathematicians as it might demm expedient.
The Committee would be expected to decide on the ones to whom the
awards should be made thirty months in advance of the following
Congress. Its decisions would be communicated to the President and
Secretary of the Organizing Committee of the Congress, this
Committee having the duty of communicating to the Prime Minister of
Canada the names of the recipients in order that the medal might be
prepared in time and forwarded to the president of the Organizing
Committee. Immediately on the appointment of the Executive Committee
of the Congress the medals would be handed over to its President.
The presentation of the medals would constitute a special feature at
some general meeting of the Congress.
In the above arrangements the role of the Organizing Committee might
be taken over by the Executive of the International Mathematical
Union at some time in the future when that organization has benn
generally accepted.
In coming to its decision the hands of the IC should be left as free
as possible. It would be understood, however, that in making the
awards while it was in recognition of work already done it was at
the same time intended to be an encouragement for further
achievement on the part of the recipients and a stimulus to renewed
effort on the part of others.
In commenting on the work of the medalists it might be well to be
conservative in one's statements to avoid envidious comparisons
explicit or implied. The Committee might ease matters by saying they
have decided to make the awards along certain lines not alone
because of the outstanding character of the achievement but also
with a view to encouraging further development along these lines. In
this connection the Committe might say that they had elected to
select subjects in Analysis, in Geometry, in the Theory of Groups,
in the Theory of Numbers etc. as the case might be. When the
Committee had come to an agreement in this sense the claims for
recognition of work done along the special lines in question could
be considered in detail by two smaller groups or subcommittees with
specialized qualifications who would have authority to take into
consultation or add to the subcommittees other mathematicians of
specialized knowledge.
With regard to the medals themselves, I might say that they should
each contain at least 200 dollars worth of gold and be of a fair
size, probably 7.5 centimeters in diameter. Because of the
international character the language to be employed it would seem
should be Latin or Greek? The design has still to be definitely
determined. It will have to be decided on by artists in consultation
with mathematicians. The suggestions made in the preceding are
tentative and open to consideration on the part of mathematicians.
It is not contemplated to make an award until 1936 at the Congress
following that at Zurich during which an international Medal
Committee should be named.
The above programme means a new departure in the matter of
international scientific cooperation and is likely to be the
precursor of moves along like lines in other sciences than
mathematics.
One would hear again emphasized the fact that the medals should be
of a character as purely international and impersonal as possible.
There should not be attached to them in any way the name of any
country, institution or person.
Perhaps provision could be made as soon as possible after the
appointment of the Executive of the Zurich Congress for the
consideration by it of the subject of the medals, and the
appointment without undue delay of a Commitee and the awards of the
medals to be made in connection with the Congress of 1936.
Suggestions with regard to the design of the medals will be welcome.
(signed) J.C. Fields Research Professor of Mathematics University of
Toronto
_________________________________________________________________
Table of Awardees
Table of Awardees
Year Name Birthplace Country Age
1936 Ahlfors, Lars Helsinki Finland 29
1936 Douglas, Jesse New York, NY USA 39
1950 Schwartz, Laurent Paris France 35
1950 Selberg, Atle Langesund Norway 33
1954 Kodaira, Kunihiko Tokyo Japan 39
1954 Serre, Jean-Pierre Bages France 27
1958 Roth, Klaus Breslau Germany 32
1958 Thom, Rene Montbeliard France 35
1962 Hormander, Lars Mjallby Sweden 31
1962 Milnor, John Orange, NJ USA 31
1966 Atiyah, Michael London UK 37
1966 Cohen, Paul Long Branch NJ USA 32
1966 Grothendieck, Alexander Berlin Germany 38
1966 Smale, Stephen Flint, MI USA 36
1970 Baker, Alan London UK 31
1970 Hironaka, Heisuke Yamaguchi-ken Japan 39
1970 Novikov, Serge Gorki USSR 32
1970 Thompson, John Ottawa, KA USA 37
1974 Bombieri, Enrico Milan Italy 33
1974 Mumford, David Worth, Sussex UK 37
1978 Deligne, Pierre Brussels Belgium 33
1978 Fefferman, Charles Washington DC USA 29
1978 Margulis, Gregori Moscow USSR 32
1978 Quillen, Daniel Orange, NJ USA 38
1982 Connes, Alain Draguignan France 35
1982 Thurston, William Washington DC USA 35
1982 Yau, Shing-Tung Kwuntung China 33
1986 Donaldson, Simon Cambridge UK 27
1986 Faltings, Gerd Germany 32
1986 Freedman, Michael Los Angeles USA 35
1990 Drinfeld, Vladimir Kharkov USSR 36
1990 Jones, Vaughan Gisborne N Zealand 38
1990 Mori, Shigefumi Nagoya Japan 39
1990 Witten, Edward Baltimore USA 38
Year Name Institution Country
1936 Ahlfors, Lars Harvard University USA
1936 Douglas, Jesse MIT USA
1950 Schwartz, Laurent Universite de Nancy France
1950 Selberg, Atle Princeton/Inst. of Advanced Studies USA
1954 Kodaira, Kunihiko Princeton University USA
1954 Serre, Jean-Pierre College de France France
1958 Roth, Klaus University of London UK
1958 Thom, Rene University of Strasbourg France
1962 Hormander, Lars University of Stockholm Sweden
1962 Milnor, John Princeton University USA
1966 Atiyah, Michael Oxford University UK
1966 Cohen, Paul Stanford University USA
1966 Grothendieck, Alex University of Paris France
1966 Smale, Stephen University of California at Berkeley USA
1970 Baker, Alan Cambridge University UK
1970 Hironaka, Heisuke Harvard University USA
1970 Novikov, Serge Moscow University USSR
1970 Thompson, John University of Chicago USA
1974 Bombieri, Enrico Univeristy of Pisa Italy
1974 Mumford, David Harvard University USA
1978 Deligne, Pierre IHES France
1978 Fefferman, Charles Princeton University USA
1978 Margulis, Gregori InstPrblmInfTrans USSR
1978 Quillen, Daniel MIT USA
1982 Connes, Alain IHES France
1982 Thurston, William Princeton University USA
1982 Yau, Shing-Tung IAS USA
1986 Donaldson, Simon Oxford University UK
1986 Faltings, Gerd Princeton University USA
1986 Freedman, Michael University of California at San Diego USA
1990 Drinfeld, Vladimir Phys.Inst.Kharkov USSR
1990 Jones, Vaughan University of California at Berkeley USA
1990 Mori, Shigefumi University of Kyoto? Japan
1990 Witten, Edward Princeton/Institute of Advanced Studies USA
References
International Mathematical Congresses, An Illustrated History
1893-1986. Donald J.Alberts, G. L. Alexanderson and Constance Reid.
Revised Edition, Including 1986, Springer Verlag, 1987.
Tropp, Henry S. The origins and history of the Fields Medal. Historia
Mathematica, 3(1976), 167-181.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/FLT/Fermat
Last-modified: December 8, 1994
Version: 6.1
Did Fermat prove this theorem?
No he did not. Fermat claimed to have found a proof of the theorem at
an early stage in his career. Much later he spent time and effort
proving the cases n = 4 and n = 5 . Had he had a proof to his theorem,
there would have been no need for him to study specific cases.
Fermat may have had one of the following ``proofs'' in mind when he
wrote his famous comment.
* Fermat discovered and applied the method of infinite descent,
which, in particular can be used to prove FLT for n = 4 . This
method can actually be used to prove a stronger statement than FLT
for n = 4 , viz, x^4 + y^4 = z^2 has no non-trivial integer
solutions. It is possible and even likely that he had an incorrect
proof of FLT using this method when he wrote the famous
``theorem''.
* He had a wrong proof in mind. The following proof, proposed first
by Lame' was thought to be correct, until Liouville pointed out
the flaw, and by Kummer which latter became and expert in the
field. It is based on the incorrect assumption that prime
decomposition is unique in all domains.
The incorrect proof goes something like this:
We only need to consider prime exponents (this is true). So
consider x^p + y^p = z^p . Let r be a primitive p -th root of
unity (complex number)
Then the equation is the same as:
(x + y)(x + ry)(x + r^2y)...(x + r^(p - 1)y) = z^p
Now consider the ring of the form:
a_1 + a_2 r + a_3 r^2 + ... + a_(p - 1) r^(p - 1)
where each a_i is an integer
Now if this ring is a unique factorization ring (UFR), then it is
true that each of the above factors is relatively prime. From this
it can be proven that each factor is a pth power and from this FLT
follows.
The problem is that the above ring is not an UFR in general.
Another argument for the belief that Fermat had no proof -and,
furthermore, that he knew that he had no proof- is that the only place
he ever mentioned the result was in that marginal comment in Bachet's
Diophantus. If he really thought he had a proof, he would have
announced the result publicly, or challenged some English
mathematician to prove it. It is likely that he found the flaw in his
own proof before he had a chance to announce the result, and never
bothered to erase the marginal comment because it never occurred to
him that anyone would see it there.
Some other famous mathematicians have speculated on this question.
Andre Weil, writes:
Only on one ill-fated occasion did Fermat ever mention a curve of
higher genus x^n + y^n = z^n , and then hardly remain any doubt that
this was due to some misapprehension on his part [for a brief moment
perhaps [he must have deluded himself into thinking he had the
principle of a general proof.
Winfried Scharlau and Hans Opolka report:
Whether Fermat knew a proof or not has been the subject of many
speculations. The truth seems obvious ...[Fermat's marginal note]
was made at the time of his first letters concerning number theory
[1637]...as far as we know he never repeated his general remark, but
repeatedly made the statement for the cases n = 3 and 4 and posed
these cases as problems to his correspondents [he formulated the
case n = 3 in a letter to Carcavi in 1659 [All these facts indicate
that Fermat quickly became aware of the incompleteness of the
[general] ``proof" of 1637. Of course, there was no reason for a
public retraction of his privately made conjecture.
However it is important to keep in mind that Fermat's ``proof"
predates the Publish or Perish period of scientific research in which
we are still living.
References
From Fermat to Minkowski: lectures on the theory of numbers and its
historical development. Winfried Scharlau, Hans Opolka. New York,
Springer, 1985.
Basic Number Theory. Andre Weil. Berlin, Springer, 1967
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/FLT/status
Last-modified: December 8, 1994
Version: 6.1
What is the current status of FLT?
Theorem 1 (Fermat's Last Theorem) There are no positive integers x, y,
z, and n>2 such that x^n + y^n = z^n.
Andrew Wiles, a researcher at Princeton, claims to have found a proof.
The proof was presented in Cambridge, UK during a three day seminar to
an audience which included some of the leading experts in the field.
The proof is long and cumbersome. At this point the reviewers have
found a glitch in the argument. Late last year, Prof. Wiles
acknowledged that a gap existed. On October 25th, 1994. Prof. Andrew
Wiles released two preprints, Modular elliptic curves and Fermat's
Last Theorem, by Andrew Wiles, and Ring theoretic properties of
certain Hecke algebras, by Richard Taylor and Andrew Wiles.
The first one (long) announces a proof of, among other things,
Fermat's Last Theorem, relying on the second one (short) for one
crucial step.
The argument described by Wiles in his Cambridge lectures had a
serious gap, namely the construction of an Euler system. After trying
unsuccessfully to repair that construction, Wiles went back to a
different approach he had tried earlier but abandoned in favor of the
Euler system idea. He was able to complete his proof, under the
hypothesis that certain Hecke algebras are local complete
intersections. This and the rest of the ideas described in Wiles'
Cambridge lectures are written up in the first manuscript. Jointly,
Taylor and Wiles establish the necessary property of the Hecke
algebras in the second paper.
The new approach turns out to be significantly simpler and shorter
than the original one, because of the removal of the Euler system. (In
fact, after seeing these manuscripts Faltings has apparently come up
with a further significant simplification of that part of the
argument.)
The preprints were submitted to The Annals of Mathematics. According
to the New York Times the new proof has been vetted by four
researchers already, who have found no mistake.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/FLT/Wrong
Last-modified: December 8, 1994
Version: 6.1
If not, then what?
FLT is usually 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 7.568x10^(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. More recently, Fermat's Last Theorem has been
proved true up to exponent 4,000,000 in the general case. The method
used was essentially that of Wagstaff: enumerating and eliminating
irregular primes by Bernoulli number computations. The computations
were performed on a set of NeXT computers by Richard Crandall et al.
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)
1. 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
by Serge Lang 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 that
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 implies
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.
2. 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].
3. Conjectures arising from some conjectured inequalities involving
Chern classes and some other deep results/conjectures in
arithmetic algebraic geometry.
This results are quite deep. Contact Barry Mazur [or Serre, or
Faltings, or Ribet, or ...]. Actually the set of people who DO
understand this stuff is fairly small.
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.
References
[1] J.P.Butler, R.E.Crandall,&R.W.Sompolski, Irregular Primes to One
Million. Math. Comp., 59 (October 1992) pp. 717-722.
Fermat's Last Theorem, A Genetic Introduction to Algebraic Number
Theory. H.M. Edwards. Springer Verlag, New York, 1977.
Thirteen Lectures on Fermat's Last Theorem. P. Ribenboim. Springer
Verlag, New York, 1979.
Number Theory Related to Fermat's Last Theorem. Neal Koblitz, editor.
Birkhduser Boston, Inc., 1982, ISBN 3-7643-3104-6
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/FLT/history
Last-modified: December 8, 1994
Version: 6.1
History of Fermat's Last Theorem
Pierre de Fermat (1601-1665) was a lawyer and amateur mathematician.
In about 1637, he annotated his copy (now lost) of Bachet's
translation of Diophantus' Arithmetika with the following statement:
Cubem autem in duos cubos, aut quadratoquadratum in duos
quadratoquadratos, et generaliter nullam in infinitum ultra
quadratum potestatem in duos ejusdem nominis fas est dividere:
cujus rei demonstrationem mirabilem sane detexi. Hanc marginis
exiguitas non caparet.
In English, and using modern terminology, the paragraph above reads
as:
There are no positive integers such that x^n + y^n = z^n for n > 2 .
I've found a remarkable proof of this fact, but there is not enough
space in the margin [of the book] to write it.
Fermat never published a proof of this statement. It became to be
known as Fermat's Last Theorem (FLT) not because it was his last piece
of work, but because it is the last remaining statement in the
post-humous list of Fermat's works that needed to be proven or
independently verified. All others have either been shown to be true
or disproven long ago.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/specialnumbers/fxtofxeqx
Last-modified: December 8, 1994
Version: 6.1
Name for f(x)^(f(x)) = x
Solving for f one finds a "continued fraction"-like answer
f(x)=log(x)/(log(log(x)/log(log(x)/(log...)))
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) or
Lambert's W function. MAPLE considers a solution in terms of W(x) as a
closed form (like the erf function). W is defined as W(x)e^(W(x)) = x
.
Notice that f(x) = exp(W(log(x))) is the solution to f(x)^f(x) = x
An extensive treatise on the known facts of Lambert's W function is
available for anonymous ftp at dragon.uwaterloo.ca at
/cs-archive/CS-93-03/W.ps.Z.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/hilbert
Last-modified: December 8, 1994
Version: 6.1
WHICH ARE THE 23 HILBERT PROBLEMS?
The original was published in German in a couple of places. A
translation was published by the AMS in 1902.
D. Hilbert. Mathematical problems. Lecture delivered before the
International Congress of Mathematicians at Paris in 1900. Bulletin
of the American Mathematical Society, 8:437-479, 1902.
The AMS Symposium mentioned at the end contains a series of papers on
the then-current state of most of the Problems.
Incidentally, the bibliography information above is taken from Yuri
Matiyasevich's book "Hilbert's Tenth Problem", published last year by
MIT Press. The full bibliography is available via anonymous ftp from
theory.lcs.mit.edu in the directory pub/hilbert10 (which also contains
some other information about the book).
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/montyhall
Last-modified: December 8, 1994
Version: 6.1
The Monty Hall problem
This problem has rapidly become part of the mathematical folklore.
The American Mathematical Monthly, in its issue of January 1992,
explains this problem carefully. The following are excerpted from that
article.
Problem:
A TV host shows you three numbered doors (all three equally likely),
one hiding a car and the other two hiding goats. You get to pick a
door, winning whatever is behind it. Regardless of the door you
choose, the host, who knows where the car is, then opens one of the
other two doors to reveal a goat, and invites you to switch your
choice if you so wish. Does switching increases your chances of
winning the car?
If the host always opens one of the two other doors, you should
switch. Notice that 1/3 of the time you choose the right door (i.e.
the one with the car) and switching is wrong, while 2/3 of the time
you choose the wrong door and switching gets you the car.
Thus the expected return of switching is 2/3 which improves over your
original expected gain of 1/3 .
Even if the hosts switches only part of the time, it pays to switch.
Only in the case where we assume a malicious host (i.e. a host who
entices you to switch based in the knowledge that you have the right
door) would it pay not to switch.
References
L. Gillman The Car and the Goats American Mathematical Monthly,
January 1992, pp. 3-7.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/mastermind
Last-modified: December 8, 1994
Version: 6.1
Master Mind
For the game of Master Mind it has been proven that no more than five
moves are required in the worst case.
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 number of moves - on average - but can take
six guesses on certain combinations.
In 1994, Kenji Koyama and Tony W. Lai found, by exhaustive search that
5625/1296~= 4.340 is the optimal strategy in the expected case. This
strategy may take six guesses in the worst case. A strategy that uses
at most five guesses in the worst case is also shown. This strategy
requires 5626/1296~= 4.341 guesses.
References
Donald E. Knuth. The Computer as Master Mind. J. Recreational
Mathematics, 9 (1976-77), 1-6.
Kenji Koyama, Tony W. Lai. An optimal Mastermind Strategy. J.
Recreational Mathematics, 1994.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/numbers
Last-modified: December 8, 1994
Version: 6.1
WHAT ARE NUMBERS?
_________________________________________________________________
* Introduction
* Construction of the Number System
* Construction of N
* Construction of Z
* Construction of Q
* Construction of R
* Construction of C
* Rounding things up
* What's next?
_________________________________________________________________
Introduction
Informally:
* \N = { 0,1,... } or \N = { 1,2,... }
Wether 0 is in \N depends on where you live and what is your field
of interest. At the informal level it is a religious topic.
* \Z = { ..., - 1,0,1,... }
* \Q = { p/q | p, q in \Z and q != 0 }
* \R = { d_0.d_1d_2... | d_0 in \Z and 0 <= d_i <= 9 for i > 0 }
* \C = { a + b o i | a, b in \R and i^2 = -1 }
_________________________________________________________________
Construction of the Number System
Formally (following the main stream in math) the numbers are
constructed from scratch out of the axioms of Zermelo Fraenkel set
theory (a.k.a. ZF set theory) [Enderton77, Henle86, Hrbacek84]. The
only things that can be derived from the axioms are sets with the
empty set at the bottom of the hierarchy. This will mean that any
number is a set (it is the only thing you can derive from the axioms).
It doesn't mean that you always have to use set notation when you use
numbers: just introduce the ``informal'' numbers as an abbreviation of
the formal counterparts.
The construction starts with \N and algebraically speaking, \N with
its operations and order is quite a weak structure. In the following
constructions the structures will be strengthen one step at the time:
\Z will be an integral domain, \Q will be a field, for the field \R
the order will be made complete, and field \C will be made
algebraically complete.
Before we start, first some notational stuff:
* a pair (a,b) = { { a } , { a,b } } ,
* an equivalence class [a] = { b | a == b } ,
* the successor of a is s(a) = a U { a } .
_________________________________________________________________
Construction of N
* { } in \N
* if a in \N then s(a) in \N
* \N is the smallest possible set such that the preceding rules
hold.
Informally n = { 0,...,n - 1 } (thus 0 = { } , 1 = { 0 } , 2 = { 0,1 }
, 3 = { 0,1,2 } ). We will refer to the elements of \N by giving them
a subscript _n . The relation <_n on \N is defined as: a_n <_n b_n iff
a_n in b_n . We can define +_n as follows:
* a_n +_n 0_n = a_n
* a_n +_n s(b_n) = s(a_n +_n b_n)
Define *_n as:
* a_n *_n 0_n = 0_n
* a_n *_n s(b_n) = (a_n *_n b_n) +_n a_n
_________________________________________________________________
Construction of Z
We define an equivalence relation on \N x \N : (a_n,b_n) ==_z(c_n,d_n)
iff a_n +_n d_n = c_n +_n b_n . Note that ==_z ``simulates'' a
subtraction in \N . \Z = { [(a_n,b_n)]_z | a_n, b_n in \N } . We will
refer to the elements of \Z by giving them a subscript _z . The
elements of \N can be embedded as follows: embed_n : \N --> \Z such
that embed_n(a_n) = [(a_n,0_n)]_z . Furthermore we can define:
* [(a_n,b_n)]_z <_z [(c_n,d_n)]_z iff a_n +_n d_n <_n c_n +_n b_n
* [(a_n,b_n)]_z +_z [(c_n,d_n)]_z = [(a_n +_n c_n, b_n +_n d_n)]_z
* [(a_n,b_n)]_z *_z [(c_n,d_n)]_z =
[((a_n *_n c_n) +_n (b_n *_n d_n), (a_n *_n d_n) +_n (c_n *_n
b_n))]_z
_________________________________________________________________
Construction of Q
We define an equivalence relation on \Z x (\Z \ { 0_z }) : (a_z,b_z)
==_q (c_z,d_z) iff a_z *_z d_z = c_z *_z b_z . Note that ==_q
``simulates'' a division in \Z . \Q = { [(a_z,b_z)]_q | a_z in \Z and
b_z in \Z \ { 0_z } } . We will refer to the elements of \Q by giving
them a subscript _q . The elements of \Z can be embedded as follows:
embed_z : \Z --> \Q such that embed_z(a_z) = [(a_z,1_z)]_q .
Furthermore we can define:
* [(a_z,b_z)]_q <_q [(c_z,d_z)]_q iff a_z *_z d_z <_z c_z *_z b_z
when 0_z <_z b_z and 0_z <_z d_z
* [(a_z,b_z)]_q +_q [(c_z,d_z)]_q = [((a_z *_z d_z) +_z (c_z *_z
b_z), b_z *_z d_z)]_q
* [(a_z,b_z)]_q *_q [(c_z,d_z)]_q = [(a_z *_z c_z, b_z *_z d_z)]_q
_________________________________________________________________
Construction of R
The construction of \R is different (and more awkward to understand)
because we must ensure that the cardinality of \R is greater than that
of \Q .
Set c is a Dedekind cut iff
* { } subset c subset \Q (strict inclusions!)
* c is closed downward:
if a_q in c and b_q <_q a_q then b_q in c
* c has no largest element:
there isn't an element a_q in c such that b_q <_q a_q for all b_q
!= a_q in c
You can think of a cut as taking a pair of scissors and cutting \Q in
two parts such that one part contains all the small numbers and the
other part contains all large numbers. If the part with the small
numbers was cut in such a way that it doesn't have a largest element,
it is called a Dedekind cut. \R = { c | c is a Dedekind cut } . We
will refer to the elements of \R by giving them a subscript _r . The
elements of \Q can be embedded as follows: embed_q : \Q --> \R such
that embed_q(a_q) = { b_q | b_q <_q a_q } . Furthermore we can define:
* a_r <_r b_r iff a_r subset b_r (strict inclusion!)
* a_r +_r b_r = { c_q +_q d_q | c_q in a_r and d_q in b_r }
* -_r a_r = ; { b_q | there exists an c_q in \Q such that b_q <_q
c_q and (-1)_q *_q c_q \not in a_r }
* |a_r|_r = a_r U -_r a_r
* *_r is defined as:
+ if not a_r <_r 0_r and not b_r <_r 0_r
then a_r *_r b_r = 0_r U { c_q *_q d_q | c_q in a_r and d_q
in b_r }
+ if a_r <_r 0_r and b_r <_r 0_r then a_r *_r b_r = |a_r|_r *_r
|b_r|_r
+ otherwise a_r *_r b_r = -_r (|a_r|_r *_r |b_r|_r)
There exists an alternative definition of \R using Cauchy sequences: a
Cauchy sequence is a s : \N --> \Q such that s(i_n) +_q((-1)_q *_q
s(j_n)) can be made arbitrary near to 0_q for all sufficiently large
i_n and j_n . We will define an equivalence relation ==_r on the set
of Cauchy sequences as: r ==_r s iff r(m_n) +_q((-1)_q *_q s(m_n)) can
be made arbitrary close to 0_q for all sufficiently large m_n . \R = {
[s]_r | s is a Cauchy sequence } . Note that this definition is close
to ``decimal'' expansions.
_________________________________________________________________
Construction of C
\C = \R x \R . We will refer to the elements of \C by giving them a
subscript _c . The elements of \R can be embedded as follows: embed_r
: \R --> \C such that embed_r(a_r) = (a_r,0_r) . Furthermore we can
define:
* (a_r,b_r) +_c (c_r,d_r) = (a_r +_r c_r, b_r +_r d_r )
* (a_r,b_r) *_c (c_r,d_r) = ((a_r *_r c_r) +_r -_r (b_r * d_r), (a_r
*_r d_r) +_r (b_r *_r c_r))
There exists an elegant alternative definition using ideals. To be a
bit sloppy: \C = \R[x]/< (x *_r x) +_r 1_r > , i.e. \C is the
resulting quotient ring of factoring ideal < (x *_r x) +_r 1_r > out
of the ring \R[x] of polynomials over \R . The sloppy part is that we
need to define concepts like quotient ring, ideal, and ring of
polynomials. Note that this definition is close to working with i^2 =
-1 : (x *_r x) +_r 1_r = 0_r can be rewritten as (x *_r x) = (-1)_r .
_________________________________________________________________
Construction of C
\C = \R x \R . We will refer to the elements of \C by giving them a
subscript _c . The elements of \R can be embedded as follows: embed_r
: \R --> \C such that embed_r(a_r) = (a_r,0_r) . Furthermore we can
define:
* (a_r,b_r) +_c (c_r,d_r) = (a_r +_r c_r, b_r +_r d_r )
* (a_r,b_r) *_c (c_r,d_r) = ((a_r *_r c_r) +_r -_r (b_r * d_r), (a_r
*_r d_r) +_r (b_r *_r c_r))
There exists an elegant alternative definition using ideals. To be a
bit sloppy: \C = \R[x]/< (x *_r x) +_r 1_r > , i.e. \C is the
resulting quotient ring of factoring ideal < (x *_r x) +_r 1_r > out
of the ring \R[x] of polynomials over \R . The sloppy part is that we
need to define concepts like quotient ring, ideal, and ring of
polynomials. Note that this definition is close to working with i^2 =
-1 : (x *_r x) +_r 1_r = 0_r can be rewritten as (x *_r x) = (-1)_r .
_________________________________________________________________
Rounding things up
At this moment we don't have that \N is a subset of \Z , \Z of \Q ,
etc. But we can get the inclusions if we look at the embedded copies
of \N , \Z , etc. Let
* \N' = ran embed_r o embed_q o embed_z o embed_n
* \Z' = ran embed_r o embed_q o embed_z
* \Q' = ran embed_r o embed_q
* \R' = ran embed_r
For these sets we have \N' subseteq \Z' subseteq \Q' subseteq \R'
subseteq \C . Furthermore these sets have all the properties that the
``informal'' numbers have.
_________________________________________________________________
This section of the FAQ is (C) Hans de Vreught. Send comments and or
corrections relating to this part to:
hdev@cp.tn.tudelft.nl (Hans de Vreught)
Archive-Name: sci-math-faq/nobel
Last-modified: December 8, 1994
Version: 6.1
_________________________________________________________________
Next: Who is N. Up: Mathematical Oddities Previous: Erdos Number
_________________________________________________________________
Why is there no Nobel in mathematics?
Nobel prizes were created by the will of Alfred Nobel, a notable
Swedish chemist.
One of the most common -and unfounded- reasons as to why Nobel decided
against a Nobel prize in math is that [a woman he proposed to/his
wife/his mistress] [rejected him because of/cheated him with] a famous
mathematician. Gosta Mittag-Leffler is often claimed to be the guilty
party.
There is no historical evidence to support the story.
For one, Mr. Nobel was never married.
There are more credible reasons as to why there is no Nobel prize in
math. Chiefly among them is simply the fact he didn't care much for
mathematics, and that it was not considered a practical science from
which humanity could benefit (a chief purpose for creating the Nobel
Foundation).
Here are some relevant facts:
* Nobel never married, hence no ``wife''. (He did have a mistress, a
Viennese woman named Sophie Hess.)
* Gosta Mittag-Leffler was an important mathematician in Sweden in
the late 19th-early 20th century. He was the founder of the
journal Acta Mathematica, played an important role in helping the
career of Sonya Kovalevskaya, and was eventually head of the
Stockholm Hogskola, the precursor to Stockholms Universitet.
However, it seems highly unlikely that he would have been a
leading candidate for an early Nobel Prize in mathematics, had
there been one - there were guys like Poincare and Hilbert around,
after all.
* There is no evidence that Mittag-Leffler had much contact with
Alfred Nobel (who resided in Paris during the latter part of his
life), still less that there was animosity between them for
whatever reason. To the contrary, towards the end of Nobel's life
Mittag-Leffler was engaged in ``diplomatic'' negotiations to try
to persuade Nobel to designate a substantial part of his fortune
to the Hogskola. It seems hardly likely that he would have
undertaken this if there was prior bad blood between them.
Although initially Nobel seems to have intended to do this,
eventually he came up with the Nobel Prize idea - much to the
disappointment of the Hogskola, not to mention Nobel's relatives
and Fraulein Hess.
* According to the very interesting study by Elisabeth Crawford,
``The Beginnings of the Nobel Institution'', Cambridge Univ.
Press, 1984, pages 52-53:
Although it is not known how those in responsible positions at the
Hogskola came to believe that a *large* bequest was forthcoming,
this indeed was the expectation, and the disappointment was keen
when it was announced early in 1897 that the Hogskola had been left
out of Nobel's final will in 1895. Recriminations followed, with
both Pettersson and Arrhenius [academic rivals of Mittag-Leffler in
the administration of the Hogskola] letting it be known that Nobel's
dislike for Mittag-Leffler had brought about what Pettersson termed
the `Nobel Flop'. This is only of interest because it may have
contributed to the myth that Nobel had planned to institute a prize
in mathematics but had refrained because of his antipathy to
Mittag-Leffler or -in another version of the same story- because of
their rivalry for the affections of a woman....
* A final speculation concerning the psychological element. Would
Nobel, sitting down to draw up his testament, presumably in a mood
of great benevolence to mankind, have allowed a mere personal
grudge to distort his idealistic plans for the monument he would
leave behind?
Nobel, an inventor and industrialist, did not create a prize in
mathematics simply because he was not particularly interested in
mathematics or theoretical science. His will speaks of prizes for
those ``inventions or discoveries'' of greatest practical benefit to
mankind. (Probably as a result of this language, the physics prize has
been awarded for experimental work much more often than for advances
in theory.)
However, the story of some rivalry over a woman is obviously much more
amusing, and that's why it will probably continue to be repeated.
References
Mathematical Intelligencer, vol. 7 (3), 1985, p. 74.
The Beginnings of the Nobel Institution. Elisabeth Crawford. Cambridge
Univ. Press, 1984.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/specialnumbers/lawPieq3
Last-modified: December 8, 1994
Version: 6.1
Indiana bill sets the value of pi to 3
The bill House Bill No. 246, Indiana State Legislature, 1897,
reportedly set the value of pi to an incorrect rational approximation.
The following is the text of the bill:
HOUSE BILL NO. 246
"A bill for an act introducing a new mathematical truth and offered
as a contribution to education to be used only by the State of
Indiana free of cost by paying any royalties whatever on the same,
provided it is accepted and adopted by the official action of the
legislature of 1897.
"Section 1. Be it enacted by the General Assembly of the State of
Indiana: It has been found that a circular area is to the square on
a line equal to the quadrant of the circumference, as the area of an
equilateral rectangle is to the square on one side. The diameter
employed as the linear unit according to the present rule in
computing the circle's area is entirely wrong, as it represents the
circles area one and one-fifths times the area of a square whose
perimeter is equal to the circumference of the circle. This is
because one-fifth of the diameter fils to be represented four times
in the circle's circumference. For example: if we multiply the
perimeter of a square by one-fourth of any line one-fifth greater
than one side, we can, in like manner make the square's area to
appear one fifth greater than the fact, as is done by taking the
diameter for the linear unit instead of the quadrant of the circle's
circumference.
"Section 2. It is impossible to compute the area of a circle on the
diameter as the linear unit without tresspassing upon the area
outside the circle to the extent of including one-fifth more area
than is contained within the circle's circumference, because the
square on the diameter produces the side of a square which equals
nine when the arc of ninety degrees equals eight. By taking the
quadrant of the circle's circumference for the linear unit, we
fulfill the requirements of both quadrature and rectification of the
circle's circumference. Furthermore, it has revealed the ratio of
the chord and arc of ninety degrees, which is as seven to eight, and
also the ratio of the diagonal and one side of a square which is as
ten to seven, disclosing the fourth important fact, that the ratio
of the diameter and circumference is as five-fourths to four; and
because of these facts and the further fact that the rule in
prresent use fails to work both ways mathematically, it should be
discarded as wholly wanting and misleading in its practical
applications.
"Section 3. In further proof of the value of the author's proposed
contribution to education, and offered as a gift to the State of
Indiana, is the fact of his solutions of the trisection of the
angle, duplication of the cube and quadrature having been already
accepted as contributions to science by the American Mathematical
Monthly, the leading exponent of mathematical thought in this
country. And be it remembered that these noted problems had been
long since given up by scientific bodies as unsolvable mysteries and
above man's ability to comprehend."
Will E. Edington in an article published in the Proceedings of the
Indiana Academy of Science describes the fate of the bill in the
committees of the Indiana legislature. First it was referred to the
House Committee on Canals, which was also referred to as the Committee
on Swamp Lands. Notices of the bill appeared in the Indianapolis
Journal and the Indianapolis Sentinel on Jan. 19,1897, both of which
described it a a bill telling how to square circles. On the same day,
"Representative M.B.Butler, of Steuben County, chairman of the
Committee on Canals, submitted the following report:
"Your Committee on Canals, to which was referred House Bill No.246,
entitled an act for the introduction of a mathematical truth, etc.,
has had the same under consideration and begs leave to report the
same back to the House with the recommendation that said bill be
referred to the Committee on Education."
The next day, the following article appeared in the Indianapolis
Sentinel:
"To SQUARE THE CIRCLE
"Claims Made That This Old Problem Has Been Solved. "The bill
telling how to square a circle, introduced in the House by
Mr.Record, is not intended to be a hoax. Mr. Record knows nothing of
the bill with the exception that he introduced it by request of
Dr.Edwin Goodwin of Posey County, who is the author of the
demonstration. The latter and State Superintendent of Public
Instruction Geeting believe that it is the long-sought solution of
the problem, and they are seeking to have it adopted by the
legislature. Dr. Goodwin, the author, is a mathematician of note. He
has it copyrighted and his proposition is that if the legislature
will indorse the solution, he will allow the state to use the
demonstration in its textbooks free of charge. The author is
lobbying for the bill."
On "February 2, 1897, ...Representative S.E. Nicholson, of Howard
County, chairman of the Committee on Education, reported to the
House.
"Your Committee on Education, to which was referred House Bill
No.246, entitled a a bill for an act entitled an act introducing a
new mathematical truth, has had same under consideration, and begs
leave to report the same back to the House with the recommendation
that said bill do pass.
"The report was cincurred in, and on February 8, 1897, it was
brought up for the second reading, following which it was considered
engrossed. Then 'Mr. Nicholson moved that the consitutional rule
requiring bills to be read on three days be suspended, that the bill
may be read a third time now.' The constitutional rule was suspended
by a vote of 72 to 0 and the bill was then read a third time. It was
passed by a vote of 67 to 0, and the Clerk of the House was directed
to inform the Senate of the passage of the bill."
The newspapers reported the suspension of the consitutional rules and
the unanimous passage of the bill matter-of-factly, except for one
line in the Indianapolis Journal to the effect that "this is the
strangest bill that has ever passed an Indiana Assembly."
The bill was referred to the Senate on Feb.10,1897 and was read for
the first time on Feb.11 and referred to the Committee on Temperance.
"On Feb.12 Senator Harry S. New, of Marion County, Chairman of the
Committee on Temperance, made the following report to the Senate:
"Your committee on Temperance, to which was referred House Bill
No.246, introduced by Mr.Record, has had the same under
consideration and begs leave to report the same back to the Senate
with the recommendation that said bill do pass."
The Senate Journal mentions only that the bill was read a second time
on Feb.12, 1897, that there was an unsuccessful attempt to amend the
bill by strike out the enacting clause, and finally it was postponed
indefinitely. That the bill was killed appears to be a matter of dumb
luck rather than the superior education or wisdom of the Senate. It is
true that the bill was widely ridiculed in Indiana and other states,
but what actually brought about the defeat of the bill is recorded by
Prof. C.A. Waldo in an article he wrote for the Proceedings of the
Indiana Academy of Science in 1916. The reason he knows is that he
happened to be at the State Capitol lobbying for the appropriation of
the Indiana Academy of Science, on the day the Housed passed House
Bill 246. When he walked in the found the debate on House Bill 246
already in progress. In his article, he writes (according to
Edington):
"An ex-teacher from the eastern part of the state was saying: 'The
case is perfectly simple. If we pass this bill which establishes a
new and correct value for pi , the author offers to our state
without cost the use of his discovery and its free publication in
our school text books, while everyone else must pay him a royalty.'
"
The roll was then called and the bill passed its third and final
reading in the lower house. A member then showed the writer [i.e.
Waldo] a copy of the bill just passed and asked him if he would like
an introduction to the learned doctor, its author. He declined the
courtesy with thanks remarking that he was acquainted with as many
crazy people as he cared to know.
"That evening the senators were properly coached and shortly
thereafter as it came to its final reading in the upper house they
threw out with much merriment the epoch making discovery of the Wise
Man from the Pocket."
The bill implies four different values for pi and one for sqrt(2) , as
follows: pi' = 16/sqrt(3) , 2 sqrt(5 pi/6) , 16 sqrt(2)/7 , 16/5
(~9.24 , ~3.236 , ~3.232 , 3.2 respectively.) sqrt(2)' = 10/7.
It has been found that a circular area is to the square on a line
equal to the quadrant of the circumference, as the area of an
equilateral rectangle is to the square on one side.
pi' = ((pi'/2))^2 = sqrt(3)/4 : 1 i.e. pi' = 16/sqrt(3) ~= 9.24 .
The diameter employed as the linear unit according to the present
rule in computing the circle's area is entirely wrong, as it
represents the circles area one and one-fifths times the area of a
square whose perimeter is equal to the circumference of the circle.
This is because one-fifth of the diameter fils to be represented
four times in the circle's circumference.
Bit tricky to decipher, but it seems to say ((2 pi'/4))^26/5 = pi i.e.
pi' = 2 sqrt(5 pi/6) ~= 3.236
Furthermore, it has revealed the ratio of the chord and arc of
ninety degrees, which is as seven to eight,
sqrt(2) : pi/2 = 7 : 8 i.e. pi = 16 sqrt(2)/7 ~= 3.232
and also the ratio of the diagonal and one side of a square which is
as ten to seven
i.e. sqrt(2) = 10/7 ~= 1.429
that the ratio of the diameter and circumference is as five-fourths
to four
i.e. pi = 16/5 = 3.2
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/pointers
Last-modified: December 8, 1994
Version: 6.1
The Sci.Math FAQ may be consulted at, which as usual contains the
latest additions and modifications, which eventually make it to
the posted version.
URL: http://daisy.uwaterloo.ca/~alopez-o/math-faq/math-faq.html
This brings up an introductory page. Users with character only
browsers should select "the Text Table of Contents"
or go directly to it via
URL: http//daisy.uwaterloo.ca/~alopez-o/math-faq/mathtext/math-faq.html
All other users (i.e. Mosaic/Air Mosaic/Netscape/etc) just follow the links
at the top of the page.
Text equations for the lynx version were generated automagically from
the LaTeX source by LaTexT coming soon to a latex2html distibution
near you.
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/proyectiveplane
Last-modified: December 8, 1994
Version: 6.1
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?
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. American Mathematical Monthly, 98 (1991), 305-318.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/primes
Last-modified: December 8, 1994
Version: 6.1
PRIME NUMBERS
Contents:
_________________________________________________________________
* Largest known Mersenne prime
* Largest known prime
* Largest known twin primes
* Largest Fermat number with known factorization
* Algorithms to factor integer numbers
* List of record numbers
* What is the current status on Mersenne primes?
* Formulae to compute prime numbers
_________________________________________________________________
Largest known Mersenne prime
2^(859433) - 1 is prime. It was discovered in 1994.
_________________________________________________________________
Largest known prime
The largest known prime is the Mersenne prime described above. The
largest known non-Mersenne prime, is 391581*2^(216193) - 1 .
Throughout history, the largest known prime has almost always been a
Mersenne prime; the period between Brown et al's discovery in Aug 1989
and Slowinski & Gage's in March 1992 is one of the few exceptions.
References
Brown, Noll, Parady, Smith, Smith, and Zarantonello. Letter to the
editor. American Mathematical Monthly, vol. 97, 1990, p. 214.
_________________________________________________________________
Largest known twin primes
The largest known twin primes are 1691232 * 1001 * 10^(4020) +- 1 ,
which is a number with 4030 digits, found by H. Dubner. The second
largest known twin primes are 4650828 * 1001 * 10^(3429) +- 1 . They
were found by H. Dubner as well.
References
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
F_(11) = (2^(2^(11))) + 1 which was factored by Brent &Morain in 1988.
F_9 = (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
F_(10) is not known.
_________________________________________________________________
Algorithms to factor integer numbers
There are several know algorithms that have subexponential estimated
running time, to mention just a few:
* Continued fraction algorithm.
* Class group method.
* Quadratic sieve algorithm.
* Class Group method.
* Elliptic curve algorithm.
* Number fi eld sieve.
* Dixon's random squares algorithm.
* Valle's two-thirds algorithm.
* Seysen's class group algorithm.
References
A.K. Lenstra, H.W. Lenstra Jr. Algorithms in Number Theory. J. van
Leeuwen (ed.), Handbook of Theoretical Computer Science, Volume A:
Algorithms and Complexity Elsevier, pp. 673-715, 1990.
_________________________________________________________________
List of record numbers
Chris Caldwell maintains a list called The Largest Known Primes.
Finger primes@math.utm.edu for a few record primes and ways to get the
lists by e-mail, the list is also available by e-mail. He can e-mail
you the lists, but prefers you use the other methods. different forms
of this list. Currently the list is available by anonymous ftp to
math.utm.edu (directory /pub/math/primes) and by gopher to
unix1.utm.edu (directory 1/user/Public_FTP/pub/math/primes). If
nothing else works, you may send e-mail to Chris for a copy of the
lists. (caldwell@utm.edu or caldwell@utmartn.bitnet).
_________________________________________________________________
What is the 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.
Number p Year Discoverer
1-4 2,3,5,7 pre-1500
5 13 1461 Anonymous
6-7 17,19 1588 Cataldi
8 31 1750 Euler
9 61 1883 I.M. Pervushin
10 89 1911 Powers
11 107 1914 Powers
12 127 1876 Lucas
13-14 521,607 1952 Robinson
15-17 1279,2203,2281 1952 R. M. Robinson
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
30 132049 1983 Slowinski
31 216091 1985 Slowinski
32? 756839 1992 Slowinski & Gage
33? 859433 1994 Slowinski
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,
360K-386K, and 430K - 520K.
References
An introduction to the theory of numbers. G.H. Hardy, E.M. Wright.
Fifth edition, 1979, Oxford.
_________________________________________________________________
Formulae to compute prime numbers
Is there a polynomial which gives all the prime numbers?
No, there is not. This is a simple exercise to prove.
Is there a non-constant polynomial that only takes on prime values?
It has been proved that no such polynomial exists. The proof is simple
enough that an high school student could probably discover it. See,
for example, Ribenboim's book The Book of Prime Number Records.
Note, however, by the work of Jones, Sato, Wada, and Wiens, there is a
polynomial in 26 variables such that the set of primes coincides with
the set of positive values taken by this polynomial. See Ribenboim,
pp. 147-150.
But most people would object to the term ``formula" restricted to mean
polynomial. Can we not use summation signs, factorial, and the floor
function in our ``formula"? If so, then indeed, there are formulas for
the prime numbers. Some of them are listed below.
If we can't, then exactly what operations do you allow and why?
Indeed, as I have previously argued, a reasonable interpretation of
the word ``formula" is simply ``Turing machine that halts on all
inputs". Under this interpretation, there certainly are halting Turing
machines which compute the n -th prime number. However, nobody knows
how to compute the n -th prime in time polynomial in log n . That's
still an open question.
Herb Wilf has addressed the question, ``What is a formula?" in his
delightful article, ``What is an answer?" which appeared in the
American Mathematical Monthly, 89 (1982), 289-292. He draws the
distinction between ``formula" and ``good formula". Anyone who claims
``there is no formula for the prime numbers" should read this article.
Here are just a few articles that discuss ``formulas" for primes.
Almost all of these do not require computation of the primes "ahead of
time". Most of them rely on standard mathematical functions such as
summation, factorial, greatest integer function, etc.
References
C. Isenkrahe. Math. Annalen 53 (1900), 42-44.
W. H. Mills. Bulletin of the American Mathematical Society 53 (1947),
604.
L. Moser. Mathematics Magazine 23 (1950), 163-164.
E. M. Wright. American Mathematical Monthly 58 (1951), 616-618.
(Correction, 59 (1952), 99.)
E. M. Wright. Journal of the London Mathematical Society 29 (1954),
63-71.
B. R. Srinivasan. Journal of the Indian Mathematical Society 25
(1961), 33-39.
C. P. Willans. Mathematics Gazette 48 (1964), 413-415.
V. C. Harris. Nordisk Mat. Tidskr. 17 (1969), 82.
U. Dudley. American Mathematical Monthly 76 (1969), 23-28.
C. Vanden Eynden. American Mathematical Monthly 79 (1972), 625.
S. W. Golomb. American Mathematical Monthly 81 (1974), 752-754.
Algorithmic Number Theory J.O. Shallit, E. Bach. (to be published, MIT
Press).
_________________________________________________________________
Archive-Name: sci-math-faq/Quaternions
Last-modified: December 8, 1994
Version: 6.1
THEORY OF QUATERNIONIC ANALYTIC FUNCTIONS
Four-dimensional analog to the theory of complex analytic functions.
It 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.
References
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.
Clifford analysis. F. Brackx and R. Delanghe and F. Sommen. Pitman,
1983.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/surfaceSphere
Last-modified: December 8, 1994
Version: 6.1
Formula for the Surface Area of a sphere in Euclidean N -Space
This is equivalent to the volume of the N -1 solid which comprises the
boundary of an r^N (pi^(N/2))/((N/2)!) -Sphere.
The volume of a ball is the easiest formula to remember: It's x! =
Gamma (x + 1) . The only hard part is taking the factorial of a
half-integer. The real definition is that (1/2 + n)! = sqrt(pi) ((2n +
2)!)/((n + 1)!4^(n + 1)) , but if you want a formula, it's:
N (pi^(N/2))/((N/2)!r^(N - 1)) To get the surface area, you just
differentiate to get e^(-x^2) .
There is a clever way to obtain this formula using Gaussian integrals.
First, we note that the integral over the line of sqrt(pi) is N .
Therefore the integral over e^(-x_1^2 - x_2^2 - ... - x_N^2) -space of
sqrt(pi)^n is Vr^(N - 1)e^(-r^2) . Now we change to spherical
coordinates. We get the integral from 0 to infinity of V , where n 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''.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/symboliccomp
Last-modified: December 8, 1994
Version: 6.1
SYMBOLIC COMPUTATION PACKAGES
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
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/unsolvedproblems
Last-modified: December 8, 1994
Version: 6.1
UNSOLVED PROBLEMS
_________________________________________________________________
* Does there exist a number that is perfect and odd?
* Collatz Problem
_________________________________________________________________
Does there exist a number that is perfect and odd?
A given number is perfect if it is equal to the sum of all its proper
divisors. This question was first posed by Euclid in ancient Greece.
This question is still open. Euler proved that if N is an odd perfect
number, then in the prime power decomposition of N , exactly one
exponent is congruent to 1 mod 4 and all the other exponents are even.
Furthermore, the prime occurring to an odd power must itself be
congruent to 1 mod 4. A sketch of the proof appears in Exercise 87,
page 203 of Underwood Dudley's Elementary Number Theory, 2nd ed. It
has been shown that there are no odd perfect numbers < 10^(300) .
_________________________________________________________________
Collatz Problem
Take any natural number m > 0 .
n : = m;
repeat
if ( n is odd) then n : = 3*n + 1 ; else n : = n/2 ;
until ( n = = 1 )
Conjecture. For all positive integers m, the program above terminates.
The conjecture has been verified up to 7 * 10^(11) .
References
Unsolved Problems in Number Theory. Richard K Guy. Springer, Problem
E16.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994
Archive-Name: sci-math-faq/FLT/Wiles
Last-modified: December 8, 1994
Version: 6.1
Wiles' line of attack
Here is an outline of the first proposed proof.
From Ken Ribet:
Here is a brief summary of what Wiles said in his three
lectures.
The method of Wiles borrows results and techniques from lots
and lots of people. To mention a few: Mazur, Hida, Flach,
Kolyvagin, yours truly, Wiles himself (older papers by Wiles),
Rubin... The way he does it is roughly as follows. Start with a
mod p representation of the Galois group of Q which is known to
be modular. You want to prove that all its lifts with a certain
property are modular. This means that the canonical map from
Mazur's universal deformation ring to its maximal Hecke algebra
quotient is an isomorphism. To prove a map like this is an
isomorphism, you can give some sufficient conditions based on
commutative algebra. Most notably, you have to bound the order
of a cohomology group which looks like a Selmer group for Sym^2
of the representation attached to a modular form. The
techniques for doing this come from Flach; you also have to use
Euler systems a la Kolyvagin, except in some new geometric
guise.
If you take an elliptic curve over Q , you can look at the
representation of Gal on the 3-division points of the curve. If
you're lucky, this will be known to be modular, because of
results of Jerry Tunnell (on base change). Thus, if you're
lucky, the problem I described above can be solved (there are
most definitely some hypotheses to check), and then the curve
is modular. Basically, being lucky means that the image of the
representation of Galois on 3-division points is GL(2,Z/3Z) .
Suppose that you are unlucky, i.e., that your curve E has a
rational subgroup of order 3. Basically by inspection, you can
prove that if it has a rational subgroup of order 5 as well,
then it can't be semistable. (You look at the four non-cuspidal
rational points of X_0(15) .) So you can assume that E[5] is
``nice''. Then the idea is to find an E' with the same
5-division structure, for which E'[3] is modular. (Then E' is
modular, so E'[5] = E[5] is modular.) You consider the modular
curve X which parameterizes elliptic curves whose 5-division
points look like E[5] . This is a twist of X(5) . It's
therefore of genus 0, and it has a rational point (namely, E ),
so it's a projective line. Over that you look at the
irreducible covering which corresponds to some desired
3-division structure. You use Hilbert irreducibility and the
Cebotarev density theorem (in some way that hasn't yet sunk in)
to produce a non-cuspidal rational point of X over which the
covering remains irreducible. You take E' to be the curve
corresponding to this chosen rational point of X .
From the previous version of the FAQ:
(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 conjecture.
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].
From Karl Rubin:
Theorem 2 If E is a semistable elliptic curve defined over Q,
then E is modular.
It has been known for some time, by work of Frey and Ribet,
that Fermat follows from this. If u^q + v^q + w^q = 0 , then
Frey had the idea of looking at the (semistable) elliptic curve
y^2 = x(x - a^q)(x + b^q) . If this elliptic curve comes from a
modular form, then the work of Ribet on Serre's conjecture
shows that there would have to exist a modular form of weight 2
on Gamma_0(2) . But there are no such forms.
To prove the Theorem, start with an elliptic curve E , a prime
p and let rho_p : Gal(\bar(Q)/Q) --> GL_2(Z/pZ) be the
representation giving the action of Galois on the p -torsion
E[p] . We wish to show that a certain lift of this
representation to GL_2(Z_p) (namely, the p -adic representation
on the Tate module T_p(E) ) is attached to a modular form. We
will do this by using Mazur's theory of deformations, to show
that every lifting which ``looks modular'' in a certain precise
sense is attached to a modular form.
Fix certain ``lifting data'', such as the allowed ramification,
specified local behavior at p , etc. for the lift. This defines
a lifting problem, and Mazur proves that there is a universal
lift, i.e. a local ring R and a representation into GL_2(R)
such that every lift of the appropriate type factors through
this one.
Now suppose that rho_p is modular, i.e. there is some lift of
rho_p which is attached to a modular form. Then there is also a
hecke ring T , which is the maximal quotient of R with the
property that all modular lifts factor through T . It is a
conjecture of Mazur that R = T , and it would follow from this
that every lift of rho_p which ``looks modular'' (in particular
the one we are interested in) is attached to a modular form.
Thus we need to know 2 things:
(a)
rho_p is modular
(b)
R = T .
It was proved by Tunnell that rho_3 is modular for every
elliptic curve. This is because PGL_2(Z/3Z) = S_4 . So (a) will
be satisfied if we take p = 3 . This is crucial.
Wiles uses (a) to prove (b) under some restrictions on rho_p .
Using (a) and some commutative algebra (using the fact that T
is Gorenstein, basically due to Mazur) Wiles reduces the
statement T = R to checking an inequality between the sizes of
2 groups. One of these is related to the Selmer group of the
symmetric square of the given modular lifting of rho_p , and
the other is related (by work of Hida) to an L -value. The
required inequality, which everyone presumes is an instance of
the Bloch-Kato conjecture, is what Wiles needs to verify.
He does this using a Kolyvagin-type Euler system argument. This
is the most technically difficult part of the proof, and is
responsible for most of the length of the manuscript. He uses
modular units to construct what he calls a geometric Euler
system of cohomology classes. The inspiration for his
construction comes from work of Flach, who came up with what is
essentially the bottom level of this Euler system. But Wiles
needed to go much farther than Flach did. In the end, under
certain hypotheses on rho_p he gets a workable Euler system
and proves the desired inequality. Among other things, it is
necessary that rho_p is irreducible.
Suppose now that E is semistable.
Case 1. rho_3 is irreducible.
Take p = 3. By Tunnell's theorem (a) above is true. Under these
hypotheses the argument above works for rho_3 , so we conclude
that E is modular.
Case 2. rho_3 is reducible. Take p = 5 . In this case rho_5
must be irreducible, or else E would correspond to a rational
point on X_0(15) . But X_0(15) has only 4 noncuspidal rational
points, and these correspond to non-semistable curves. If we
knew that rho_5 were modular, then the computation above would
apply and E would be modular.
We will find a new semistable elliptic curve E' such that
rho_(E,5) = rho_(E',5) and rho_(E',3) is irreducible. Then by
Case I, E' is modular. Therefore rho_(E,5) = rho_(E',5) does
have a modular lifting and we will be done.
We need to construct such an E' . Let X denote the modular
curve whose points correspond to pairs (A, C) where A is an
elliptic curve and C is a subgroup of A isomorphic to the group
scheme E[5] . (All such curves will have mod-5 representation
equal to rho_E .) This X is genus 0, and has one rational point
corresponding to E , so it has infinitely many. Now Wiles uses
a Hilbert Irreducibility argument to show that not all rational
points can be images of rational points on modular curves
covering X , corresponding to degenerate level 3 structure
(i.e. im(rho_3) != GL_2(Z/3) ). In other words, an E' of the
type we need exists. (To make sure E' is semistable, choose it
5-adically close to E . Then it is semistable at 5, and at
other primes because rho_(E',5) = rho_(E,5) .)
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Sun Nov 20 20:45:48 EST 1994