Here is a little more detail;
the chapters are now fleshed out with 3-4 papers apiece.
Some of these paper-topics are relevant to some of the issues
we have been arguing about.  If mitch already owns this
book then he can be counted on to quote parts of them out
of context in mistaken attacks on other people's positions.
I can only hope to be able to check this book out when
the library re-opens after Xmas break; then I should be able
to quote right back, in rebuttal.  But even feeling that one
needs to quote authorities is ridiculous.  My main reason
for making this threat is the hope that mitch can be deterred
in advance from excerpting these papers in dishonest and misleading
ways.
Introduction: Logic, Philosophy, and Philosophical Logic: Dale Jacquette 
(Pennsylvania State University).
1. Ancient Greek Philosophical Logic: Robin Smith (Texas A&M University).
3. The Rise of Modern Logic: Rolf George (University of Waterloo) and James 
Van Evra (University of Waterloo).
Part II: Symbolic Logic and Ordinary Language:
4. Language, Logic, and Form: Kent Bach (San Francisco State University).
5. Puzzles About Intensionality: Nathan Salmon (University of California, 
Santa Barbara).
6. Symbolic Logic and Natural Language: Emma Borg (University of Reading) 
and Ernest Lepore (Rutgers University).
Part III: Philosophical Dimensions of Logical Paradoxes:
7. Logical Paradoxes: James Cargile (University of Virginia).
9. Philosophical Implications of Logical Paradoxes: Roy A. Sorensen 
(Dartmouth College).
Part IV: Truth and Definite Description in Semantic Analysis:
10. Truth, the Liar, and Tarski's Semantics: Gila Sher (University of 
California, San Diego).
11. Truth, the Liar, and Tarskian Truth Definition: Greg Ray (University of 
Florida).
12. Descriptions and Logical Form: Gary Ostertag (New York University).
13. Russell's Theory of Definite Descriptions as a Paradigm for Philosophy: 
Gregory Landini (University of Iowa).
Part V: Concepts of Logical Consequence:
14. Necessity, Meaning, and Rationality: The Notion of Logical Consequence: 
Stewart Shapiro (Ohio State University).
15. Varieties of Consequence : B.G. Sundholm (Leiden University).
16. Modality of Deductively Valid Inference : Dale Jacquette (Pennsylvania 
State University).
Part VI Logic, Existence, and Ontology:
17. Quantifiers, Being and Canonical Notation: Paul Gochet (University of 
Li̬ge).
19. Putting Language First: The Liberation of Logic from Ontology: 
Ermanno Bencivenga (University of California, Irvine).
Part VII: Metatheory and the Scope and Limits of Logic:
20. Metatheory: Alasdair Urquhart (University of Toronto).
21. Metatheory of Logics and the Characterization Problem: Jan Wolenski 
(Jagiellonian University).
22. Logic in Finite Structures: Definability, Complexity, and Randomness: 
Scott Weinstein (University of Pennsylvania).
Part VIII: Logical Foundations of Set Theory and Mathematics:
23. Logic and Ontology: Numbers and Sets: Jos̩ 
Benardete (Syracuse University).
24. Logical Foundations of Set Theory and Mathematics: Mary Tiles 
(University of Hawaii) .
25. Property-Theoretic Foundations of Mathematics: Michael Jubien 
(University of California, Davis).
Part IX: Modal Logics and Semantics:
26. Modal Logic: Johan van Benthem (University of Amsterdam).
27. First Order Alethic Modal Logic: Melvin Fitting (City University of New 
York).
28. Proofs and Expressiveness in Alethic Modal Logic: Maarten de Rijke 
(University of Amsterdam) and Heinrich Wansing (Dresden University of 
Technology).
29. Alethic Modal Logics and Semantics: Gerhard Schurz (University of 
Erfurt).
30. Epistemic Logic: Nicholas Rescher (University of Pittsburgh).
Part X: Intuitionistic, Free, and Many-Valued Logics:
32. Intuitionism: Dirk van Dalen (University of Utrecht) and Mark van Atten 
(University of Utrecht).
33. Many-Valued, Free, and Intuitionistic Logics: Richard Grandy (Rice 
University).
34. Many-Valued Logic: Grzegorz Malinowski (University of Lodz).
Part XI: Inductive, Fuzzy, and Quantum Probability Logics:
35. Inductive Logic : Stephen Glaister (University of Washington).
36. Heterodox Probability Theory: Peter Forrest (University of New 
England).
37. Why Fuzzy Logic?: Petr HÌÁjek (Academy of 
Sciences of the Czech Republic).
Part XII: Relevance and Paraconsistent Logics:
38. Relevance Logic: Edwin Mares (Victoria University of Wellington).
39. Paraconsistency: Bryson Brown (University of Lethbridge).
40. Logicians Setting Together Contradictories: A Perspective on Relevance, 
Paraconsistency, and Dialetheism: Graham Priest (University of Melbourne).
Part XIII: Logic, Machine Theory, and Cognitive Science:
41. The Logical and the Physical: Andrew W. Hodges (Wadham College, Oxford 
University).
42. Modern Logic and its Role in the Study of Knowledge: Peter A. Flach 
(University of Bristol).
43. Actions and Normative Positions: A Modal-Logical Approach : Robert 
Demolombe (Toulouse Center) and Andrew J.I. Jones (University of Oslo).
Part XIV: Mechanization of Logical Inference and Proof Discovery:
44. The Automation of Sound Reasoning and Successful Proof Finding: Larry 
Wos (Argonne National Laboratory) and Branden Fitelson (Yale University).
45. A Computational Logic for Applicative Common LISP: J. Strother Moore 
(University of Texas) and Matt Kaufmann (Advanced Micro Devices, Inc).
46. Sampling Labelled Deductive Systems: D.M. Gabbay (King's College).
Resources for Further Study.
-- 
 ---
  It's difficult ... you need to be united to have any 
   strength, but internal issues have to  be addressed.
 --- E. Ray Lewis, on  liberalism in America
====
 Mitch ought to really love chapters 5-19 since each of them can
> say WHY the treatment of its topic in ways other than the
> standard/default/classical treatment afforded by classical FOL
> might be IMPORTANT.  This is of course something that mitch himself
> has never been able to say in his own words.
Right George.  I showed up with years of thinking about set theory as a 
foundation for mathematics and axioms that formally
expressed my thoughts on the nature of identity in mathematics.
I did not show up as an expert on logics.
I am not an advocate of intuitionism.  But, I spent a lot of time thinking 
about the relationship of apartness as the appropriate
mathematical criterion for distinctness (hence, identity).
I could probably answer questions like Why Kant? or Why geometry?  
But it really wouldn't matter, now would it?
You should look at the end of the chapter on negation,
Therefore, intuitionistic negation and any representation
of predicate term negation as a unary connective are
bound to be imperfect in Avron's sense.  Indeed, the
only negation among the unary connectives that emerges
as perfect from both the syntactic and semantic point of
view is the Boolean negation of classical logic.
Now for a stupidfuck math person like me, Avron's terms describing 
implication as internal and external are completely
meaningless.  But, I sure do know why I expect a criterion that identifies 
the strength of Boolean negation.  Mostly, it is
because I understand it with respect to geometric reflection.
For what it is worth, thanks.
====
This is a variation of the same proof I give in
Cardinality of Computable Numbers.
I define two Turing machines.
The first machine (TM1) has these instructions:
1) Write a 1
2) Move right one position
Repeat
Assume we give this machine a tape that has
an infinite string of 0's. It would seem that
TM1 will output an infinite string of 1's.
Instructions for TM2:
1) Scan right until a 0 is found
2) Scan right until a second 0 is found
3) Backup and write a 1 on the previous 0
Repeat
Assume we give TM2 a tape that contains
an infinite string of 0's.
Even if we assume that TM2 performs an
infinite number of operations, the tape
produced by TM2 will contain an initial
segment with a finite number of 1's
followed by a 0.
TM2 is incapable of writing an infinite
number of 1's. If TM2 can not write
an infinite number of 1's, we can not
assume that TM1 does.
No TM can write an infinite string of symbols.
Russell
- 2 many 2 count
====
> This is a variation of the same proof I give in
> Cardinality of Computable Numbers.
> 
> I define two Turing machines.
> 
> The first machine (TM1) has these instructions:
> 
> 1) Write a 1
> 2) Move right one position
> Repeat
> 
Ok.  You could be more specific though, saying what the tape alphabet
is
and what the states are and exactly what the set of quadruples (or 
quintuples) are.  But I can guess, I suppose.
> Assume we give this machine a tape that has
> an infinite string of 0's. It would seem that
> TM1 will output an infinite string of 1's.
> 
No, TM1 doesn't halt, and therefore it doesn't ouput anything. 
However,
it will continue printing out 1's forever with no finite bound on the
number of 1's printed.
> Instructions for TM2:
> 
> 1) Scan right until a 0 is found
> 2) Scan right until a second 0 is found
> 3) Backup and write a 1 on the previous 0
> Repeat
> 
Again, I can guess what the tape alphabet, the states, and the
specific
quadruples are.
> Assume we give TM2 a tape that contains
> an infinite string of 0's.
> 
> Even if we assume that TM2 performs an
> infinite number of operations, the tape
> produced by TM2 will contain an initial
> segment with a finite number of 1's
> followed by a 0.
No, again it is obvious that TM2 will not halt and that it will
continue
printing 1's forever.  I don't see why you would claim otherwise.  If
you
make such a shocking claim, perhaps you should provide us with some
sort of proof or argument as to why you believe it to be true, so that
we
can point out where you went wrong.
> 
> TM2 is incapable of writing an infinite
> number of 1's.
Of course; it never halts.
> If TM2 can not write
> an infinite number of 1's, we can not
> assume that TM1 does.
> 
Well, seems like a non-sequitur, but I do grant you that TM1 cannot
write an infinite number of 1's either.  TM1 never halts and never
accomplishes writing an infinite number of 1's.
> No TM can write an infinite string of symbols.
I am surprised that you can come to this correct conclusion despite
the
profoundly fundamental flaws in your reasoning.
Let me suggest a remedy:
You believe that there is an *actual* infinity, a place that you can
get
to by repeatedly adding 1's enough times.   You believe an infinite
task
can be completed.  This is what your intuition tells you.  Intuition
about
infinity can be very misleading.  That is why it is important to make
your thoughts mathematically rigorous.  Until you do, it is pointless
arguing with mathematicians about the mathematics of infinity.  But if
you
do try to make your point of view rigorous, you will see that your
current
beliefs are inconsistent.
In mathematics, there is no *actual* infinity (in a sense).  
Infinite processes are never
completed.  Infinity is only *potential*.  Some would say, wait a
minute,
Cantor gave us the actually infinite.  The infinite ordinals are
perfectly rigorous objects.  But this doesn't tell the whole story. 
Modern set
theory *postulates* the existence of omega and then proceeds to
manipulate
it like every other mathematical object.  Omega is never
*constructed*.
This is fine.  ZF is quite a rigorous and fascinating theory.  Also
note
that to speak about omega and the other transfinite numbers, once has
to lift
oneself beyond the natural numbers.  Within the ordinary arithmetic of
the natural numbers (I am speaking about the standard model of the
Peano
postulates), there is no omega, no infinity.  It's not there. 
Therefore,
you will always be wrong when you argue in the way you've done above.
It's funny that since I started reading these newsgroups a year or two
ago,
I've encountered at least half a dozen people arguing exactly the same
point of view as you, but in different contexts.  Now we're in the
context
of Turing Machines.  That's very humorous to me.  I didn't know people
with your beliefs existed, and now I find that there are at least
several
of you.  Have you heard of this guy Phil who used to post absurd
things like
the statement that all natural numbers have finitely many digits?
It's quite fascinating.  I could write a book about it.  
> 
> 
> Russell
> - 2 many 2 count
====
[re: Russell Easterly]
> It's funny that since I started reading these newsgroups a year or two
> ago, I've encountered at least half a dozen people arguing exactly the
> same point of view as you, but in different contexts.  Now we're in the
> context of Turing Machines.  That's very humorous to me.  I didn't know
> people with your beliefs existed, and now I find that there are at least
> several of you.
  I can easily understand how some people don't get infinity.  I
myself still don't quite get ordinal numbers, although I've got
the cardinals down pretty well now. :)
  What I don't understand is how some people who don't get infinity
seem to compulsively post *wrong* statements to the Internet, rather
than trying to understand *right* ones.  And how a guy like Russell
can seem to have such a reasonable grasp of what a Turing machine
is, without having even a basic conception of the properties of the
integer numbers!
>  Have you heard of this guy Phil who used to post absurd
> things like the statement that all natural numbers have finitely
> many digits?  It's quite fascinating.  I could write a book about it.
  I remember Phil.  But I must point out that you forgot to complete
that thought: All natural numbers *do* have finitely many digits!
But Phil made a leap from that true statement to the false statement
that *the number of* natural numbers was finite -- and stuck to it --
and that's what was absurd.
-Arthur
====
>   I can easily understand how some people don't get infinity.  I
> myself still don't quite get ordinal numbers, although I've got
> the cardinals down pretty well now. :)
Eek.  I find the ordinals far more comprehensible than cardinals;
maybe I've done too much set theory.
Thomas
====
> This is a variation of the same proof I give in
> Cardinality of Computable Numbers.
> 
> I define two Turing machines.
> 
> The first machine (TM1) has these instructions:
> 
> 1) Write a 1
> 2) Move right one position
> Repeat
> 
> Assume we give this machine a tape that has
> an infinite string of 0's. It would seem that
> TM1 will output an infinite string of 1's.
No, you wouldn't have an output at all.  This machine would never
halt.
> 
> Instructions for TM2:
> 
> 1) Scan right until a 0 is found
> 2) Scan right until a second 0 is found
> 3) Backup and write a 1 on the previous 0
> Repeat
> 
> Assume we give TM2 a tape that contains
> an infinite string of 0's.
> 
> Even if we assume that TM2 performs an
> infinite number of operations, the tape
> produced by TM2 will contain an initial
> segment with a finite number of 1's
> followed by a 0.
> 
> TM2 is incapable of writing an infinite
> number of 1's. If TM2 can not write
> an infinite number of 1's, we can not
> assume that TM1 does.
> 
> No TM can write an infinite string of symbols.
I'm not sure why you think this is particularly important.  By
construction, at any given time-step, a Turing machine has modified
only finitely many cells, but there is no upper bound on the number of
cells a Turing machine can modify.  (There is a glaringly obvious
analogy with the natural numbers here).
'cid 'ooh
====
>This is a variation of the same proof I give in
>Cardinality of Computable Numbers.
I define two Turing machines.
The first machine (TM1) has these instructions:
1) Write a 1
>2) Move right one position
>Repeat
Assume we give this machine a tape that has
>an infinite string of 0's. It would seem that
>TM1 will output an infinite string of 1's.
 No, you wouldn't have an output at all.  This machine would never
> halt.
I don't know why everyone is so worried about a TM halting.
The word halt does not appear in Turing's paper.
This is the definition of a computable number given by Turing:
<quote>
Computing machines.
If an a-machine prints two kinds of symbols, of which the first kind 
(called
figures) consists entirely of 0 and 1 (the others being called symbols of
the second kind), then the machine will be called a computing machine. If
the machine is supplied with a blank tape and set in motion, starting from
the correct initial m-configuration, the subsequence of the symbols printed
by it which are of the first kind will be called the sequence computed by
the machine. The real number whose expression as a binary decimal is
obtained by prefacing this sequence by a decimal point is called the number
computed by the machine.
At any stage of the motion of the machine, the number of the scanned 
square,
the complete sequence of all symbols on the tape, and the m-configuration
will be said to describe the complete configuration at that stage. The
changes of the machine and tape between successive complete configurations
will be called the moves of the machine.
{233}
Circular and circle-free machines.
symbols of the first kind it will be called circular. Otherwise it is said
to be circle-free.
A machine will be circular if it reaches a configuration from which there 
is
no possible move, or if it goes on moving, and possibly printing symbols of
the second kind, but cannot print any more symbols of the first kind. The
significance of the term circular will be explained in ¤8.
Computable sequences and numbers.
A sequence is said to be computable if it can be computed by a circle-free
machine. A number is computable if it differs by an integer from the number
computed by a circle-free machine.
We shall avoid confusion by speaking more often of computable sequences 
than
of computable numbers.
</quote>
According to this definition, any TM that halts is circular and does NOT
produce a computable sequence.
I don't know if Turing allows symbols of the first kind to be overwritten.
In my proof, let 1 be the only symbol of the first kind and
substitute blank for 0.
TM1 as I define it is circle free and  produces the computable sequence:
.11111... (base 2)
>Instructions for TM2:
1) Scan right until a 0 is found
>2) Scan right until a second 0 is found
>3) Backup and write a 1 on the previous 0
>Repeat
Assume we give TM2 a tape that contains
>an infinite string of 0's.
Even if we assume that TM2 performs an
>infinite number of operations, the tape
>produced by TM2 will contain an initial
>segment with a finite number of 1's
>followed by a 0.
Using Turing's definition, TM2 produces a computable sequence
that represents the largest rational number less than 1.
.111...1110 (base 2)
>TM2 is incapable of writing an infinite
>number of 1's. If TM2 can not write
>an infinite number of 1's, we can not
>assume that TM1 does.
No TM can write an infinite string of symbols.
It is impossible to determine if TM2 is circle free.
Otherwise, TM2 is circle free.
This is essentially the same reason Turing gives why
the diagonal argument doesn't work with computable numbers.
The problem can be converted into determining whether
every TM is circular or not.
Turing proves this is impossible
TM2 is not an arbitrary TM. It is easily specified.
If we can not determine if TM2 is circle free,
how can we say that any TM is circle free?
> I'm not sure why you think this is particularly important.  By
> construction, at any given time-step, a Turing machine has modified
> only finitely many cells, but there is no upper bound on the number of
> cells a Turing machine can modify.  (There is a glaringly obvious
> analogy with the natural numbers here).
I am showing there is an upper bound.
A TM can't write infinitely many unique representations.
A TM can not compute an irrational number if it can only write finitely 
many
cells.
Russell
- 2 many 2 count
    <DpadnVJoh5H6SGmiRVn-uQ@comcast.com>
====

>> Assume we give this machine a tape that has
>> an infinite string of 0's. It would seem that
>> TM1 will output an infinite string of 1's.
No, you wouldn't have an output at all.  This machine would never
>halt.
 I don't know why everyone is so worried about a TM halting.
  Because that's the way you originally phrased the problem.
You referred to the idea that a TM could output a number, and
in traditional programming jargon, the only way you can see a
program's output is to wait for it to halt, and then look at
what it's produced.
> The word halt does not appear in Turing's paper.
  I'm willing to bet that a majority of participants in this
discussion have not read Turing's paper in the last 20 or 30
years.  I certainly have never read it.  I get my knowledge
second-hand, from people who can explain ideas more clearly (one
hopes) than the original geniuses who came up with them.  (Let's
hear it for Martin Gardner! :)
  But I am glad you posted a bit of the paper you're discussing,
because it is very important to figure out what we're talking
about, here, specifically.
> This is the definition of a computable number given by Turing:
 <quote Computing machines.
> If an a-machine prints two kinds of symbols, of which the first kind 
(called
> figures) consists entirely of 0 and 1 (the others being called symbols of
> the second kind), then the machine will be called a computing machine. If
> the machine is supplied with a blank tape and set in motion, starting 
from
> the correct initial m-configuration,
  Please define m-configuration, as defined in Turing's paper.
> the subsequence of the symbols printed
> by it which are of the first kind will be called the sequence computed by
> the machine. The real number whose expression as a binary decimal is
> obtained by prefacing this sequence by a decimal point is called the 
number
> computed by the machine.
  Okay.  Here Turing is apparently assuming that the sequence printed
by the machine will have a beginning, though not necessarily an end.
It's not clear how he defines the beginning of the sequence, though --
is it left-to-right order? or chronological?  Left-to-right has the
advantage of intuitiveness, but chronological makes more sense
mathematically to me.  Please clarify this point: briefly, what does
Turing mean by the word prefacing?
> At any stage of the motion of the machine, the number of the scanned 
square,
> the complete sequence of all symbols on the tape, and the m-configuration
> will be said to describe the complete configuration at that stage. The
> changes of the machine and tape between successive complete 
configurations
> will be called the moves of the machine.
 {233}
> Circular and circle-free machines.
> symbols of the first kind it will be called circular. Otherwise it is 
said
> to be circle-free.
  All right.  This is fairly bizarre terminology, IMHO -- do you have
any idea why Turing chose these particular words to describe the two
kinds of machines?  Perhaps a quote from section 8 would be in order.
> A machine will be circular if it reaches a configuration from which there 
is
> no possible move, or if it goes on moving, and possibly printing symbols 
of
> the second kind, but cannot print any more symbols of the first kind. The
> significance of the term circular will be explained in §8.
 Computable sequences and numbers.
 A sequence is said to be computable if it can be computed by a 
circle-free
> machine. A number is computable if it differs by an integer from the 
number
> computed by a circle-free machine.
 We shall avoid confusion by speaking more often of computable sequences 
than
> of computable numbers.
> </quote
> According to this definition, any TM that halts is circular and does NOT
> produce a computable sequence.
  Correct.  Now that we know our definitions, or at least some of
them, we can conclusively say that for instance your TM1 computes
the sequence 11111..., which implies that all numbers differing from
the natural number 1 by an integer amount are computable.
> I don't know if Turing allows symbols of the first kind to be 
overwritten.
  Nor do I.  As you're the one with the paper, I suggest you try
to settle this question.
> In my proof, let 1 be the only symbol of the first kind and
> substitute blank for 0.
 TM1 as I define it is circle free and  produces the computable sequence:
> .11111... (base 2)
  Incorrect.  The computable sequence is 11111..., an infinite sequence
of 1's.  The real number corresponding to that sequence is .11111...
(base 2), or the real number 1.
>> Instructions for TM2:
>> 1) Scan right until a 0 is found
>> 2) Scan right until a second 0 is found
>> 3) Backup and write a 1 on the previous 0
>> Repeat
> Using Turing's definition, TM2 produces a computable sequence
  I doubt it.  This depends heavily on the definition of the word
prefacing in Turing's paper.
> that represents the largest rational number less than 1.
  Blatantly false.  No such number exists, computable or otherwise.
That's like saying that your machine computes the number of digits
in pi, or a recipe for granite cheesecake.
> .111...1110 (base 2)
  This is not correct notation.  It reminds me very strongly of
Phil's ramblings, and I really do suggest you take a look at
Google Groups for sci.math, and search on rational numbers
countable, largest integer, and terms of that nature.
> This is essentially the same reason Turing gives why
> the diagonal argument doesn't work with computable numbers.
> The problem can be converted into determining whether
> every TM is circular or not.
> Turing proves this is impossible
  While it is certainly impossible to determine whether Turing
Machine X is circular, for some value of X, it doesn't necessarily
follow that the computable numbers are uncountable.  For that,
you'd need to actually give a reference to Turing's proof, so
that we could look at it and see whether it proves what you think
it does.
  It may very well prove what you think it does.  Before you
posted this quote, I think most participants assumed you were
talking about a different sort of computable number altogether,
one which I won't rehash here since it's irrelevant now.
> TM2 is not an arbitrary TM. It is easily specified.
> If we can not determine if TM2 is circle free,
> how can we say that any TM is circle free?
  The question of whether TM2 is circle-free depends entirely on
Turing's definition of m-configuration.  The question of whether
TM2 computes a number depends entirely on Turing's definitions of
sequence and of prefacing.
>I'm not sure why you think this is particularly important.  By
>construction, at any given time-step, a Turing machine has modified
>only finitely many cells, but there is no upper bound on the number of
>cells a Turing machine can modify.  (There is a glaringly obvious
>analogy with the natural numbers here).
 I am showing there is an upper bound.
> A TM can't write infinitely many unique representations.
> A TM can not compute an irrational number if it can only write finitely 
many
> cells.
compute the following irrational number, though I have not bothered
to write out its state transitions:
  .10110111011110111110111111011111110111111110111111111011111111110...
  That number, which is approximately 0.71673, is computable, but
certainly not rational!  Also computable: pi and e, among many
others.
-Arthur
====

>>Assume we give this machine a tape that has
>>an infinite string of 0's. It would seem that
>>TM1 will output an infinite string of 1's.
>> No, you wouldn't have an output at all.  This machine would never
>> halt.
I don't know why everyone is so worried about a TM halting.
   Because that's the way you originally phrased the problem.
> You referred to the idea that a TM could output a number, and
> in traditional programming jargon, the only way you can see a
> program's output is to wait for it to halt, and then look at
> what it's produced.
That is not how Turing defined them.
Of course, I had to read the paper to figure that out.
>The word halt does not appear in Turing's paper.
   I'm willing to bet that a majority of participants in this
> discussion have not read Turing's paper in the last 20 or 30
> years.  I certainly have never read it.  I get my knowledge
> second-hand, from people who can explain ideas more clearly (one
> hopes) than the original geniuses who came up with them.  (Let's
> hear it for Martin Gardner! :)
I had never read the paper until a few days ago.
Someone suggested that I read it since Turing had
addressed the very question I was examining:
Can the diagonal argument be appied to computable numbers.
I also wanted to make sure that the TMs I was describing
were compatable with Turing's definition.
>   But I am glad you posted a bit of the paper you're discussing,
> because it is very important to figure out what we're talking
> about, here, specifically.
This is the definition of a computable number given by Turing:
<quoteComputing machines.
>If an a-machine prints two kinds of symbols, of which the first kind
(called
>figures) consists entirely of 0 and 1 (the others being called symbols
of
>the second kind), then the machine will be called a computing machine.
If
>the machine is supplied with a blank tape and set in motion, starting
from
>the correct initial m-configuration,
   Please define m-configuration, as defined in Turing's paper.
I think he means what is now called the state transition table.
The TM's instructions.
>the subsequence of the symbols printed
>by it which are of the first kind will be called the sequence computed
by
>the machine. The real number whose expression as a binary decimal is
>obtained by prefacing this sequence by a decimal point is called the
number
>computed by the machine.
   Okay.  Here Turing is apparently assuming that the sequence printed
> by the machine will have a beginning, though not necessarily an end.
> It's not clear how he defines the beginning of the sequence, though --
> is it left-to-right order? or chronological?  Left-to-right has the
> advantage of intuitiveness, but chronological makes more sense
> mathematically to me.  Please clarify this point: briefly, what does
> Turing mean by the word prefacing?
Turing assumes that all TMs start in a defined initial state at the
beginning (leftmost) position of a blank tape.
Turing was not a very good programmer.
(this is like saying Gregor Mendel wasn't a very good microbiologist).
It is obvious that TMs were just a means to an end.
As the title says, Turing was more interested in talking about
Godel's Entscheidungsproblem than in creating the
foundations of modern computers.
I think Turing assumes the output tape will be read from left to right.
Later in the paper, Turing adopts the convention of only writing
symbols of the first kind on every other square.
the 0s and 1's. He states that these other symbols are removed
at some point, but isn't very specific about when or how this happens.
Prefacing means putting a decimal point in front of a binary string.
Turing is trying to show that a TM can generate any real number.
>At any stage of the motion of the machine, the number of the scanned
square,
>the complete sequence of all symbols on the tape, and the
m-configuration
>will be said to describe the complete configuration at that stage. The
>changes of the machine and tape between successive complete
configurations
>will be called the moves of the machine.
{233}
>Circular and circle-free machines.
>symbols of the first kind it will be called circular. Otherwise it is
said
>to be circle-free.
   All right.  This is fairly bizarre terminology, IMHO -- do you have
> any idea why Turing chose these particular words to describe the two
> kinds of machines?  Perhaps a quote from section 8 would be in order.
I have no idea why Turing defines computable numbers this way.
He may have been worried that someone would claim a TM
can't write an infinitely long string. If so, he was right to be worried.
This is exactly what I am claiming.
>A machine will be circular if it reaches a configuration from which
there is
>no possible move, or if it goes on moving, and possibly printing 
symbols
of
>the second kind, but cannot print any more symbols of the first kind.
The
>significance of the term circular will be explained in 
¤8.
Computable sequences and numbers.
A sequence is said to be computable if it can be computed by a
circle-free
>machine. A number is computable if it differs by an integer from the
number
>computed by a circle-free machine.
We shall avoid confusion by speaking more often of computable sequences
than
>of computable numbers.
></quote
>According to this definition, any TM that halts is circular and does 
NOT
>produce a computable sequence.
   Correct.  Now that we know our definitions, or at least some of
> them, we can conclusively say that for instance your TM1 computes
> the sequence 11111..., which implies that all numbers differing from
> the natural number 1 by an integer amount are computable.

>I don't know if Turing allows symbols of the first kind to be
overwritten.
   Nor do I.  As you're the one with the paper, I suggest you try
> to settle this question.
I suspect the answer in no.
Symbols of the second kind can be erased or overwritten.
The paper is available on the internet:
http://www.abelard.org/turpap2/tp2-ie.asp
>In my proof, let 1 be the only symbol of the first kind and
>substitute blank for 0.
TM1 as I define it is circle free and  produces the computable 
sequence:
>.11111... (base 2)
   Incorrect.  The computable sequence is 11111..., an infinite sequence
> of 1's.  The real number corresponding to that sequence is .11111...
> (base 2), or the real number 1.
OK
>>Instructions for TM2:
>>1) Scan right until a 0 is found
>>2) Scan right until a second 0 is found
>>3) Backup and write a 1 on the previous 0
>>Repeat
Using Turing's definition, TM2 produces a computable sequence
   I doubt it.  This depends heavily on the definition of the word
> prefacing in Turing's paper.
Prefacing just means putting a decimal point in front of the string.
>that represents the largest rational number less than 1.
   Blatantly false.  No such number exists, computable or otherwise.
> That's like saying that your machine computes the number of digits
> in pi, or a recipe for granite cheesecake.
This sequence may not represent a real number, but it is computable.
>.111...1110 (base 2)
   This is not correct notation.  It reminds me very strongly of
> Phil's ramblings, and I really do suggest you take a look at
> Google Groups for sci.math, and search on rational numbers
> countable, largest integer, and terms of that nature.
I don't know who Phil is, but I have started several
threads about the largest natural number.
I don't know why you find the idea so bizarre.
The idea that there is a finite number of natural numbers
is certainly not as strange as the idea that there are
more real numbers than natural numbers.
Several people have suggested that this proof says
there is a finite number of natural numbers.
This is incorrect. This proof shows that no set
can contain every natural number.
This is not the same as saying there is a largest natnum.
Just the opposite.
A set can't contain every natnum precisely because
there in no largest natmun.
>This is essentially the same reason Turing gives why
>the diagonal argument doesn't work with computable numbers.
>The problem can be converted into determining whether
>every TM is circular or not.
>Turing proves this is impossible
   While it is certainly impossible to determine whether Turing
> Machine X is circular, for some value of X, it doesn't necessarily
> follow that the computable numbers are uncountable.  For that,
> you'd need to actually give a reference to Turing's proof, so
> that we could look at it and see whether it proves what you think
> it does.
>   It may very well prove what you think it does.  Before you
> posted this quote, I think most participants assumed you were
> talking about a different sort of computable number altogether,
> one which I won't rehash here since it's irrelevant now.
TM2 is not an arbitrary TM. It is easily specified.
>If we can not determine if TM2 is circle free,
>how can we say that any TM is circle free?
   The question of whether TM2 is circle-free depends entirely on
> Turing's definition of m-configuration.  The question of whether
> TM2 computes a number depends entirely on Turing's definitions of
> sequence and of prefacing.
m-configuration means state table.
It is easy to define the state table for TM2.
It only requires three states.
I give the state table for TM2 in Cardinality of Computable Number.
A sequence is an infinitely long string of 0's and/or 1's.
>> I'm not sure why you think this is particularly important.  By
>> construction, at any given time-step, a Turing machine has modified
>> only finitely many cells, but there is no upper bound on the number 
of
>> cells a Turing machine can modify.  (There is a glaringly obvious
>> analogy with the natural numbers here).
I am showing there is an upper bound.
>A TM can't write infinitely many unique representations.
>A TM can not compute an irrational number if it can only write finitely
many
>cells.
>
No it doesn't.
At least, no one can prove that it does.
It is simple to show that the number of 1's
written by TM1 is some multiple of the number
of 1's written by TM2.
>A Turing machine can certainly
> compute the following irrational number, though I have not bothered
> to write out its state transitions:
   .10110111011110111110111111011111110111111110111111111011111111110...
This is the same sequence I use in Cardinality of Computable Numbers.
Turing gives a similar string as an example of the output of a TM.
He provides a state table to produce the string 001011011101111...
If you let TM2 read this tape it will produce a sequence, of 1's
followed by a 0, that is longer than any such sequence on the initial tape.
>   That number, which is approximately 0.71673, is computable, but
> certainly not rational!  Also computable: pi and e, among many
> others.
These numbers are computable only if you can show there is
a circle free TM that computes the relevant infinite sequence.
I doubt any TM can be shown to be circle free as Turing defines it.
Russell
- 2 many 2 count
====
> I don't know why everyone is so worried about a TM halting.
> The word halt does not appear in Turing's paper.
Because in your fallacious proof, you are considering
situations that can only hold _after_ the run of the TM,
not _during_ the run, such as: an infinite number of 1s
being written by the first TM.  At any time _during_ the
run, only a _finite_ number of 1s has been written by
_either_ of the machines you described.
You are being told (yet again) that you cannot consider
what happens after a run, for a TM run which does not
halt. It doesn't _have_ an after; that's what does
not halt _means_, and that's why your description of
the behavior of TM2 makes no sense at all. You are
describing the situation after a finite number of steps
as if it were the situation after the run, but again,
there _is_ no after, so your description is not merely
incorrect, it is _meaningless_, just like an argument
based on characteristics of members of the empty set.
Please don't post _at all_ again about infinite behaviors
until you can understand this simple objection to your
methods at the most profound level.  Intuitive arguments
don't _work_ for infinities, that's why mathematically
sound arguments are the only appropriate tools for discussing
infinite behaviors.
xanthian.
-- 
====

>No TM can write an infinite string of symbols.

> say X is the time unit
 say TM2 is at position log X
> and TM1 is at position X.
 even if X->oo, TM2 is always < TM1
 in other words, why cannot we assume TM1 has infinite operations also.
Let x be the number of 1's written by TM2.
The number of 1's written TM1 must be
cx where c is a constant.
You are saying that cx = infinity where
c and x are both finite numbers.
Russell
- 2 many 2 count
====

> No TM can write an infinite string of symbols.
say X is the time unit
say TM2 is at position log X
>and TM1 is at position X.
even if X->oo, TM2 is always < TM1
in other words, why cannot we assume TM1 has infinite operations also.

> Let x be the number of 1's written by TM2.
> The number of 1's written TM1 must be
> cx where c is a constant.
 You are saying that cx = infinity where
> c and x are both finite numbers.
>
Depends what framework you are arguing in.  i wouldn't make
those 2 statements in the same context.  I'd put cx->oo.
You asserted TM2 has an infinite run length, so you're allowing
infinity into your experiment.  All your methods are constructable,
you have to consider a multiple level processor queue now.
Herc
old OS lecture floods back to memory
====

> 
>> No TM can write an infinite string of symbols.
say X is the time unit
say TM2 is at position log X
>and TM1 is at position X.
even if X->oo, TM2 is always < TM1
in other words, why cannot we assume TM1 has infinite operations also.
> 
> 
> Let x be the number of 1's written by TM2.
> The number of 1's written TM1 must be
> cx where c is a constant.
> 
> You are saying that cx = infinity where
> c and x are both finite numbers.
> 
> 
> Russell
> - 2 many 2 count
> 
> 
Each of the numbers of ones generated corresponds to a natural number, 
and vice-versa. 
According  to your analyses, there can only be finitely many natural 
numbers.
It must be quite frustrating to live in such a limited world.
====

>>No TM can write an infinite string of symbols.
>> say X is the time unit
>> say TM2 is at position log X
>> and TM1 is at position X.
>> even if X->oo, TM2 is always < TM1
>> in other words, why cannot we assume TM1 has infinite operations 
also.
>> TM2 will fail to terminate, and reads an infinite number (and

>Let x be the number of 1's written by TM2.
>The number of 1's written TM1 must be
>cx where c is a constant.
You are saying that cx = infinity where
>c and x are both finite numbers.

>Russell
>- 2 many 2 count

 Each of the numbers of ones generated corresponds to a natural number,
> and vice-versa.
 According  to your analyses, there can only be finitely many natural
> numbers.
Not really.
I have argued that in the past.
This is more about what can be represented by a TM.
My TM2 can be thought of as a machine that counts
how many 1's it has written.
Even if the input tape contains an infinite number of 0's,
TM2 will think the tape contains a finite number.
TM2 is incapable of writing a number that represents
an infinite number of 1's.
Russell
- 2 many 2 count
====
Also available at http://math.ucr.edu/home/baez/week200.html
This Week's Finds in Mathematical Physics - Week 200
John Baez     
Happy New Year!  
I'm making some changes in my life.  For many years I've dreamt 
of writing a book on higher-dimensional algebra that will explain
n-categories and their applications to homotopy theory, representation 
theory, quantum physics, combinatorics, logic - you name it!  It's an 
intimidating goal, because every time I learn something new about these 
subjects I want to put it in this imaginary book, so it keeps getting 
longer and longer in my mind!  Actually writing it will require heroic 
acts of pruning.  But, I want to get started.  
It'll be freely available online, and it'll show up here as it
materializes - but so far I've just got a tentative outline:
http://math.ucr.edu/home/baez/hda.html
Unfortunately, I'm very busy these days.  As you get older, duties 
accumulate like barnacles on a whale if you're not careful!  When I 
started writing This Week's Finds a bit more than ten years ago, I 
was lonely and bored with plenty of time to spare.  My life is very 
different now: I've got someone to live with, a house and a garden 
that seem to need constant attention, a gaggle of grad students, and 
too many invitations to give talks all over the place.
In short, the good news is I'm never bored and there's always something 
fun to do.  The bad news is there's always TOO MUCH to do!  So, a while 
ago I decided to shed some duties and make more time for things I consider 
really important: thinking, playing the piano, writing this book... 
and yes, writing This Week's Finds. 
 
First I quit working for all the journals I helped edit.  Then I started 
job it's really fun to quit.  But doing so didn't free up nearly enough 
time.  
So now I've also decided to stop moderating the newsgroup 
This is painful, because I've learned so much from this newsgroup over 
the last 10 years, met so many interesting people, and had such fun. 
I thank everyone on the group.  I'll miss you!  I'll probably be back
whenever I get lonely or bored.
Ahem.  Before I get weepy and nostalgic, I should talk about some math.  
This November in Florence there was a conference in honor of the 40th 
anniversary of Bill Lawvere's Ph.D. thesis - a famous thesis called
Functorial Semantics of Algebraic Theories, which explored the 
applications of category theory to algebra, logic and physics.  
There are videos of all the talks on the conference website:
2) Ramifications of Category Theory, http://ramcat.scform.unifi.it/
but right now this website seems to be down.
This conference was organized and funded by Michael Wright, a businessman 
with a great love of mathematics and philosophy, so it was appropriate 
that it was held in the old city of Cosimo de Medici, Renaissance banker 
and patron of scholars.  And since there were talks both by mathematicians
and philosophers - especially Alberto Peruzzi, a philosopher at the
University of Florence who helped run the show - I couldn't help but 
remember Cosimo's Platonic Academy, which spearheaded the rebirth of 
classical learning in Renaissance Italy.  When not attending talks, I 
spent a lot of time roaming around twisty old streets, talking category 
theory at wonderful restaurants, reading The Rise and Fall of the House 
ofMedici, and desperately trying to soak up the overabundance of incredible
art and architecture: the Ponte Vecchio, the Piazza del Duomo, the Santa
Croce where everyone from Galileo to Dante to Machiavelli is buried....
Ahem.  Math!
What was Lawvere's thesis about?  It's never been published, so I've 
never read it - though I hear it's going to be.   So, my impression of 
its contents comes from gossip, rumors and later research that refers to 
his work.
Lawvere started out as a student of Clifford Truesdell, working 
on continuum mechanics, which is the very practical branch of field 
theory that deals with fluids, elastic bodies and the like.  In the
process, Lawvere got very interested in the foundations of physics, 
particularly the notions of continuum and physical theory.  
Somehow he decided that only category theory could give him the tools
to really make progress in understanding these notions.  After all, this
was the 1960s, and revolution was in the air.  So, he somehow got himself 
sent to Columbia University to learn category theory from Sam Eilenberg, 
     In my own education I was fortunate to have two teachers who used 
     the term foundations in a common-sense way (rather than in the 
     speculative way of the Bolzano-Frege-Peano-Russell tradition).  
     This way is exemplified by their work in Foundations of Algebraic 
     The Mechanical Foundations of Elasticity and Fluid Mechanics, 
     published in the same year by Truesdell. The orientation of these 
     works seemed to be concentrate the essence of practice and in turn
     use the result to guide practice. 
It may seem like a big jump from the down-to-earth world of continuum 
mechanics to category theory, but to Lawvere the connection made perfect 
sense - and while I've always found his writings inpenetrable, after 
hearing him give four long lectures in Florence I think it makes sense 
to me too!  Let's see if I can explain it. 
Lawvere first observes that in the traditional approach to physical 
theories, there are two key players.  First, there are concrete 
particulars - like specific ways for a violin string to oscillate, 
or specific ways for the planets to move around the sun.  Second, 
there are abstract generals: the physical laws that govern the motion
of the violin string or the planets.  
In traditional logic, an abstract general is called a theory, while a 
concrete particular is called a model of this theory.  A theory is 
usually presented by giving some mathematical language, some rules of 
deduction, and then some axioms.  A model is typically some sort of map 
that sends everything in the theory to something in the world of sets and
truth values, in such a way that all the axioms get mapped to true. 
Since theories involve playing around with symbols according to fixed
rules, the study of theories is often called syntax.  Since the
meaning of a theory is revealed when you look at its models, the
study of models is called semantics.  The details vary a lot depending 
on what you want to do, and physicists rarely bother to formulate their
theories axiomatically, but this general setup has been regarded as the 
ideal of rigor ever since the work of Bolzano, Frege, Peano and Russell 
around the turn of the 20th century.
And this is what Lawvere wanted to overthrow!  
Actually, I'm sort of kidding.  He didn't really want to overthrow this 

setup: he wanted to radically build on it.  First, he wanted to free the 
notion of model from the chains of set theory.  In other words, he 
wanted to consider models not just in the category of sets, but in other
categories as well.  And to do this, he wanted a new way of describing
theories, which is less tied up in the nitty-gritty details of syntax.
To see what Lawvere did, we need to look at an example.  But there
are so many examples that first I should give you a vague sense of the
*range* of examples.  
You see, in logic there are many levels of what you might call strength 

or expressive power, ranging from wimpy languages that don't let you say 

very much and deduction rules that don't let you prove very much, to 
ultra-powerful ones that let you do all sorts of marvelous things.  Near 
the bottom of this hierarchy there's the propositional calculus where 
we only get to say things like
((P implies Q) and (not Q)) implies (not P)
Further up there's the first-order predicate calculus, where we get 
to say things like
for all x (for all y ((x = y and P(x)) implies P(y)))
Even further up, there's the second-order predicate calculus where 
we get to quantify over predicates and say things like
for all x (for all y (for all P (P(x) iff P(y)) implies x = y))
Etcetera...  
And, while you might think it's always best to use the most powerful 
form of logic you can afford, this turns out not to be true!
One reason is that the more powerful your logic is, the fewer categories
theories expressed in this logic can have models in.  This point may
sound esoteric, but the underlying principle should be familiar.  Which
is better: a hand-operated drill, an electric drill, or a drill press?  
A drill press is the most powerful.  But I forgot to mention: you're 
using it to board up broken windows after a storm.  You can't carry a
drill press around, so now the electric drill sounds best.  But another
thing: this is in rural Ghana!  With no electricity, now the hand-operated 
drill is your tool of choice.
In short, there's a tradeoff between power and flexibility.  Specialized
tools can be powerful, but they only operate in a limited context.  
These days we're all painfully aware of this from using computers: fancy 
software only works in a fancy environment!  
Lawvere has even come up with a general theory of how this tradeoff works 
in mathematical logic... he called this the theory of doctrines.  But 
I'm getting way ahead of myself!  He came up with doctrines in 1969, 
and I'm still trying to explain his 1963 thesis.
Just like traditional logic, Lawvere's new approach to logic has been 
studied at many different levels in the hierarchy of strength.  He began 
fairly near the bottom, in a realm traditionally occupied by something 
called universal algebra, developed by Garrett Birkhoff in 1935.  The 
idea here was that a bunch of basic mathematical gadgets can be defined 
using very simple axioms that only involve n-ary operations on some set 
and equations between different ways of composing these operations.  A 
theory like this is called an algebraic theory.  The axioms for an 
algebraic theory aren't even allowed to use words like and, or, 
not 
or implies.  Just equations.
Okay, now for an example.
A good example is the algebraic theory of groups.  A group is a set 
equipped with a binary operation called multiplication, a unary 
operation called inverse, and a nullary operation (that is, a 
constant) called the unit, satisfying these equational laws:
  (gh)k = g(hk)                ASSOCIATIVITY
     1g = g                    LEFT UNIT LAW
                               
     g1 = g                    RIGHT UNIT LAW
g^{-1}g = 1                    LEFT INVERSE LAW
                               
gg^{-1} = 1                    RIGHT INVERSE LAW
Such a primitive gadget is robust enough to survive in very rugged 
environments... it's more like a stone tool than a drill press!
Lawvere noticed that we can talk about models of these axioms not just 
in the category of sets, but in any category with finite products.  
The point is that to talk about an n-ary operation, we just need to be 
able to take the product of an object G with itself n times and consider 
a morphism
f: G x  ...  x G -> G
   |- n times -|
For example, the category of smooth manifolds has finite products, 
so we can talk about a group object in this category, which is just 
a *Lie group*.   The category of topological spaces has finite products, 
so we can talk about a group object in this category too: it's a 
*topological group*.  And so on.  
But Lawvere's really big idea was that there's a certain category
with finite products whose only goal in life is to contain a group
object.  To build this category, first we put in an object 
G
 
Since our category has finite products this automatically means
it gets objects 1, G, G x G, G x G x G, and so on.  Next, we put in 
a binary operation called multiplication, namely a morphism
m: G x G -> G
We also put in a unary operation called inverse:
inv: G -> G
and a nullary operation called the unit:
i: 1 -> G
And then we say a bunch of diagrams commute, which express all
the axioms for a group listed above.
Lawvere calls this category the theory of groups, Th(Grp).  The object 
G is just like a group - but not any *particular* group, since its 
operations only satisfy those equations that hold in *every* group!
By calling this category a theory, Lawvere is suggesting that like a 
theory of the traditional sort, it can have models - and indeed
it can!  A model of theory of groups in some category X with finite
products is just a product-preserving functor
F: Th(Grp) -> X
By the way things are set up, this gives us an object
F(G)
in C, together with morphisms
F(m): F(G) x F(G) -> F(G)
F(inv): F(G) -> F(G)
F(i): F(1) -> F(G)
that serve as the multiplication, inverse and identity element
for F(G)... all making a bunch of diagrams commute, that express
the axioms for a group!
So, a model of the theory of groups in X is just a group object in X.
Whew.  So far I've just explained the *title* of Lawvere's PhD thesis: 
Functorial Semantics of Algebraic Theories.  In Lawvere's approach, 
an algebraic theory is given not by writing down a list of axioms, 
but by specifying a category C with finite products.  And the semantics 
of such theories is all about product-preserving functors F: C -> X.
Hence the term functorial semantics.
Lawvere did a lot starting with these ideas.  Let me just briefly 
summarize, and then move on to his work on topos theory and mathematical 
physics.  
Wise mathematicians are interested not just in models, but also the 
homomorphisms between these.  So, given an algebraic theory C,
Lawvere defined its category of models in X, say Mod(C,X), to have 
product-preserving functors F: C -> X as objects and natural 
transformations between these as morphisms.  For example, taking 
C to be the theory of groups and X to be the category of sets, we get 
the usual category of groups:
Mod(Th(Grp),Set) = Grp
That's reassuring, and that's how it always works.  What's less obvious, 
though, is that one can always recover C from Mod(C,Set) together with 
its forgetful functor to the category of sets.  
In other words: not only can we get the models from the theory, but we 
can also get back the theory from its category of models!
I explained how this works in week136 so I won't do so again here.  
This result actually generalizes an old theorem of Birkhoff on universal 
algebra.  But fans of the Tannaka-Krein reconstruction theorem for 
quantum groups will recognize this duality between theories and their
category of models as just another face of the duality between 
algebras and their category of representations - the classic 
example being the Fourier transform and inverse Fourier transform!  
And this gives me an excuse to explain another bit of Lawvere's jargon: 
while a theory is an abstract general, and particular model of it
is a concrete particular, he calls the category of *all* its models 
in some category a concrete general.  For example, Th(Grp) is an 
abstract general, and any particular group is a concrete particular, but 
Grp is a concrete general.  I mention this mainly because Lawvere flings 
around this trio of terms quite a bit, and some people find them 
off-putting.  There are lots of reasons to find his work daunting, but 
this need not be one.
In short, we have this kind of setup:
            ABSTRACT GENERAL              CONCRETE GENERAL
            theory                        models
            syntax                        semantics
and a precise duality between the two columns!
I would love to dig deeper in this direction - I've really just
scratched the surface so far, and I'm afraid the experts will be
disappointed... but I'm even more afraid that if I went further,
the rest of you readers would drop like flies.  So instead, let me 
say a bit about Lawvere's work on topos theory and physics.  
Most practical physics makes use of logic that's considerably stronger 
than that of algebraic theories, but still considerably weaker than 
what most of us have been brainwashed into accepting as our default
setting, namely Zermelo-Fraenkel set theory with the axiom of choice.  
So if we want, we can do physics in a context less general than an 
arbitrary category with finite products, while still not restricting 
ourselves to the category of sets.  This is where topoi come in - 
they're a lot like the category of sets, but vastly more general. 
Topos theory was born when Grothendieck decided to completely rewrite 
algebraic geometry as part of a massive plan to prove the Weil 
conjectures.  Grothendieck was another revolutionary of the early 1960s, 
and he arrived at his concept of topos sometime around 1962.  In 1963, 
Lawvere and Myles Tierney took this concept - now called a Grothendieck 
topos - and made it both simpler and more general, arriving at the 
present definition.   Briefly put, a topos is a category with finite 
limits, exponentials, and a subobject classifier.  But instead of saying 
what these words mean, I'll just say that this lets you do most of what 
you normally want to do in mathematics, but without the law of excluded 
middle or the axiom of choice.  
One of the many reasons this middle ground is so attractive is that it 
lets you do calculus with infinitesimals the way physicists enjoy doing
it!  Lawvere started doing this in 1967 - he called it synthetic 
differential geometry.   Basically, he cooked up some axioms on a topos
that let you do calculus and differential geometry with infinitesimals.  
The most famous topos like this is the topos of schemes - algebraic
geometers use this one a lot.  The usual category of smooth manifolds is 
not even a topos, but there are topoi that can serve as a substitute, 
which have infinitesimals.
I won't list the axioms of synthetic differential geometry, but the
main idea is that our topos needs to contain an object T called the 
infinitesimal arrow.  This is a rigorous version of those little 
arrows physicists like to draw when talking about vectors:  
                             ----->
 
The usual problem with these little arrows is that they need to be
really tiny, but still point somewhere.  In other words, the head
can't be at a finite distance from the tail - but they can't be at the 
same place, either!  This seems like a paradox, but one can neatly 
sidestep it by dropping the law of excluded middle - or in technical
jargon, working with a non-Boolean topos. 
That sounds like a drastic solution - a cure worse than the disease, 
perhaps! - but it's really not so bad.  Indeed, algebraic geometers 
are perfectly comfortable with the topos of schemes, and they don't 
even raise an eyebrow over the fact that this topos is non-Boolean - 
mainly because you're allowed to use ordinary logic to reason *about*
a topos, even if its internal logic is funny.
But enough logic!  Let's do some geometry!  Let's say we're in some
topos with an infinitesimal arrow object, T.  I'll call the objects
of this topos smooth spaces and the morphisms smooth maps.  How
does geometry work in here?
It's very nice.  The first nice thing is that given any smooth space X, 
a tangent vector in X is just a smooth map 
f: T -> X
that is, a way of drawing an infinitesimal arrow in X.  In general, the
maps from any object A of a topos to any other object B form an object
called B^A - this is part of what we mean when we say a topos has 
exponentials.  So, the space of all tangent vectors in X is X^T.  
And this is what people usually call the tangent bundle of X!   
So, the tangent bundle is pathetically simple in this setup: it's just
a space of maps.  This means we can compose a tangent vector f: T -> X 
with any smooth map g: X -> Y to get a tangent vector gf: T -> Y.  This 
is what people usually call pushing forward tangent vectors.   This 
trick gives a smooth map between tangent bundles, the differential of g, 

which it makes sense to call
g^T: X^T -> Y^T
Moreover, it's pathetically easy to check the chain rule:
(gh)^T = g^T h^T
And so far we haven't used *any* axioms about the object T - just basic 
stuff about how maps work!
We can also define higher derivatives using T.  For second derivatives
we start with T x T, which looks like an infinitesimal square.  Then
we mod out by the map
S_{T,T}: T x T -> T x T
that switches the two factors.  You should visualize this map as 
reflection across the diagonal.  When we mod out by it, we get 
a quotient space that deserves the name
T^2/2!
and if we now use some axioms about T, it turns out that a smooth map
f: T^2/2! -> X
picks out what's called a second-order jet in X.  This is a concept
familiar from traditional geometry, but not as familiar as it should be.
The information in a second-order jet consists of a point in X, the 
first derivative of a curve through X, and also the *second* derivative 
of a curve through X.   Or in physics lingo: position, velocity and 
acceleration!  
We can go ahead and define nth-order jets using T^n/n! in a perfectly
analogous way, and the visual resemblance to Taylor's theorem is by no
means an accident... but let me stick to second derivatives, since I'm
trying to get to Newton's good old F = ma.
Just as the space of all tangent vectors in X is the tangent bundle X^T, 
the space of all 2nd-order jets in X is the 2nd-order jet bundle
X^{T^2/2!}
Using some axioms about T, we can show there is a smooth map 
T^2/2! -> T
which throws out the second-order infinitesimal data and just
keeps the first-order part.   This gives a smooth map
p_X: X^{T^2/2!} -> X^T
from the 2nd-order jet bundle to the tangent bundle.  Intuitively
you can think of this as sending any position-velocity-acceleration
triple, say (q,q',q), to the pair (q,q').   
Now for the fun part: Lawvere defines a dynamical law to be a smooth 
map going the other way:
s_X: X^T -> X^{T^2/2!}
such that s_X followed by p_X is the identity.  In other words, 
it's a way of mapping any position-velocity pair (q,q') to a triple 
(q,q',q).  So, it's a formula for acceleration in terms of position 
and velocity!  
There is a category where an object is a smooth space equipped 
with a dynamical law and a morphism is a lawful motion: that
is, a smooth map
f: X -> Y
that makes the obvious diagram commute:
                
                    s_X
         X^T -------------> X^{T^2/2!}
          |                     |
          |                     |
          |                     |
      f^T |                     | f^{T^2/2!}
          |                     |
          |                     |
          |                     |
          V         s_Y         V
         Y^T -------------> Y^{T^2/2!}
In particular, if we take R to be the real numbers - time - and equip 
it with the law saying 
q = 0 
meaning that time ticks at an unchanging rate, then a lawful motion
f: R -> X
is precisely a trajectory in X that follows the law, meaning that 
the acceleration of the trajectory is the desired function of position
and velocity.  This example is a setup for the classical mechanics
by replacing R by a higher-dimensional space.
I'm sure many of you have the same impression that I had when seeing
this stuff, namely that it's a bit quixotic for a high-powered 
mathematician
to be reformulating the foundations of classical mechanics here at the turn
of the 21st century, instead of working on something cutting-edge like
string theory.  Even if Lawvere's approach is better, one can't help but 
wonder if it gives truly *new* insights, or just a clearer formulation
of existing ones.  And either way, one can't help wonder: does he actually 
expect enough people to learn this stuff to make a difference?  Does he 
really think topos theory can break the Microsoft-like grip that ordinary 
set theory has on mathematics?  
(Note the software analogy raising its ugly head again.  Zermelo-Fraenkel 
set theory is a bit like the Windows operating system: once you're locked 
into it, it's hard to imagine breaking out.  You use it because everyone
else does and you're too lazy to do anything about it.  Topos theory is 
more like the open source movement: you're welcome and even expected to 

keep tinkering with the code.)
I have some sense of the answer to these questions.  First of all, Lawvere 
wants to do math the right way regardless of whether it's popular.  But 
secondly, he's been hard at work trying to make the subject accessible 
to beginners.  He's recently written a couple of textbooks you don't 
need a degree in math to read:
3) F. William Lawvere and Steve Schanuel, Conceptual Mathematics: A First 
Introduction to Categories, Cambridge U. Press, Cambridge, 1997. 
4) F. William Lawvere and Robert Rosebrugh, Sets for Mathematics,
Cambridge U. Press, Cambridge, 2002. 
And third, the great thing about topos theory is that you don't
need to accept it to profit from it.  In math, what really matters
is not believing the axioms but coming up with good ideas.  Topos 
theory is full of good ideas, and these are bound to propagate.
I'll finish off with some references to help you learn more about
this stuff.
Alas, I believe Lawvere's thesis is still lurking in the stacks at 
Columbia University:
5) F. W. Lawvere, Functorial semantics of algebraic theories, 
Dissertation, Columbia University, 1963.
and so far he's only gotten around to publishing a brief summary:
6) F. William Lawvere, Functorial semantics of algebraic theories,
Proceedings, National Academy of Sciences, U.S.A. 50 (1963), 869-872.
But, you can find expositions of his work on algebraic theories here
and there.  Here's a gentle one geared towards computer scientists:
7) Roy L. Crole, Categories for Types, Cambridge U. Press, Cambridge,
1993.
A considerably more macho one is available free online:
8) Michael Barr and Charles Wells, Toposes, Triples and Theories. 
Springer-Verlag, New York, 1983. Available for free electronically at 
http://www.cwru.edu/artsci/math/wells/pub/ttt.html 
This book also talks about sketches, which are a way of syntactically
presenting a category with finite products.  It also serves as an 
introduction to topoi... umm, or at least toposes.  I used to find it
fearsomely difficult and dry.  Now I don't, which is sort of scary.
A really beautiful more advanced treatment of algebraic theories and
also essentially algebraic theories can be found here:
9) Maria Cristina Pedicchio, Algebraic Theories, in Textos de Matematica:
School on Category Theory and Applications, Coimbra, July 13-17, 1999,
pp. 101-159.  
Someone should urge her to make this available online - it's already
in TeX, and it deserves to be easier to get!
Shortly after his thesis, Lawvere tackled topoi in this paper:
10) F. William Lawvere, Elementary theory of the category of sets, 
Proceedings of the National Academy of Science 52 (1964), 1506-1511.
the like:
11) F. William Lawvere, Algebraic theories, algebraic categories, 
and algebraic functors, in Theory of Models, North-Holland, Amsterdam 
(1965), 413-418.
12) F. William Lawvere, Functorial semantics of elementary theories, 
Journal of Symbolic Logic, Abstract, 31 (1966), 294-295.
13) F. William Lawvere, The category of categories as a foundation 
for mathematics, in La Jolla Conference on Categorical Algebra, 
Springer, Berlin 1966, pp. 1-20.
14) F. William Lawvere, Some algebraic problems in the context of
functorial semantics of algebraic theories, in Reports of the Midwest
Category Seminar, eds. Jean Benabou et al, Springer Lecture Notes in
Mathematics No. 61, Springer, Berlin 1968, pp. 41-61.
Then came his work on doctrines, which I vaguely alluded to a while
back:
15) F. William Lawvere, Ordinal sums and equational doctrines, 
Springer Lecture Notes in Mathematics No. 80, Springer, Berlin,
1969, pp. 141-155.
I think he first published on synthetic differential geometry in
Lawvere started publishing his ideas on mathematical physics in the 
late 1970s, though he must have been thinking about them all along:
17) F. William Lawvere, Categorical dynamics, in Proceedings 
of Aarhus May 1978 Open House on Topos Theoretic Methods in 
Geometry, Aarhus/Denmark (1979).
18) F. William Lawvere, Toward the description in a smooth topos 
of the dynamically possible motions and deformations of a continuous 
body, Cahiers de Topologie et Geometrie Differentielle Categorique 
21 (1980), 337-392.
In 1981, Anders Kock came out with a textbook on synthetic differential
geometry:
19) Anders Kock, Synthetic Differential Geometry, Cambridge U. Press, 
Cambridge, 1981. 
More recently, Lawvere came out with a book on applications of 
category theory to physics:
19) F. William Lawvere and S. Schanuel, editors, Categories in 
Continuum Physics, Springer Lecture Notes in Mathematics No. 1174,
Springer, Berlin, 1986. 
The quote about Lawvere's teachers is from:
20) F. William Lawvere, Foundations and applications: axiomatization and
at http://www.math.ucla.edu/~asl/bsl/0902/0902-006.ps
and this gives a good overview of his ideas, though not easy to read!
Finally, Colin McLarty - whom I was delighted to meet in Florence - has
a nice quick introduction to synthetic differential geometry in
his textbook on categories and topos theory:
21) Colin McLarty, Elementary Categories, Elementary Toposes, 
Clarendon Press, Oxford, 1995. 
Along with Lawvere's books Conceptual Mathematics and Sets for
Mathematics, this is the one reference that's really good for
beginners!
 
Okay... now that everyone is gone except the people who are absolutely
nuts about category theory, let me say a bit more about doctrines and 
theory-model duality.  The nuts who are still reading are probably 
disappointed that I kept everything very gentle and expository and 
didn't drop any mind-blowing bombshells of abstraction, which is what 
they like about category theory!  So, let's turn up the abstraction a
few notches.
What's a doctrine?
Well, in week89 I described a monad in an arbitrary 2-category.  
But most of the time when people talk about monads they mean monads 
in Cat, the 2-category of all categories.  These are the most important
monads - but I've never really said what they're good for!  I need to
come clean and explain this now, since a doctrine is a categorified
version of a monad.  
What monads are good for is to describe how objects in one category 
can be regarded as objects of some other category equipped with extra 
structure.  This theme pervades mathematics, and is of the utmost 
importance.  For example: groups are sets equipped with extra structure, 
abelian groups are groups equipped with extra structure, rings are 
abelian groups equipped with extra structure, and so on.  We keep building 
up fancier gadgets from simpler ones.  And pretty much whenever we
do, there's a monad lurking in the background, running the show! 
Suppose we've got two categories C and D, and the objects of D are
objects of C equipped with extra structure.  Then we get a pair of 
adjoint functors:
R: D -> C
L: C -> D
The right adjoint R sends each D-object to its underlying C-object, 
and the left adjoint L sends each C-object to the free D-object on 
it.  Often R is called a forgetful functor.  For example, if
C = Set
and
D = Grp
then we can take the underlying set of any group, and the
free group on any set.
We get a monad on C by letting
T = LR: C -> C
Then, we can use facts about adjoint functors to get natural 
transformations called multiplication
m: TT => T
and the unit
i: 1_C => T
Using more facts about adjoint functors, we can check that these 
satisfy associativity and the left and right unit laws.   I did
all this in week92 so I won't do it again here.  The upshot is
that T is a lot like a monoid - which is why Mac Lane dubbed it a 
monad.   
Now, monoids like to *act* on things, and the same is true for
monads.  It turns out that a monad T on C can act on any object 
of C.  When this happens, we call that object an algebra of T,
or a T-algebra for short.   And when our monad comes from a pair
of adjoint functors as above, the main way we get T-algebras is
from objects of D.  And in nice cases, T-algebras are the *same*
as objects of D.
So, for example, we can describe groups as T-algebras where T is
some monad on the category of sets.  And we can describe abelian
groups as T-algebras where T is some monad on the category of groups.
And we can describe rings as T-algebras where T is some monad on
the category of abelian groups.  And so on!
To really see how this works, we'd need to look at a few examples.
I remember when James Dolan was first teaching me this stuff in a 
little coffeeshop here in Riverside, which has since gone out of
business.  I considered monads too abstract and dug my heels in 
like a stubborn mule, refusing to learn about them - until I went 
through a bunch of examples and saw that *yes*, this monad business 
really *does* capture the essence of what it means to build up 
fancy gadgets from simple ones by adding extra structure!  And 
by now I'm completely sold on it.  One reason is the relation to 
topology, which I explained in part N of week118, and also week174.
But alas, I'm too eager to get to the *really* cool stuff to work 
through examples right now.  So if you're a complete novice at monads, 
you'll have to work out some examples yourself.  Right now, I'll just 
say a bit of fancier stuff to fill in a couple gaps for the semi-experts.
First, when I said in nice cases, I really meant that the category of 
T-algebras is equivalent to D when the forgetful functor R: D -> C is 
monadic.  A bit more precisely: for any monad T on C there's a category
of T-algebras, which is usually called C^T for some silly reason.
And, whenever we have a pair of adjoint functors R: D -> C and L: C -> D, 
we get a monad T = LR and a functor from D to C^T.  This is just a 
careful way of saying that any D-object gives us a T-algebra.  And 
finally, we say that R is monadic if this functor from D to C^T is 
an equivalence of categories.  There's a theorem by Beck that says 
how to tell when a functor is monadic, just by looking at it.
Second, to make the analogy between monoids and monads precise,
we just need to realize that a monad on C is a monoid object in 
the monoidal category hom(C,C).  I already explained this in week92,
in even greater generality than we need here, but we need this now
because I'm about to categorify monads and get doctrines.
Okay: so, monads are good for describing objects equipped with extra
structure and properties.  But now suppose we want to describe 
*categories* equipped with extra structure and properties!  For 
example, the categories with finite products that I was talking 
about earlier, or topoi.  There are LOTS of different interesting
kinds of categories equipped with extra structure and properties, and
each of them gives a different kind of *logic*: the logic that works
inside this kind of category!  The more structure and properties our
category has, the more powerful logic we can use inside it.  This is
what gives the hierarchy of expressive power I was talking about.
So, it pays to have a good general way to describe categories equipped
with extra structure and properties.  
And this is what Lawvere's doctrines do!
I've said how monads on a category C are good for describing 
objects of C equipped with extra structure and properties.  But 
there's a certain category called Cat whose objects are categories!   
So, let's take C = Cat!  A monad on Cat will describe categories
equipped with extra structure and properties.
And this is the simplest definition of doctrine: a monad on Cat.
However, those of you familiar with n-categories will realize that
it's odd to talk about the category of all categories.  Not because 
of Russell's paradox - though that's a problem too, forcing us to talk
about the category of *small* categories - but because what's really
important is the 2-CATEGORY of all categories.  It's best to think
of Cat as a 2-category.  But this suggests that we should work with
a categorified, *weakened* version of monad when defining doctrines.
For this, we need to categorify and weaken the concept of monad.
People have done this, and the result is sometimes called a pseudomonad, 

but I prefer to call it a weak 2-monad, since I have dreams of 
categorifying further, and I don't want my notation to become
ridiculous.  I'd rather talk about weak 3-monads than 
pseudopseudomonads,
wouldn't you?  Furthermore, if you look up pseudomonad in the
dictionary you'll get this:
     PSEUDOMONAD: bacterium usually producing greenish fluorescent 
     water-soluble pigment; some pathogenic for plants and animals.
Yuck!  So, let's be very general and sketch how to define a weak 2-monad 
in any weak 3-category (aka tricategory).   
Given a weak 3-category C and an object c of C, a weak 2-monad on c 
is just a weak monoidal category object in hom(c,c).  
Huh?  Well, hom(c,c) is a weak monoidal 2-category, which is precisely 
the right environment in which to define a weak monoidal category 
object, and that's what we're doing here.  Start with the usual 
definition of a weak monoidal category, which is a gadget living in 
Cat.  Cat is an example of a weak monoidal 2-category, and we can 
write down the same definition in *any* weak monoidal 2-category X,
getting the concept of weak monoidal category object in X.  Then,
take X = hom(c,c).  
(Of course I'm lying slightly here: Cat is more strict than your
average weak monoidal 2-category, so it may not be immediately obvious
how to generalize the concept of weak monoidal category as I'm 
suggesting.  Still, I claim it's not hard if you know about this stuff.)
Now that you know how to define a weak 2-monad on any object c of a 
3-category C, you can take c to be Cat and C to be 2Cat... and this 
is what we really should call a doctrine.
Unsurprisingly, people often consider stricter versions of the
concept of 2-monad and doctrine.  For example, most people 
define their pseudomonads not in a weak 3-category but just a 
semistrict one, also known as a Gray-category - since 2Cat is one
of these.  For more details, try these papers:
22) R. Blackwell, G. M. Kelly, and A. J. Power, Two-dimensional monad
theory, Jour. Pure Appl. Algebra 59 (1989), 1-41.
23) Brian Day and Ross Street, Monoidal bicategories and Hopf algebroids, 
Adv. Math. 129 (1997) 99-157.  
24) F. Marmolejo, Doctrines whose structure forms a fully faithful 
adjoint string, Theory and Applications of Categories 3 (1997), 23-44.
Available at http://www.tac.mta.ca/tac/volumes/1997/n2/3-02abs.html
23) S. Lack, A coherent approach to pseudomonads, Adv. Math. 152 (2000),
179-202.  Also available at 
http://www.maths.usyd.edu.au:8000/u/stevel/papers/psm.ps.gz
Anyway, suppose T is a doctrine.  Then we get a 2-category of 
T-algebras Cat^T, whose objects we should think of as categories 
equipped with extra structure of type T.  The classic example would 
be categories with finite products.  Just as Lawvere thought of 
these as algebraic theories, we can think of *any* T-algebra as a 
theory of type T, and define its category of models: given T-algebras 
C and D, the category of models of C in D is hom(C,D), where the hom 
is taken in Cat^T.   
Depending on what doctrine T we consider, we get many different forms 
of logic, and I'll just list a few to whet your appetite:
     Cat^T = categories with finite products = algebraic theories
     gives what one might call algebraic logic - purely equational
     reasoning about n-ary operations.  The theory of groups, or 
     abelian groups, or rings lives here.
     Cat^T = symmetric monoidal categories gives a sort of logic that
     allows for theories known as operads and PROPs - see 
week191
     for more.  This doctrine is weaker than the previous one, since
     we can only use equations where all the same variables appear on both 
     sides, with no duplications or deletions.  If we go further in this
     direction we obtain various sorts of quantum logic.
     Cat^T = categories with finite limits = essentially algebraic
     theories gives what one might call essentially algebraic logic.
     This doctrine is strong than that of algebraic theories, since it
     allows partially defined operations that are defined only when 
     some equations hold.  The theory of categories lives here, since 
     composition of morphisms is an operation of this sort.
     Cat^T = regular categories gives regular logic.  This doctrine
     is even stronger, since it allows for theories that involve 
     relations as well as n-ary operations.
     Cat^T = cartesian closed categories gives the typed 
lambda-calculus.
     This allows for operations on operations on operations... etc.
     
     Cat^T = topoi gives topos logic.
The typed lambda-calculus is very popular in theoretical computer
science, and I recommend Crole's book cited above for more about how
it's related to cartesian closed categories.  A good introduction to
topos logic is McLarty's book cited above.  For an exhaustive study 
of many other sorts of logic that should be on this list but aren't, 
I recommend part D of this book:
24) Peter Johnstone, Sketches of an Elephant: a Topos Theory
Compendium, Oxford U. Press, Oxford. Volume 1, comprising Part A:
Toposes as Categories, and Part B: 2-categorical Aspects of Topos
Theory, 720 pages, 2002. Volume 2, comprising Part C: Toposes as 
Spaces, and Part D: Toposes as Theories, 880 pages, 2002.
We can do a lot of fun stuff with all these different forms of logic,
and people have indeed done so... but I think I'll stop here.  My
point is merely that higher category theory and logic go hand-in-glove, 
and there is plenty of room for exploration here, especially if we keep 
categorifying - and also keep trying to craft our logic to real-world
applications, both in quantum theory and computer science.
I wish you all a Happy New Year, and good luck on all your adventures.
-----------------------------------------------------------------------
mathematics and physics, as well as some of my research papers, can be
obtained at
http://math.ucr.edu/home/baez/
For a table of contents of all the issues of This Week's Finds, try
http://math.ucr.edu/home/baez/twf.html
A simple jumping-off point to the old issues is available at
http://math.ucr.edu/home/baez/twfshort.html
If you just want the latest issue, go to
http://math.ucr.edu/home/baez/this.week.html
====
> 
> Also available at http://math.ucr.edu/home/baez/week200.html
> 
> This Week's Finds in Mathematical Physics - Week 200
> John Baez
> 
> Happy New Year!
[snip]
Every presidential election year has been a leap year, except for one.
> So now I've also decided to stop moderating the newsgroup
> This is painful, because I've learned so much from this newsgroup over
> the last 10 years, met so many interesting people, and had such fun.
> I thank everyone on the group.  I'll miss you!  I'll probably be back
> whenever I get lonely or bored.
[snip]
The best way to play is to attempt great things wrapped in others'
fears of failure.  At worst it is only playing.  At best you get to
build a new playground for everyone to enjoy - and finance your
expense accounts with the gate.  Quid pro quo.
  
The sound barrier was never there!  The transsonic approach,
however, was lethally bumpy.
> The usual problem with these little arrows is that they need to be
> really tiny, but still point somewhere.  In other words, the head
> can't be at a finite distance from the tail - but they can't be at the
> same place, either!  This seems like a paradox, but one can neatly
> sidestep it by dropping the law of excluded middle - or in technical
> jargon, working with a non-Boolean topos.
Direction without magnitude?  Have a background that is homogeneous
but not isotropic.  It doesn't solve the problem, but it does spread
it out.  Who says every real cow has to be spherical, homogeneous, and
isotropic?  Sometimes simplicity leads to impossibility.  Some
structures are emergent, requiring an unavoidable  threshhold of
complexity.  Stuff should be simple, but not too simple.
[snip]
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
Quis custodiet ipsos custodes?  The Net!
====
[I'm piggybacking this post.]
>> 
>> Also available at http://math.ucr.edu/home/baez/week200.html
>> 
>> This Week's Finds in Mathematical Physics - Week 200
>> John Baez
>> 
<snip> So now I've also decided to stop moderating the newsgroup
>> This is painful, because I've learned so much from this newsgroup over
>> the last 10 years, met so many interesting people, and had such fun.
>> I thank everyone on the group.  I'll miss you!  I'll probably be back
>> whenever I get lonely or bored.
Shit.  [emoticon doffs its hat in honor of work done and submits
a fervent thank you].
<snip>
/BAH
====
Littlemanwearingbigboypants misstates yet again:
> Every presidential election year has been a leap year, except for one.
Bzzzzzzt. Wrong! There have been three.
1)
1800 was not a leap year but had a presidential election.
Jefferson, Burr, Adams, Pinckney, and Jay ran for office.
Burr and Jefferson received the same number of electoral votes
and the matter was settled by the House.
2)
1900 was not a leap year but had a presidential election
Mckinley, Bryan, Woolley and Bebs ran for office.
McKinley won
And of course, the first US presidential election:
3)
1789 was not a leap year but had a presidential election
Washington, Adams, Jay, Harrison and Rutledge ran for office.
Washington won.
Schwartz, please check your facts before posting nonsense to usenet.
====
> 
> Littlemanwearingbigboypants misstates yet again:
> 
>Every presidential election year has been a leap year, except for one.
> 
> Bzzzzzzt. Wrong! There have been three.
[snip]
http://scienceworld.wolfram.com/astronomy/LeapYear.html
Gardyloo.
-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
Quis custodiet ipsos custodes?  The Net!
====
Littlemanwearingbigboypants misstates yet again:
> Every presidential election year has been a leap year, except for 
one.
Bzzzzzzt. Wrong! There have been three.
> [snip]
> 
> http://scienceworld.wolfram.com/astronomy/LeapYear.html
> 
> Gardyloo.
When you throw the bucket straight up, move.
This site supports Bill Vajk, supposing that he is correct that there
have been presidential elections in 1789, 1800 & 1900, none of which
were leap years.
====
Joan Baez:
I'm an aspiring dice setter in the casino game of craps, so I'm
interested in physics, but I am largely uneducated in math and
physics.  So, as I read your post, I searched for some hint that
something you're talking about might be useful.
Having read your post in its entirety, and having found nothing
whatsoever that I can use to any advantage, I at least wanted to post
a reply saying so.
I read your post, and you've read mine, and now we're even.
Speaking of vectors, what do think of the idea that the randomization
of dice depends on a near-equality of possible vectors at some point
and that such randomization can be avoided by having an overriding
vector at every point of the hitting and bouncing process?
Very Respectfully,
Ray
====
reformulating mathematics, logic and physics in categorical terms:
 We can do a lot of fun stuff with all these different forms of logic,
> and people have indeed done so... but I think I'll stop here.  My
> point is merely that higher category theory and logic go hand-in-glove,
> and there is plenty of room for exploration here, especially if we keep
> categorifying - and also keep trying to craft our logic to real-world
> applications, both in quantum theory and computer science.
>
====
Oops, I just accidentally sent an incomplete message. Here I complete it.
 reformulating mathematics, logic and physics in categorical terms:

>We can do a lot of fun stuff with all these different forms of logic,
>and people have indeed done so... but I think I'll stop here.  My
>point is merely that higher category theory and logic go hand-in-glove,
>and there is plenty of room for exploration here, especially if we keep
>categorifying - and also keep trying to craft our logic to real-world
>applications, both in quantum theory and computer science.

Being basically a pure mathematician I love all this stuff.  But I am
curious if so far these reformulations have led to anything new in physics?
...any new conjectures about the structure of our universe?
--Edwin Clark
====
>> reformulating mathematics, logic and physics in categorical terms:
>>We can do a lot of fun stuff with all these different forms of logic,
>>and people have indeed done so... but I think I'll stop here.  My
>>point is merely that higher category theory and logic go 
hand-in-glove,
>>and there is plenty of room for exploration here, especially if we 
keep
>>categorifying - and also keep trying to craft our logic to real-world
>>applications, both in quantum theory and computer science.
>Being basically a pure mathematician I love all this stuff.  But I am
>curious if so far these reformulations have led to anything new in 
physics?
>...any new conjectures about the structure of our universe?
Not new theories of physics yet; so far just new ways of thinking about
existing theories.  One reason may be that few *physicists* know topos
theory; mathematicians rarely come up with new theories of physics.
Of course, I have my own hopes and dreams.
====
> reformulating mathematics, logic and physics in categorical terms:
>We can do a lot of fun stuff with all these different forms of logic,
>and people have indeed done so... but I think I'll stop here.  My
>point is merely that higher category theory and logic go 
hand-in-glove,
>and there is plenty of room for exploration here, especially if we 
keep
>categorifying - and also keep trying to craft our logic to real-world
>applications, both in quantum theory and computer science.
>>Being basically a pure mathematician I love all this stuff.  But I am
>>curious if so far these reformulations have led to anything new in 
physics?
>>...any new conjectures about the structure of our universe?
>Not new theories of physics yet; so far just new ways of thinking about
>existing theories.  One reason may be that few *physicists* know topos
>theory; mathematicians rarely come up with new theories of physics.
Actually, this reply was a bit hasty.  It's pretty much true that
nobody has formulated new theories of physics using topoi, despite 
some work on topos theory and physics by Chris Isham and Fotini
Markopoulou (which you can find on the physics arXiv).  
But if we move over to other doctrines, such as the doctrine of 
symmetric monoidal categories, the story changes.  Most importantly, 
Graeme Segal's definition of a conformal field theory is formulated 
in this doctrine, and conformal field theory is a big chunk of the 
mathematical infrastructure of string theory!  My own spin foam 
models also live in this doctrine.  The basic point is that symmetric
monoidal categories are better than topoi as a context for quantum
physics.
(Here I'm using doctrine in the technical sense described in 
week200.
It means roughly a kind of category, in which a certain kind of logic 
holds, which one can use to formulate a certain class of theories.)
I'll fix the version on my website to make this a bit clearer.  
I got worn out before I reached that particular punchline!
====
I've talked of an error in core mathematics that comes about from
the limitation of the ring of algebraic integers, and thinking back on
discussions, I think there's been a lot of confusion on just what I
mean.
Context helps so I'll mention that the idea of different types of
integers includes Gauss considering numbers of the form a+bi, where a,
and b are integers, which are called gaussian integers in his honor. 
Later there are algebraic integers, which include gaussian integers,
but they are defined to be the roots of monic polynomials with integer
coefficients.
The problem then is that mathematicians thought they were done, but my
result shows they are not.  Understanding how that's possible isn't
really difficult, and an easy way to see it, is to consider gaussian
integers and algebraic integers again.
For instance 2 is a gaussian integer, as well as just an integer.  It
is also an algebraic integer.  However, consider the following
equation:
x^2 = 2
which is outside of the ring of gaussian integers.
Now sqrt(2) is well-known *today* so I think that's a good example for
how a ring can be limited.
What I've found is similar to that, in that I've used a
*decomposition* to show numbers outside of the ring of algebraic
integers.
With my example here x^2 = 2, the decomposition is of 2 into equal
factors.
While 2 is an integer, and a gaussian integer, that decomposition
leads to a result that's neither, though it's an algebraic integer.
I use a polynomial decomposition to show the limitation of the ring of
algebraic integers.
Once I had my result showing that the ring of algebraic integers, like
that of gaussian integers, and of integers before them was still too
small, I found a definition for a fully inclusive ring: the uber ring,
which I call the Object Ring or object ring.
The Object Ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime.
That definition isolates the key property of the numbers in question,
and includes integer, gaussian integers, algebraic integers, and
beyond.
Now mathematicians as a group apparently were unaware that the ring of
algebraic integers was so limited, and proceeded for quite some time
assuming that they'd found the most inclusive ring, which is the
error.
So, in case you're wondering, no, I don't think the definition for
algebraic integers needs to be changed any more than the definition
for gaussian integers needs to be changed.  What's needed is a
recognition of the limitations of the ring.
Want more?  Then go to my blog archives:
James Harris
====
|Once I had my result showing that the ring of algebraic integers, like
|that of gaussian integers, and of integers before them was still too
|small, I found a definition for a fully inclusive ring: the uber ring,
|which I call the Object Ring or object ring.
|
|The Object Ring is a commutative ring that includes all numbers such
|that -1 and 1 are the only members that are both a unit and an
|integer, where no non-unit member is a factor of any two integers that
|are coprime.
|
|That definition isolates the key property of the numbers in question,
|and includes integer, gaussian integers, algebraic integers, and
|beyond.
i generally don't try very hard to understand what you're trying to
say about mathematics, because i find that the language you use is
almost always much too ambiguous to be understood clearly; and
because, even more importantly, you seem to have the wrong idea about
what it means when people have trouble understanding you.  so often
you seem to take it as evidence that you've demonstrated your
superiority over the people who can't understand you, when instead you
should be considering the possibility that you've done a bad job of
explaining things.
in this case, when i read your definition of the object ring above,
i find that as usual it seems impossible to figure out what you really
mean, but i wonder whether what you're really trying to do is to
define what an object is, rather than what the object ring is.
before discussing this any further, though, i'd like to at least
temporarily change your terminology from object to h-number, since
for various reasons i don't think object is a good choice of name
here.
so suppose we define h-number as follows:
  an h-number is an algebraic number such that the sub-ring of the
  ring of algebraic numbers that it generates doesn't contain the
  reciprocals of any integers other than 1 and -1.
then does that definition agree with what you're really trying to say?
or would you perhaps prefer a slightly different version that uses
complex numbers instead of algebraic numbers, as follows:
  an h-number is a complex number such that the sub-ring of the
  ring of complex numbers that it generates doesn't contain the
  reciprocals of any integers other than 1 and -1.
if you decide to at least temporarily accept one of the above
definitions of h-number, then this raises an obvious question: do
the h-numbers form a sub-ring of the larger ring?
so, can you prove that the h-numbers form a sub-ring of the larger
ring?  (or perhaps resolve the question in some other way?)
if the h-numbers as defined above form a sub-ring of the larger ring,
then of course you could start talking about the ring of h-numbers
or the h-number ring and people would probably understand what you
were talking about.  do you agree though that if the h-numbers _don't_
form a sub-ring of the larger ring, then it would be difficult to
figure out what someone might mean by the h-number ring?
-- 
====
> 
> |Once I had my result showing that the ring of algebraic integers, like
> |that of gaussian integers, and of integers before them was still too
> |small, I found a definition for a fully inclusive ring: the uber ring,
> |which I call the Object Ring or object ring.
> |
> |The Object Ring is a commutative ring that includes all numbers such
> |that -1 and 1 are the only members that are both a unit and an
> |integer, where no non-unit member is a factor of any two integers that
> |are coprime.
> |
> |That definition isolates the key property of the numbers in question,
> |and includes integer, gaussian integers, algebraic integers, and
> |beyond.
> 
> i generally don't try very hard to understand what you're trying to
> say about mathematics, because i find that the language you use is
> almost always much too ambiguous to be understood clearly; and
> because, even more importantly, you seem to have the wrong idea about
> what it means when people have trouble understanding you.  so often
> you seem to take it as evidence that you've demonstrated your
> superiority over the people who can't understand you, when instead you
> should be considering the possibility that you've done a bad job of
> explaining things.
Ok.
James Harris
====
> I've talked of an error in core mathematics that comes about from
> the limitation of the ring of algebraic integers, and thinking back on
> discussions, I think there's been a lot of confusion on just what I
> mean.
That's an understatement. But there is no error in core mathematics. 
The
error is in your argument.[snip]
> What I've found is similar to that, in that I've used a
> *decomposition* to show numbers outside of the ring of algebraic
> integers.
 With my example here x^2 = 2, the decomposition is of 2 into equal
> factors.
 While 2 is an integer, and a gaussian integer, that decomposition
> leads to a result that's neither, though it's an algebraic integer.
 I use a polynomial decomposition to show the limitation of the ring of
> algebraic integers.
It isn't necessary to show any limitation of the ring of algebraic
integers. It  is a subset of a larger ring -- a fact that has been known
since it was first defined. Any definition of any entity which includes
some elements and excludes others is limited in the sense you use the
word. So what? You can point to the quantity 1/2 to show the limitation 
of
the ring of algebraic integers, too.
[snip trivial and worthless discuss of Object rings]
--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
> 
>I've talked of an error in core mathematics that comes about from
>the limitation of the ring of algebraic integers, and thinking back on
>discussions, I think there's been a lot of confusion on just what I
>mean.
> 
> That's an understatement. But there is no error in core mathematics. 
The
> error is in your argument.[snip]
There is no error in my argument. 
>What I've found is similar to that, in that I've used a
>*decomposition* to show numbers outside of the ring of algebraic
>integers.
With my example here x^2 = 2, the decomposition is of 2 into equal
>factors.
While 2 is an integer, and a gaussian integer, that decomposition
>leads to a result that's neither, though it's an algebraic integer.
I use a polynomial decomposition to show the limitation of the ring of
>algebraic integers.
> 
> It isn't necessary to show any limitation of the ring of algebraic
> integers. It  is a subset of a larger ring -- a fact that has been 
known
> since it was first defined. Any definition of any entity which includes
> some elements and excludes others is limited in the sense you use the
> word. So what? You can point to the quantity 1/2 to show the limitation 
of
> the ring of algebraic integers, too.
> 
> [snip trivial and worthless discuss of Object rings]
If you're going to objectively discuss the issues at hand, then it
doesn't help to delete out the relevant mathematics, and then call it
trivial and worthless.
What I found is a key property of rings, like the ring of integers,
ring of gaussian integers, and ring of algebraic integers, is that in
those rings the only unit integers are -1 and 1.
My research shows that there must be another more inclusive ring
beyond algebraic integers with the same property i.e. that -1 and 1
are the only integers in that ring that are units.
Now then, if you C. Bond are simply too emotional to objectively
discuss the mathematics which backs up my claims, then you're probably
just going to frustrate yourself, and waste bandwith!!!
James Harris
====
> I've talked of an error in core mathematics that comes about from
>> the limitation of the ring of algebraic integers, and thinking back 
on
>> discussions, I think there's been a lot of confusion on just what I
>> mean.
That's an understatement. But there is no error in core mathematics. 
The
>error is in your argument.[snip]
 There is no error in my argument.
On the contrary. You have been thoroughly refuted multiple times. Your 
so-called 'proof' is
riddled with errors.
>> What I've found is similar to that, in that I've used a
>> *decomposition* to show numbers outside of the ring of algebraic
>> integers.
>> With my example here x^2 = 2, the decomposition is of 2 into equal
>> factors.
>> While 2 is an integer, and a gaussian integer, that decomposition
>> leads to a result that's neither, though it's an algebraic integer.
>> I use a polynomial decomposition to show the limitation of the ring 
of
>> algebraic integers.
It isn't necessary to show any limitation of the ring of algebraic
>integers. It  is a subset of a larger ring -- a fact that has been 
known
>since it was first defined. Any definition of any entity which includes
>some elements and excludes others is limited in the sense you use 
the
>word. So what? You can point to the quantity 1/2 to show the 
limitation of
>the ring of algebraic integers, too.
[snip trivial and worthless discuss of Object rings]
 If you're going to objectively discuss the issues at hand, then it
> doesn't help to delete out the relevant mathematics, and then call it
> trivial and worthless.
But it is trivial and worthless. Furthermore, it has been posted multiple 
times before, so it
is not only trivial and worthless, it is redundant.
> What I found is a key property of rings, like the ring of integers,
> ring of gaussian integers, and ring of algebraic integers, is that in
> those rings the only unit integers are -1 and 1.
 My research shows that there must be another more inclusive ring
> beyond algebraic integers with the same property i.e. that -1 and 1
> are the only integers in that ring that are units.
 Now then, if you C. Bond are simply too emotional to objectively
> discuss the mathematics which backs up my claims, then you're probably
> just going to frustrate yourself, and waste bandwith!!!
Well, James Harris, the mathematics which backs up your claims has 
been discussed at
length and the final score is:
James Harris = 0, Mathematicians = 100
And I can be objective about this because I am neither James Harris nor a 
mathematician.
--
There are two things you must never attempt to prove: the unprovable -- and 
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
====
>I've talked of an error in core mathematics that comes about from
>the limitation of the ring of algebraic integers, and thinking back on
>discussions, I think there's been a lot of confusion on just what I
>mean.
[Demonstration of the earth-shattering discovery: not everything
>is an algebraic integer! snipped...]
Now mathematicians as a group apparently were unaware that the ring of
>algebraic integers was so limited, and proceeded for quite some time
>assuming that they'd found the most inclusive ring, which is the
>error.
This is simply hilarious. As often happens, you're simply projecting
you errors, this time onto mathematicians in general.
Exactly what evidence do you have that mathematicians thought
the algebraic integers were the most inclusive ring?
>So, in case you're wondering, no, I don't think the definition for
>algebraic integers needs to be changed any more than the definition
>for gaussian integers needs to be changed.  
Huh. Then you're _not_ talking about an error in standard mathematics
after all! What a surprise.
>What's needed is a
>recognition of the limitations of the ring.
Where the limitation is that it contains only the algebraic
integers...
>Want more?  Then go to my blog archives:
>
Love that mathforprofit thing. Just curious: how's the paypal 
working out? So far the donations amount to how much?
>James Harris
************************
David C. Ullrich
====
> 
>I've talked of an error in core mathematics that comes about from
>the limitation of the ring of algebraic integers, and thinking back on
>discussions, I think there's been a lot of confusion on just what I
>mean.
[Demonstration of the earth-shattering discovery: not everything
>is an algebraic integer! snipped...]
Now mathematicians as a group apparently were unaware that the ring of
>algebraic integers was so limited, and proceeded for quite some time
>assuming that they'd found the most inclusive ring, which is the
>error.
> 
> This is simply hilarious. As often happens, you're simply projecting
> you errors, this time onto mathematicians in general.
> 
> Exactly what evidence do you have that mathematicians thought
> the algebraic integers were the most inclusive ring?
> 
That's not a bad question.  If in fact they don't then there shouldn't
be reason for people to argue with me further.  As the discoverer of a
more inclusive ring, I get to name it, and I have, so then discussions
can move from antagonistic to considerations of that ring and its
properties.
James Harris
====
>> 
>>I've talked of an error in core mathematics that comes about from
>>the limitation of the ring of algebraic integers, and thinking back on
>>discussions, I think there's been a lot of confusion on just what I
>>mean.
>>[Demonstration of the earth-shattering discovery: not everything
>>is an algebraic integer! snipped...]
>>Now mathematicians as a group apparently were unaware that the ring of
>>algebraic integers was so limited, and proceeded for quite some time
>>assuming that they'd found the most inclusive ring, which is the
>>error.
>> 
>> This is simply hilarious. As often happens, you're simply projecting
>> you errors, this time onto mathematicians in general.
>> 
>> Exactly what evidence do you have that mathematicians thought
>> the algebraic integers were the most inclusive ring?
>> 
That's not a bad question.  If in fact they don't then there shouldn't
>be reason for people to argue with me further.  
Nobody has ever argued with the assertion that the algebraic integers
are not the most inclusive ring. The arguments are about other
statements
>As the discoverer of a
>more inclusive ring, 
Wow. Choose anything that's not an algebraic integer. Adjoin it
to the algebraic integers. That gives a more inclusive ring.
>I get to name it, and I have, so then discussions
>can move from antagonistic to considerations of that ring and its
>properties.
First you have to give a coherent _definition_ of the ring. You've
never done that.
(One can see that the definition you give is incoherent because
of all the discussions about what you might mean by it - one
person says you mean this, another person says you mean
something else, others say the definition is simply meaningless.
If the definition you gave were coherent those discussions
would not arise - people would know what you meant just
>James Harris
************************
David C. Ullrich
====
I've talked of an error in core mathematics that comes about from
>the limitation of the ring of algebraic integers, and thinking back on
>discussions, I think there's been a lot of confusion on just what I
>mean.
[Demonstration of the earth-shattering discovery: not everything
>is an algebraic integer! snipped...]
Now mathematicians as a group apparently were unaware that the ring of
>algebraic integers was so limited, and proceeded for quite some time
>assuming that they'd found the most inclusive ring, which is the
>error.
 This is simply hilarious. As often happens, you're simply projecting
> you errors, this time onto mathematicians in general.
 Exactly what evidence do you have that mathematicians thought
> the algebraic integers were the most inclusive ring?
So, in case you're wondering, no, I don't think the definition for
>algebraic integers needs to be changed any more than the definition
>for gaussian integers needs to be changed.
 Huh. Then you're _not_ talking about an error in standard mathematics
> after all! What a surprise.
What's needed is a
>recognition of the limitations of the ring.
 Where the limitation is that it contains only the algebraic
> integers...
Want more?  Then go to my blog archives:

> Love that mathforprofit thing. Just curious: how's the paypal
> working out? So far the donations amount to how much?
Well, after taxes are figured out.................$0 *giggle*
James Harris

> ************************
 David C. Ullrich
David Moran
Chief Meteorologist
Oklahoma Storm Team
====
In sci.math, James Harris
<jstevh@msn.com>
<3c65f87.0312311751.34f1e7ef@posting.google.com>:
> I've talked of an error in core mathematics that comes about from
> the limitation of the ring of algebraic integers, and thinking back on
> discussions, I think there's been a lot of confusion on just what I
> mean.
> 
> Context helps so I'll mention that the idea of different types of
> integers includes Gauss considering numbers of the form a+bi, where a,
> and b are integers, which are called gaussian integers in his honor. 
> Later there are algebraic integers, which include gaussian integers,
> but they are defined to be the roots of monic polynomials with integer
> coefficients.
> 
> The problem then is that mathematicians thought they were done, but my
> result shows they are not.  Understanding how that's possible isn't
> really difficult, and an easy way to see it, is to consider gaussian
> integers and algebraic integers again.
> 
> For instance 2 is a gaussian integer, as well as just an integer.  It
> is also an algebraic integer.  However, consider the following
> equation:
> 
> x^2 = 2
> 
> which is outside of the ring of gaussian integers.
> 
> Now sqrt(2) is well-known *today* so I think that's a good example for
> how a ring can be limited.
> 
> What I've found is similar to that, in that I've used a
> *decomposition* to show numbers outside of the ring of algebraic
> integers.
> 
> With my example here x^2 = 2, the decomposition is of 2 into equal
> factors.
> 
> While 2 is an integer, and a gaussian integer, that decomposition
> leads to a result that's neither, though it's an algebraic integer.
> 
> I use a polynomial decomposition to show the limitation of the ring of
> algebraic integers.
> 
> Once I had my result showing that the ring of algebraic integers, like
> that of gaussian integers, and of integers before them was still too
> small, I found a definition for a fully inclusive ring: the uber ring,
> which I call the Object Ring or object ring.
> 
> The Object Ring is a commutative ring that includes all numbers such
> that -1 and 1 are the only members that are both a unit and an
> integer, where no non-unit member is a factor of any two integers that
> are coprime.
> 
> That definition isolates the key property of the numbers in question,
> and includes integer, gaussian integers, algebraic integers, and
> beyond.
> 
> Now mathematicians as a group apparently were unaware that the ring of
> algebraic integers was so limited, and proceeded for quite some time
> assuming that they'd found the most inclusive ring, which is the
> error.
> 
> So, in case you're wondering, no, I don't think the definition for
> algebraic integers needs to be changed any more than the definition
> for gaussian integers needs to be changed.  What's needed is a
> recognition of the limitations of the ring.
> 
> Want more?  Then go to my blog archives:
> 
> 
> 
> James Harris
I'm going to have to go through your mathblogs in any event, to try
to trace through the impossibly complex contortions of this so-called
proof, and the threads arguing such.  So lessee...
You give as an example
P(x) = 14706125*x^3 - 900375*x^2 - 17640*x + 1078
     = 7^2*(2401*x^3 - 147*x^2 + 3*x)*(5^3) - 3*(-1 + 49*x)*(5)*(7^2) + 7^3
This equality verifies fine, although at some point I'm
going to have to dig up where you got the '7' from, as
the equation came from somewhere else in a grander proof.
But that doesn't matter at present.
You then state
P(x) = (5*a_1(x) + 7) * (5*a_2(x) + 7) * (5*a_3(x) + 7)
where the a's are the roots of the cubic
R(a,x) = a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401 * x^3 - 147 * x^2 + 3*x)
[there is a small error in the text perhaps, as the a's
are not explicitly shown as functions of x, but that's not
horribly important; I've also inserted GP/Pari-compatible
notation for my own convenience].
This is fine too, although an intermediate step might be handy,
since the derivation is not that obvious.  However, I've verified
it in another post and won't reprise it here unless absolutely
necessary.
Now Q(x) = P(x)/49 also has integer coefficients, and it turns out
one can in fact divide, generating
Q(x) = (5/7*a_1(x) + 1) * (5/7*a_2(x) + 1) * (5*a_3(x) + 7)
in the general case.  Of course one can also generate such
things as
Q'(x) = (5*a_1(x) + 7) * (5/7*a_2(x) + 1) * (5/7*a_3(x) + 1)
or
Q(x) = (5*a_1(x) + 7) * (5/49*a_2(x) + 1/7) * (5*a_3(x) + 7)
so there are some minor issues here.
It is your contention that 5/7*a_1(x) and 5/7*a_2(x) must be
algebraic integers, at least as far as I can understand your blog.
It is my contention that one cannot conclude such
without at least an idea of the value of x, as x is a
free variable.  Certainly for x = 0,  5/7*a_1(x) and
5/7*a_2(x) are in fact algebraic integers (namely, 0).
However, it turns out Dik Winter has written up in a prior
post ( <HqJ8rn.KHG@cwi.nl> ) an extremely elegant method
he found for determining the roots of an arbitrary cubic
with monic x^3 term.  I shall reprise that method here,
with minor modifications necessitated by notation (e.g.,
his cubic uses x^3 + a*x^2 + b*x + c, which overloads
the value 'a').
We take R(a,x) = a^3 + 3*(-1 + 49*x)*a^2 - 49*(2401 * x^3 - 147 * x^2 + 
3*x)
and solve for a.  The values are as follows:
q(x) = -4 * (3*(-1 + 49*x))^2
r(x) = -108 * (-49*(2401*x^3 - 147*x^2 + 3*x)) - 8*(3*(-1 + 49*x))^3
k1(x) = cbrt(r(x) + sqrt(q(x)^3 + r(x)^2))/2
k2(x) = cbrt(r(x) - sqrt(q(x)^3 + r(x)^2))/2
w = (-1 + sqrt(-3))/2, w^2 = (-1 - sqrt(-3))/2, w^3 = 1
where cbrt(z) is the cube root of z nearest the positive real axis
in the complex plane.  (Not that it matters all that much, as
long as conjugate(cbrt(z)) = cbrt(conjugate(z)).)
The roots are therefore
a_1(x) = (-(3*(-1 + 49*x)) + w*k1(x) + w^2*k2(x))/3
a_2(x) = (-(3*(-1 + 49*x)) + w^2*k1(x) + w^1*k2(x))/3
a_3(x) = (-(3*(-1 + 49*x)) + k1(x) + k2(x))/3
It is obvious that I now have an explicit solution for
a_1(x) et al, written in a rather ugly but workable
form, and can compute them by simply inserting x and
grinding it out.
For checking purposes, setting x = 0 yields
q(0) = -36
r(0) = 216
k1(0) = cbrt(216 + sqrt((-36)^3 + 216^2))/2 = 3
k2(0) = cbrt(216 - sqrt((-36)^3 + 216^2))/2 = 3
a_1(0) = (3 + w*3 + w^2*3)/3 = 0
a_2(0) = (3 + w^2*3 + w*3)/3 = 0
a_3(0) = (3 + 3 + 3)/3 = 3
which shows I didn't make any gross computational errors.
If I set x = 1/49 I get:
q(1/49) = 0
r(1/49) = 108
k1(1/49) = cbrt(108 + sqrt(108^2))/2 = 3
k2(1/49) = cbrt(108 - sqrt(108^2))/2 = 0
a_1(0) = (0 + w*3 + w^2*0)/3 = w
a_2(0) = (0 + w^2*3 + w*0)/3 = w^2
a_3(0) = (0 + 3 + 0)/3 = 1
which is entirely consistent, as R(a,1/49) = a^3 - 1.
Unfortunately, neither 5*a_1(0)/7 nor 5*a_2(0)/7 is
an algebraic integer.  For 5/7 this is obvious.
For 5/7 * w and 5/7 * w^2 it turns out that all three
are roots of the equation
u^3 - 5^3/7^3 = 0
Since 5^3/7^3 is not an integer, 5*a_1(1/49)/7 and 5*a_2(1/49)
cannot be algebraic integers.
If you wish to prove that, given any algebraic integer x,
5/7*a_1(x) and 5/7*a_2(x) are algebraic integers, you'll
have to work hard at it -- in fact, there's an indefinite
number of counterexamples.  If one sets x=1, for example,
one gets:
R(a,1) = a^3 + 144*a^2 - 110593
This polynomial is irreducible.  The explicit roots
are actually rather messy so I'll fall back on plan B,
which is to merely show that b = 5/7*a or a = 7/5*b is
not an algebraic integer.  To do that requires the usual
substitution:
R(7/5*b,1)*5^3/7^3 = b^3 + 720/7*b^2 - 282125/7
Oops, doesn't work.
R(7/5*b,2)*5^3/7^3 = b^3 + 1455/7*b^2 - 2328250/7
also doesn't work, so x=2 is out.
R(7/5*b,-1)*5^3/7^3 = b^3 - 750/7*b^2 + 318875/7
Nope; x=-1 is out too.  In fact,
R(7/5*b,x)*5^3/7^3 = b^3 + (-15/7+105*x)*b^2-42875*x^3+2625*x^2-375/7*x
I can prove that, if x is in the rational numbers, and
this equation is irreducible (one might get lucky for
certain rational x; I can't say without a lot more work), then, for
the b's to be algebraic integers, 105*x - 15/7 and
-42875*x^3+2625*x^2-375/7*x must be integers.
The first requires x = 1/49 + z/105 for some integer z.
Plugging into the second, one gets after a bit of grinding
the coefficient -1/27*z^3 - 125/343.  This can never be an
integer for any integer z; therefore b_1(), b_2(), and b_3()
cannot be integer for rational x except for certain special
cases such as x=0, where the polynomial becomes reducible.
I can't speak for non-rational x, of course; odd things happen.
But it's clear that 5/7 * a_1(x) and 5/7 * a_2(x) are not excluded
at times; they're excluded most of the time.
I will now conclude with a diatribe on divisibility.  It is true
that P(x) can be divided by 49.  It is not true that this means much.
The polynomial
x^4 - 4*x^3 + 6*x^2 - 4*x - 1
is also divisible by 49, in the algebraic number field, yielding
of course the rather silly polynomial
x^4/49 - 4/49*x^3 + 6/49*x^2 - 4/49*x - 1
So what do we have at the end of the day?  Not a whole lot,
from what I can see.
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
====
<delete 
> It is your contention that 5/7*a_1(x) and 5/7*a_2(x) must be
> algebraic integers, at least as far as I can understand your blog.
That is false.  My point is that in general they are not which proves
there's more beyond algebraic integers in terms of rings with the
property that -1 and 1 are the only unit integers!!!
So you have it reversed.
My research involves the step beyond algebraic integers, like before
algebraic integers were a step beyond gaussian integers, while
gaussian integers were a step beyond integers.
Understand?
James Harris
====
In sci.math, James Harris
<jstevh@msn.com>
on 1 Jan 2004 08:52:53 -0800
<3c65f87.0401010852.5aa5d45b@posting.google.com>:
> <delete>

>> It is your contention that 5/7*a_1(x) and 5/7*a_2(x) must be
>> algebraic integers, at least as far as I can understand your blog.
> 
> That is false.  My point is that in general they are not which proves
> there's more beyond algebraic integers in terms of rings with the
> property that -1 and 1 are the only unit integers!!!
> 
> So you have it reversed.
> 
> My research involves the step beyond algebraic integers, like before
> algebraic integers were a step beyond gaussian integers, while
> gaussian integers were a step beyond integers.
> 
> Understand?
> 
> 
> James Harris
Ah, OK.  Mind you, everyone knows there is a next step
beyond algebraic integers -- namely, algebraic numbers.
Unless you're trying to show there's a set of numbers
with units -1 and +1 and algebraic-integer-like properties
that constitute a ring, with the algebraic integers as a
proper subset.
Using standard multiplication, however, you will run into
a problem, as *every* unit of the algebraic integers will
be a unit of your set as well.  This includes numbers
such as 4 - sqrt(15) and sqrt(2)/2 - i*sqrt(2)/2.
[x = 4 - sqrt(15) solves x^2 - 8*x + 1 = 0;
 x = sqrt(2)/2 - i*sqrt(2)/2 solves x^4 + 1 = 0.]
Please clarify this conundrum.
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
====
> In sci.math, James Harris
> <jstevh@msn.com on 1 Jan 2004 08:52:53 -0800
> <3c65f87.0401010852.5aa5d45b@posting.google.com>:
><delete
>> 
>> It is your contention that 5/7*a_1(x) and 5/7*a_2(x) must be
>> algebraic integers, at least as far as I can understand your blog.
That is false.  My point is that in general they are not which proves
>there's more beyond algebraic integers in terms of rings with the
>property that -1 and 1 are the only unit integers!!!
So you have it reversed.
My research involves the step beyond algebraic integers, like before
>algebraic integers were a step beyond gaussian integers, while
>gaussian integers were a step beyond integers.
Understand?

>James Harris
> 
> Ah, OK.  Mind you, everyone knows there is a next step
> beyond algebraic integers -- namely, algebraic numbers.
> 
The ring of algebraic numbers is a field.
There is a step beyond that is NOT a field, where only -1 and 1 are
units in the ring.
James Harris
====
In sci.math, James Harris
<jstevh@msn.com>
on 1 Jan 2004 18:21:10 -0800
<3c65f87.0401011821.10845d56@posting.google.com>:
>> In sci.math, James Harris
>> <jstevh@msn.com> on 1 Jan 2004 08:52:53 -0800
>> <3c65f87.0401010852.5aa5d45b@posting.google.com>:
>><delete 
> It is your contention that 5/7*a_1(x) and 5/7*a_2(x) must be
> algebraic integers, at least as far as I can understand your blog.
>>That is false.  My point is that in general they are not which proves
>>there's more beyond algebraic integers in terms of rings with the
>>property that -1 and 1 are the only unit integers!!!
>>So you have it reversed.
>>My research involves the step beyond algebraic integers, like before
>>algebraic integers were a step beyond gaussian integers, while
>>gaussian integers were a step beyond integers.
>>Understand?
>>James Harris
>> 
>> Ah, OK.  Mind you, everyone knows there is a next step
>> beyond algebraic integers -- namely, algebraic numbers.
>> 
> 
> The ring of algebraic numbers is a field.
> 
> There is a step beyond that is NOT a field, where only -1 and 1 are
> units in the ring.
Better clarify that phrasing; presumably you mean step beyond algebraic
integers, *not* step beyond algebraic numbers (which leads us
to transcendentals).
As it is, I have two questions:
[1] Does this step beyond contain the algebraic integers as a proper
    subset?
[2] Is the number 4 - sqrt(15) part of that step beyond?
> 
> 
> James Harris
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
====
   [.snip.]
>Ah, OK.  Mind you, everyone knows there is a next step
>beyond algebraic integers -- namely, algebraic numbers.
There are plenty of rings between the algebraic integers and the
complex numbers, most of them not fields. You can localize at any
number of places to obtain a ring that still satisfies the condition
given: the intersection with Q is equal to Z. Or you can throw in any
of a number of different transcendentals. I do not really see a next
step, but a continuum of rings lying in between.
-- 
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
In sci.math, Arturo Magidin
<magidin@math.berkeley.edu>
on Thu, 1 Jan 2004 23:13:27 +0000 (UTC)
<bt29in$mfn$1@agate.berkeley.edu>:
> 
>    [.snip.]
> 
>>Ah, OK.  Mind you, everyone knows there is a next step
>>beyond algebraic integers -- namely, algebraic numbers.
> 
> There are plenty of rings between the algebraic integers and the
> complex numbers, most of them not fields. You can localize at any
> number of places to obtain a ring that still satisfies the condition
> given: the intersection with Q is equal to Z. Or you can throw in any
> of a number of different transcendentals. I do not really see a next
> step, but a continuum of rings lying in between.
> 
An interesting point that; I could construct A[pi], for example;
I'm assuming that's what you're referring to, where A is the
ring of algebraic integers.
However, any ring containing the algebraic integers will
by necessity contain its units as well.  An object ring O
with units [-1, +1] containing A simply doesn't work.
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
====
>In sci.math, Arturo Magidin
><magidin@math.berkeley.eduon Thu, 1 Jan 2004 23:13:27 +0000 (UTC)
><bt29in$mfn$1@agate.berkeley.edu>:
>> 
>>    [.snip.]
>> 
>Ah, OK.  Mind you, everyone knows there is a next step
>beyond algebraic integers -- namely, algebraic numbers.
>> 
>> There are plenty of rings between the algebraic integers and the
>> complex numbers, most of them not fields. You can localize at any
>> number of places to obtain a ring that still satisfies the condition
>> given: the intersection with Q is equal to Z. Or you can throw in any
>> of a number of different transcendentals. I do not really see a next
>> step, but a continuum of rings lying in between.
>> 
An interesting point that; I could construct A[pi], for example;
>I'm assuming that's what you're referring to, where A is the
>ring of algebraic integers.
However, any ring containing the algebraic integers will
>by necessity contain its units as well. 
So? The only qualification is that the only unit which are ALSO
integers be 1 and -1. That is, that the intersection of the ring with
Q be equal to Z.
> An object ring O
>with units [-1, +1] containing A simply doesn't work.
I think you are misunderstanding the condition given. The condition is
that the only integers which are also units of the ring are 1 and -1,
not that the only units are 1 and -1.
-- 
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
In sci.math, Arturo Magidin
<magidin@math.berkeley.edu>
on Fri, 2 Jan 2004 05:12:38 +0000 (UTC)
<bt2uk6$rm2$1@agate.berkeley.edu>:
>>In sci.math, Arturo Magidin
>><magidin@math.berkeley.edu>on Thu, 1 Jan 2004 23:13:27 +0000 (UTC)
>><bt29in$mfn$1@agate.berkeley.edu>:
> 
>    [.snip.]
> 
>>Ah, OK.  Mind you, everyone knows there is a next step
>>beyond algebraic integers -- namely, algebraic numbers.
> 
> There are plenty of rings between the algebraic integers and the
> complex numbers, most of them not fields. You can localize at any
> number of places to obtain a ring that still satisfies the condition
> given: the intersection with Q is equal to Z. Or you can throw in any
> of a number of different transcendentals. I do not really see a 
next
> step, but a continuum of rings lying in between.
> 
>>An interesting point that; I could construct A[pi], for example;
>>I'm assuming that's what you're referring to, where A is the
>>ring of algebraic integers.
>>However, any ring containing the algebraic integers will
>>by necessity contain its units as well. 
> 
> So? The only qualification is that the only unit which are ALSO
> integers be 1 and -1. That is, that the intersection of the ring with
> Q be equal to Z.
> 
>> An object ring O
>>with units [-1, +1] containing A simply doesn't work.
> 
> I think you are misunderstanding the condition given. The condition is
> that the only integers which are also units of the ring are 1 and -1,
> not that the only units are 1 and -1.
> 
Ah yes...he's confirmed that in one of his posts today (yesterday?).
Of course that's also true of the algebraic integers; the only
units which are integers are -1 and +1.
Color me confused, but from the looks of it he might be trying
to construct A<a_1(x)*5/7,a_2(x)*5/7,(a_3(x)-3)*5/22> for some x
(where a_1(x) etc. is the root of his second polynomial).  Presumably
this leads to a perfectly reasonable if slightly esoteric ring.
There may be another parameter in in there, at all --
I'll call it 'y' -- as he's been using '7', and it looks
an awful lot like a special case of something, but I can't
remember what now.
-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
====
>In sci.math, Arturo Magidin
><magidin@math.berkeley.edu>
   [.snip.]
>> I think you are misunderstanding the condition given. The condition is
>> that the only integers which are also units of the ring are 1 and -1,
>> not that the only units are 1 and -1.
>> 
Ah yes...he's confirmed that in one of his posts today (yesterday?).
>Of course that's also true of the algebraic integers; the only
>units which are integers are -1 and +1.
Color me confused, but from the looks of it he might be trying
>to construct A<a_1(x)*5/7,a_2(x)*5/7,(a_3(x)-3)*5/22> for some x
>(where a_1(x) etc. is the root of his second polynomial).  Presumably
>this leads to a perfectly reasonable if slightly esoteric ring.
It will, in general, lead to rings which include nonintegral
rationals; for example, since a_1, a_2, and a_3 are the three
conjugate, and satisfy
   a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
for each value of x.
Therefore,
a1*a2*a3 = 49(2401 x^3 - 147x^2 + 3x)
a1*a2 + a1*a3 + a2*a3 = 0
a1 + a2 + a3 = -3(-1+49x).
So if you have a_1/7 and a_2/7, then you have
-3(-1+49x)/7 - a3/7
and other elements, which will usually lead you to noninteger
rationals. 
If you include several values of x, things get nastier. See for
example an elementary analsysi of this problem with an earlier
attempt (with adifferent factorization):
http://groups.google.com/groups?selm=9p2ada%241kva%241%40agate.berkeley.edu
-- 
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
 > For instance 2 is a gaussian integer, as well as just an integer.  It
 > is also an algebraic integer.  However, consider the following
 > equation:
 >   x^2 = 2
 > which is outside of the ring of gaussian integers.
Yup.
 > Now sqrt(2) is well-known *today* so I think that's a good example for
 > how a ring can be limited.
 > 
 > What I've found is similar to that, in that I've used a
 > *decomposition* to show numbers outside of the ring of algebraic
 > integers.
No, you have not shown that.  It has been known already a long time
that the algebraic integers are a subset of the algebraic numbers.
Moreover, there are also numbers that are not algebraic numbers.
This is neither new nor revolutionary.
 > Once I had my result showing that the ring of algebraic integers, like
 > that of gaussian integers, and of integers before them was still too
 > small, I found a definition for a fully inclusive ring: the uber ring,
 > which I call the Object Ring or object ring.
 > 
 > The Object Ring is a commutative ring that includes all numbers such
 > that -1 and 1 are the only members that are both a unit and an
 > integer, where no non-unit member is a factor of any two integers that
 > are coprime.
 > 
 > That definition isolates the key property of the numbers in question,
 > and includes integer, gaussian integers, algebraic integers, and
 > beyond.
But the definition is also incomplete.  How do I determine whether a
particular number is element of the ring or not?
 > Now mathematicians as a group apparently were unaware that the ring of
 > algebraic integers was so limited, and proceeded for quite some time
 > assuming that they'd found the most inclusive ring, which is the
 > error.
That is your error.
 > So, in case you're wondering, no, I don't think the definition for
 > algebraic integers needs to be changed any more than the definition
 > for gaussian integers needs to be changed.  What's needed is a
 > recognition of the limitations of the ring.
That has alread been recognised long ago.  So why do it again?
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> The Object Ring is a commutative ring that includes all numbers such
> that -1 and 1 are the only members that are both a unit and an
> integer, where no non-unit member is a factor of any two integers that
> are coprime.
> But the definition is also incomplete.  How do I determine whether a
> particular number is element of the ring or not?
If it is both a unit and an integer, then it has to be 1 or -1; if it is
not a unit, then for any pair of coprime integers it can not be a factor
of both.
Victor who has no idea what he has just written
-- 
homepage: cs utk edu tilde lastname
====
 > 
 >The Object Ring is a commutative ring that includes all numbers 
such
 >that -1 and 1 are the only members that are both a unit and an
 >integer, where no non-unit member is a factor of any two integers 
that
 >are coprime.
 > 
 >But the definition is also incomplete.  How do I determine whether a
 >particular number is element of the ring or not?
 > 
 > If it is both a unit and an integer, then it has to be 1 or -1; if it is
 > not a unit, then for any pair of coprime integers it can not be a factor
 > of both.
Whether something is a unit depends on the ring where you are working in.
So, you can only know whether a number is a unit when you know the complete
set of numbers that are in the ring.  To be more precise.  The ring of
algebraic integers satisfy the definition.  Given an arbitrary algebraic
number not in that ring, can we add it without contradiction?  Within the
ring of algebraic numbers it is a unit, but will it be so in the new ring
when we adjoin it to that ring?
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> 
>The Object Ring is a commutative ring that includes all numbers such
>that -1 and 1 are the only members that are both a unit and an
>integer, where no non-unit member is a factor of any two integers 
that
>are coprime.
>  
>But the definition is also incomplete.  How do I determine whether a
>particular number is element of the ring or not?
> 
> If it is both a unit and an integer, then it has to be 1 or -1; if it is
> not a unit, then for any pair of coprime integers it can not be a factor
> of both.
A key thing here is to realize that the sets called integers, gaussian
integers, and algebraic integers are ALL distinguished by the fact
that -1 and 1 are the only integers that are units in them.
In stating that I've simply abstracted out a key defining property.
James Harris
====
> For instance 2 is a gaussian integer, as well as just an integer.  It
> is also an algebraic integer.  However, consider the following
> equation:
>   x^2 = 2
> which is outside of the ring of gaussian integers.
> 
> Yup.
> 
> Now sqrt(2) is well-known *today* so I think that's a good example for
> how a ring can be limited.
> 
> What I've found is similar to that, in that I've used a
> *decomposition* to show numbers outside of the ring of algebraic
> integers.
> 
> No, you have not shown that.  It has been known already a long time
> that the algebraic integers are a subset of the algebraic numbers.
> Moreover, there are also numbers that are not algebraic numbers.
> This is neither new nor revolutionary.
Ah, but algebraic numbers are a *field* while the ring of algebraic
integers is not!!!
 
> Once I had my result showing that the ring of algebraic integers, like
> that of gaussian integers, and of integers before them was still too
> small, I found a definition for a fully inclusive ring: the uber ring,
> which I call the Object Ring or object ring.
> 
> The Object Ring is a commutative ring that includes all numbers such
> that -1 and 1 are the only members that are both a unit and an
> integer, where no non-unit member is a factor of any two integers that
> are coprime.
> 
> That definition isolates the key property of the numbers in question,
> and includes integer, gaussian integers, algebraic integers, and
> beyond.
> 
> But the definition is also incomplete.  How do I determine whether a
> particular number is element of the ring or not?
That's your problem.  The set exists and is not empty as it includes
the ring of integers.
 
> Now mathematicians as a group apparently were unaware that the ring of
> algebraic integers was so limited, and proceeded for quite some time
> assuming that they'd found the most inclusive ring, which is the
> error.
> 
> That is your error.
Nope.
 
> So, in case you're wondering, no, I don't think the definition for
> algebraic integers needs to be changed any more than the definition
> for gaussian integers needs to be changed.  What's needed is a
> recognition of the limitations of the ring.
> 
> That has alread been recognised long ago.  So why do it again?
Knowledge is a good thing Dik Winter.
James Harris
====
 >For instance 2 is a gaussian integer, as well as just an integer.  
It
 >is also an algebraic integer.  However, consider the following
 >equation:
 >  x^2 = 2
 >which is outside of the ring of gaussian integers.
 >
 >Yup.
 >
 >Now sqrt(2) is well-known *today* so I think that's a good example 
for
 >how a ring can be limited.
 >
 >What I've found is similar to that, in that I've used a
 >*decomposition* to show numbers outside of the ring of algebraic
 >integers.
 >
 >No, you have not shown that.  It has been known already a long time
 >that the algebraic integers are a subset of the algebraic numbers.
 >Moreover, there are also numbers that are not algebraic numbers.
 >This is neither new nor revolutionary.
 > 
 > Ah, but algebraic numbers are a *field* while the ring of algebraic
 > integers is not!!!
Ok.  Adjoin 1/2 to the ring of algebraic integers.  You get a new ring that
is also not a field.  Also already known a long time.  There are an
infinite number of rings that are not fields between the algebraic
integers and the algebraic numbers.  So what you state is neither new
nor revolutionary.
 >The Object Ring is a commutative ring that includes all numbers 
such
 >that -1 and 1 are the only members that are both a unit and an
 >integer, where no non-unit member is a factor of any two integers 
that
 >are coprime.
...
 >But the definition is also incomplete.  How do I determine whether a
 >particular number is element of the ring or not?
 > 
 > That's your problem.  The set exists and is not empty as it includes
 > the ring of integers.
The largest ring I can find that satisfies your definition is the ring
of algebraic integers.
 >Now mathematicians as a group apparently were unaware that the ring 
of
 >algebraic integers was so limited, and proceeded for quite some 
time
 >assuming that they'd found the most inclusive ring, which is the
 >error.
 >
 >That is your error.
 > 
 > Nope.
Yup.
 >So, in case you're wondering, no, I don't think the definition for
 >algebraic integers needs to be changed any more than the definition
 >for gaussian integers needs to be changed.  What's needed is a
 >recognition of the limitations of the ring.
 >
 >That has alread been recognised long ago.  So why do it again?
 > 
 > Knowledge is a good thing Dik Winter.
Yes, let me remind you.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
> Knowledge is a good thing Dik Winter.
> 
> 
> James Harris
Then why do you resist so strongly gaining more of it?
====
>But the definition is also incomplete.  How do I determine whether a
>particular number is element of the ring or not?
> 
> That's your problem.  The set exists and is not empty as it includes
> the ring of integers.
>  
Not so! A set must have its membership unambiguously defined.
Until you can at least show that membership in this vague object of 
yours is unambiguous, you should not call it a set.
====
>>  [...]
> 
>> Now mathematicians as a group apparently were unaware that the ring 
of
>> algebraic integers was so limited, and proceeded for quite some time
>> assuming that they'd found the most inclusive ring, which is the
>> error.
>> 
>> That is your error.
Nope.
Nope? Sounds like you're assuming that people are going to believe
happen, because so many of the things you say turn out to be false.
Give us some _evidence_ in support of your assertion that until you
came along mathematicians assumed that the algebraic integers
were the most inclusive ring.
 
>> So, in case you're wondering, no, I don't think the definition for
>> algebraic integers needs to be changed any more than the definition
>> for gaussian integers needs to be changed.  What's needed is a
>> recognition of the limitations of the ring.
>> 
>> That has alread been recognised long ago.  So why do it again?
Knowledge is a good thing Dik Winter.

>James Harris
************************
David C. Ullrich
====
> 
>>  [...]
>  
>> Now mathematicians as a group apparently were unaware that the ring 
of
>> algebraic integers was so limited, and proceeded for quite some 
time
>> assuming that they'd found the most inclusive ring, which is the
>> error.
>> 
>> That is your error.
Nope.
> 
> Nope? Sounds like you're assuming that people are going to believe
> happen, because so many of the things you say turn out to be false.
No, I simply don't find it worth my time to go explain in a lot of
detail every time some poster makes a false statement, as it happens a
lot.
 
James Harris
====
>> 
>  [...]
>>  
> Now mathematicians as a group apparently were unaware that the ring 
of
> algebraic integers was so limited, and proceeded for quite some 
time
> assuming that they'd found the most inclusive ring, which is the
> error.
> 
> That is your error.
>>Nope.
>> 
>> Nope? Sounds like you're assuming that people are going to believe
>> happen, because so many of the things you say turn out to be false.
No, I simply don't find it worth my time to go explain in a lot of
>detail every time some poster makes a false statement, as it happens a
>lot.
But people have asked you this question _many_ times, and you've
never explained, not even once.
(Which of course is no surprise, since what you're saying about what
mathematicians thought is simply nonsense.)
>James Harris
************************
David C. Ullrich
====
> I've talked of an error in core mathematics that comes about from
> the limitation of the ring of algebraic integers, and thinking back on
> discussions, I think there's been a lot of confusion on just what I
> mean.
> 
> Context helps so I'll mention that the idea of different types of
> integers includes Gauss considering numbers of the form a+bi, where a,
> and b are integers, which are called gaussian integers in his honor. 
> Later there are algebraic integers, which include gaussian integers,
> but they are defined to be the roots of monic polynomials with integer
> coefficients.
> 
> The problem then is that mathematicians thought they were done, but my
> result shows they are not.  Understanding how that's possible isn't
> really difficult, and an easy way to see it, is to consider gaussian
> integers and algebraic integers again.
> 
> For instance 2 is a gaussian integer, as well as just an integer.  It
> is also an algebraic integer.  However, consider the following
> equation:
> 
> x^2 = 2
> 
> which is outside of the ring of gaussian integers.
> 
> Now sqrt(2) is well-known *today* so I think that's a good example for
> how a ring can be limited.
> 
> What I've found is similar to that, in that I've used a
> *decomposition* to show numbers outside of the ring of algebraic
> integers.
> 
> With my example here x^2 = 2, the decomposition is of 2 into equal
> factors.
> 
> While 2 is an integer, and a gaussian integer, that decomposition
> leads to a result that's neither, though it's an algebraic integer.
> 
> I use a polynomial decomposition to show the limitation of the ring of
> algebraic integers.
> 
> Once I had my result showing that the ring of algebraic integers, like
> that of gaussian integers, and of integers before them was still too
> small, I found a definition for a fully inclusive ring: the uber ring,
> which I call the Object Ring or object ring.
> 
> The Object Ring is a commutative ring that includes all numbers such
> that -1 and 1 are the only members that are both a unit and an
> integer, where no non-unit member is a factor of any two integers that
> are coprime.
I don't think this works as a definition.  The Object Ring is a set
of numbers, OK, that's a start.  The definition is finished when, by
using it one can know whether or not a particular number is a member of
the Object Ring.  So, one wants a sentence something like this: the
object ring is the set of all numbers z such that ...(something about
z).  The 'something' might be, either z is an algebraic integer, or
else it ... (some other condition).
To be careful one would prove that the resulting set really is a ring,
that is, the sum and product of to elements are also in the set.
I am uneasy with the apparent assumption that only 1 and -1 are both
units and integers in this ring.  sqrt(2)+1 and sqrt(2)-1 are algebraic
numbers that are units in the ring of algebraic integers.  (Note that
their product is 1.)  Bringing more numbers into the ring cannot
destroy that property.  But perhaps I am misreading the above
statement.
> 
> That definition isolates the key property of the numbers in question,
> and includes integer, gaussian integers, algebraic integers, and
> beyond.
> 
> Now mathematicians as a group apparently were unaware that the ring of
> algebraic integers was so limited, and proceeded for quite some time
> assuming that they'd found the most inclusive ring, which is the
> error.
> 
> So, in case you're wondering, no, I don't think the definition for
> algebraic integers needs to be changed any more than the definition
> for gaussian integers needs to be changed.  What's needed is a
> recognition of the limitations of the ring.
> 
> Want more?  Then go to my blog archives:
> 
> 
> 
> James Harris
====
 
<deleted
>The Object Ring is a commutative ring that includes all numbers such
>that -1 and 1 are the only members that are both a unit and an
>integer, where no non-unit member is a factor of any two integers that
>are coprime.
> 
> I don't think this works as a definition.  The Object Ring is a set
> of numbers, OK, that's a start.  The definition is finished when, by
> using it one can know whether or not a particular number is a member of
> the Object Ring.  So, one wants a sentence something like this: the
> object ring is the set of all numbers z such that ...(something about
> z).  The 'something' might be, either z is an algebraic integer, or
> else it ... (some other condition).
> 
> To be careful one would prove that the resulting set really is a ring,
> that is, the sum and product of to elements are also in the set.
> 
> I am uneasy with the apparent assumption that only 1 and -1 are both
> units and integers in this ring.  sqrt(2)+1 and sqrt(2)-1 are algebraic
> numbers that are units in the ring of algebraic integers.  (Note that
> their product is 1.)  Bringing more numbers into the ring cannot
> destroy that property.  But perhaps I am misreading the above
> statement.
 
If you have 1/2 in the ring, then isn't 2 then a unit?
Understand?
What I did was sit down and figure out that what differentiates rings
like integers, gaussian integers, and algebraic integers from other
rings like rings that are fields is the fact that the only integer
units in the ring are -1 and 1.
I also have where no non-unit member is a factor of any two integers
that are coprime, as that's also interesting in its own right, and
it's another differentiating property.
Now then, it may seem like a subtle definition, but think about it for
a while.
James Harris
====
...
 > What I did was sit down and figure out that what differentiates rings
 > like integers, gaussian integers, and algebraic integers from other
 > rings like rings that are fields is the fact that the only integer
 > units in the ring are -1 and 1.
I do not know how exactly you differentiate the first rings you mention
from rings that are fields.  The ring you get when you add 1/2 to the
ring of integers is *not* a field.  It is simply a ring where 2 is a unit.
 > I also have where no non-unit member is a factor of any two integers
 > that are coprime, as that's also interesting in its own right, and
 > it's another differentiating property.
Eh?  What do you mean?  Do you mean coprime in the integers?  Or something
else?  Coprimeness depends on the ring where you are working in.
 > Now then, it may seem like a subtle definition, but think about it for
 > a while.
I think we have thought about it for at least a year, it is not yet 
clearer.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
   [.snip.]
>Eh?  What do you mean?  Do you mean coprime in the integers?  Or something
>else?  Coprimeness depends on the ring where you are working in.
Remember that the notion being used is:
(*) Let R be a ring; x and y are 'coprime in R' if and only if any
     common factor of x and y in R is a unit in R.
This notion is ring dependent, and is not inherited to subrings or to
overrings. 
The property given is vacuous (i.e., always holds). If you assume that
coprimeness if being defined with respect to the ring in question, R,
then by definition, any factor of two elements coprime in R will
necessarily be a unit, in particular for integers.
If you assume that coprime is in some intermediate ring, then in
conjunction with R intersect Q is equal to Z, you get that the
property is also vacuous.
PROP. Let S be a subring of C which intersects Q in Z, and let x and y
be two integers. If x and y are coprime in the sense of (*) in any
subring of S, then they are coprime in Z; therefore, there exist r and
s in Z such that xr+ys = 1, and x and y are coprime in the sense of
(*) in any ring containing the integers.
Proof. Assume that x and y are not coprime in Z. Then there is a
rational prime p which divides x and y in Z, and therefore in S. Since
x and y are assumed coprime in S, it follows that p must be a unit in
S. But that implies that 1/p lies in S intersect Q, which is
impossible. Therefore, x and y are coprime in Z in the sense of
(*). It is well known that condition (*) in Z implies the existence of
r and s. If R is ANY ring containing the integers, and u is an element
of R that divides both x and y in R, then x=ua, y=ub, and 
1 = xr+ys = uar + ubs = u(ar+bs),
and a,r,b,s are in R, so u is a unit in R, proving that x and y are
coprime in R in the sense of (*).
So the condition on coprimeness is completely vacuous.
-- 
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
> I don't think this works as a definition.  The Object Ring is a set
> of numbers, OK, that's a start.  The definition is finished when, by
> using it one can know whether or not a particular number is a member of
> the Object Ring.  So, one wants a sentence something like this: the
> object ring is the set of all numbers z such that ...(something about
> z).  The 'something' might be, either z is an algebraic integer, or
> else it ... (some other condition).
Both you and Dik mention that a definition ought to give a rule for
determining whether a particular number is in the set or not.  I don't
see why this is so.
I'm not defending James's definition here, but I don't see why a
definition should necessarily yield principles for determining
membership.  I don't have any examples at hand, but in principle, if I
can prove that there is a unique set satisfying some particular
property, then that property is suitable for a definiens[1], whether
or not there is an easily applicable rule determining membership.
Attack James's definition where one should: It's not at all clear
whether there is a unique set satisfying his property.  Don't make up
rules about what definitions must satisfy (like feasible membership
tests).
Footnotes: 
[1]  Golly, I hope I use that term correctly.  I got a 50/50 chance.
-- 
Jesse F. Hughes
[Lancelot] sighed, defeated. 'It is as practical to hurry an acorn
toward treeness as to urge a damsel when her mind is set.'
====
On Thu, 01 Jan 2004 15:16:57 +0100, jesse@phiwumbda.org (Jesse F.
> I don't think this works as a definition.  The Object Ring is a set
>> of numbers, OK, that's a start.  The definition is finished when, by
>> using it one can know whether or not a particular number is a member of
>> the Object Ring.  So, one wants a sentence something like this: the
>> object ring is the set of all numbers z such that ...(something about
>> z).  The 'something' might be, either z is an algebraic integer, or
>> else it ... (some other condition).

>Both you and Dik mention that a definition ought to give a rule for
>determining whether a particular number is in the set or not.  I don't
>see why this is so.
I'm not defending James's definition here, but I don't see why a
>definition should necessarily yield principles for determining
>membership.  I don't have any examples at hand, but in principle, if I
>can prove that there is a unique set satisfying some particular
>property, then that property is suitable for a definiens[1], whether
>or not there is an easily applicable rule determining membership.
Attack James's definition where one should: It's not at all clear
>whether there is a unique set satisfying his property.  Don't make up
>rules about what definitions must satisfy (like feasible membership
>tests).
Hmm. Since a set is determined by its elements there's a definite
sense in which defining a set _is_ the same as specifying what
the elements are. In a sense - this doesn't say anything about
a feasible test for membership...
So it's not so clear to me whether you have a point or not.
Would be much more compelling if you _had_ an example
in mind where a set is defined in a way that does not in
some sense give a test for membership.
>Footnotes: 
>[1]  Golly, I hope I use that term correctly.  I got a 50/50 chance.
************************
David C. Ullrich
====
On Fri, 02 Jan 2004 03:10:18 +0100, jesse@phiwumbda.org (Jesse F.
> Hmm. Since a set is determined by its elements there's a definite
>> sense in which defining a set _is_ the same as specifying what
>> the elements are. In a sense - this doesn't say anything about
>> a feasible test for membership...
>> So it's not so clear to me whether you have a point or not.
>> Would be much more compelling if you _had_ an example
>> in mind where a set is defined in a way that does not in
>> some sense give a test for membership.
Yes, it would be more compelling if I had an example.  I expect that a
>cleverer lad than I am could toss off a fixed point construction
>fairly easily in which determining whether a particular guy is an
>element of the, say, greatest fixed point is not an easy task.  If the
>construction also requires an application of the axiom of choice, one
>could imagine that the task isn't really feasible in any reasonable
>sense.
Feasibility has nothing to do with it! (Where it is the it I've been
talking about.) Say I define S = {1} if Goldbach's conjecture is
true and S = {2} if Goldbach's conjecture is false. There is no
feasible test for membership in S. I nonetheless _have_ specified
the members of S, and given a test for membership: 1 is a 
member if and only if every even number > 2 is the sum of
two primes.
>Too bad I'm not a cleverer lad than I am.  
************************
David C. Ullrich
 <87vfnvwy7a.fsf@phiwumbda.org>
 <1rh8vvsdekd46h57vq7iro8a5f9485k7kl@4ax.com>
 <874qvft81h.fsf@phiwumbda.org>
 <58ravvguog6ipaf9n8lsqsg5jqh3ev73qn@4ax.com>
====
> On Fri, 02 Jan 2004 03:10:18 +0100, jesse@phiwumbda.org (Jesse F.
 Hmm. Since a set is determined by its elements there's a definite
> sense in which defining a set _is_ the same as specifying what
> the elements are. In a sense - this doesn't say anything about
> a feasible test for membership...
 So it's not so clear to me whether you have a point or not.
> Would be much more compelling if you _had_ an example
> in mind where a set is defined in a way that does not in
> some sense give a test for membership.
>>Yes, it would be more compelling if I had an example.  I expect that a
>>cleverer lad than I am could toss off a fixed point construction
>>fairly easily in which determining whether a particular guy is an
>>element of the, say, greatest fixed point is not an easy task.  If the
>>construction also requires an application of the axiom of choice, one
>>could imagine that the task isn't really feasible in any reasonable
>>sense.
 Feasibility has nothing to do with it! (Where it is the it I've been
> talking about.) Say I define S = {1} if Goldbach's conjecture is
> true and S = {2} if Goldbach's conjecture is false. There is no
> feasible test for membership in S. I nonetheless _have_ specified
> the members of S, and given a test for membership: 1 is a 
> member if and only if every even number > 2 is the sum of
> two primes.
Are you sure we're disagreeing?
James's definition is not a legitimate definition because there is no
proof that a unique structure satisfies the definition.  Elementhood
tests (feasible or in principle or whatever) have nothing to do with
it.
I interpreted the elementhood complaint in terms of feasible tests,
just because I can't figure out what the complaint is supposed to mean
otherwise. 
-- 
So, at this time, I'd like to assure you that I am not interested in
I'll have prosecutors knocking on your doors.  I have no problem with
====
 > On Thu, 01 Jan 2004 15:16:57 +0100, jesse@phiwumbda.org (Jesse F.
...
 >Both you and Dik mention that a definition ought to give a rule for
 >determining whether a particular number is in the set or not.  I don't
 >see why this is so.
...
 > Hmm. Since a set is determined by its elements there's a definite
 > sense in which defining a set _is_ the same as specifying what
 > the elements are. In a sense - this doesn't say anything about
 > a feasible test for membership...
 > 
 > So it's not so clear to me whether you have a point or not.
 > Would be much more compelling if you _had_ an example
 > in mind where a set is defined in a way that does not in
 > some sense give a test for membership.
I think Jesse is right.  To define a set it is not necessarily true that
you need a membership test.  On the other hand it should be clear that
either an element is in the set or is not, and that should *not* be
dependent on what is already put in the set or not.  Also some
consistency checks are needed.
An example: the set of TM's that halt is (I think) a well-defined set.
There is not a clear membership test (unless you have infinite time ;-)).
On the other hand, it is not the case that you can put TM-1 in the set
if and only if TM-2 is not in the set.  (It is the case that you can
put TM-1 in the set if and only if TM-2 is in the set, but that is no
problem.)
Compare James' ring (which has more structure than a set in itself).
that in a number of cases two conjugate complex numbers can not go
together in the ring, but that one of them should go in it.  There is
no way to show which one should go in, nor is there a way to show that
your choice conflicts with other choices you make, or not, until you
may have to make a third choice that creates a conflicting situation.
But even with his current requirements (that a number of algebraic
integers divided by 7 go into the ring) the definition can not be shown
to be non-conflicting.  (And that only to show FLT for p=3...)
There actually *is* a definition, yes.  But what it entails is unclear.
And my thinking is that to clarify that is just as difficult, if not
more difficult, than proving FLT.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
>On Thu, 01 Jan 2004 15:16:57 +0100, jesse@phiwumbda.org (Jesse F.
>...
>>Both you and Dik mention that a definition ought to give a rule for
>>determining whether a particular number is in the set or not.  I don't
>>see why this is so.
>...
>Hmm. Since a set is determined by its elements there's a definite
>sense in which defining a set _is_ the same as specifying what
>the elements are. In a sense - this doesn't say anything about
>a feasible test for membership...
So it's not so clear to me whether you have a point or not.
>Would be much more compelling if you _had_ an example
>in mind where a set is defined in a way that does not in
>some sense give a test for membership.
I think Jesse is right.  To define a set it is not necessarily true that
>you need a membership test.  On the other hand it should be clear that
>either an element is in the set or is not, and that should *not* be
>dependent on what is already put in the set or not.  Also some
>consistency checks are needed.
An example: the set of TM's that halt is (I think) a well-defined set.
>There is not a clear membership test (unless you have infinite time ;-)).
There's not a membership test that one can execute; there's no
algorithmic membership test. There certainly is a membership
test in the abstract mathematical sense: If it halts it's in, 
otherwise it's out.
The point is that whether or not we feel that membership test
is the best word for this (come to think of it membership
criterion would be much better) there is no membership test
in even this sense given by the definition of the Object Ring.
Or if there is I've never seen anyone state coherently what it is.
>On the other hand, it is not the case that you can put TM-1 in the set
>if and only if TM-2 is not in the set.  (It is the case that you can
>put TM-1 in the set if and only if TM-2 is in the set, but that is no
>problem.)
Compare James' ring (which has more structure than a set in itself).
>that in a number of cases two conjugate complex numbers can not go
>together in the ring, but that one of them should go in it.  There is
>no way to show which one should go in, nor is there a way to show that
>your choice conflicts with other choices you make, or not, until you
>may have to make a third choice that creates a conflicting situation.
But even with his current requirements (that a number of algebraic
>integers divided by 7 go into the ring) the definition can not be shown
>to be non-conflicting.  (And that only to show FLT for p=3...)
There actually *is* a definition, yes.  But what it entails is unclear.
>And my thinking is that to clarify that is just as difficult, if not
>more difficult, than proving FLT.
************************
David C. Ullrich
 <1rh8vvsdekd46h57vq7iro8a5f9485k7kl@4ax.com> <HquBos.F3o@cwi.nl>
====
> To define a set it is not necessarily true that you need a
> membership test.  On the other hand it should be clear that either
> an element is in the set or is not, and that should *not* be
> dependent on what is already put in the set or not.
I was going to agree with this comment, but I got to thinking about it
a bit.
I suppose whether or not I agree depends on what we count as a
definition.  The Cantor-Bernstein(-Schroeder?) theorem constructs a
bijection in which subsequent choices depend on previous choices.  It
is perhaps just a semantic quibble whether the proof of the theorem
could be said to *define* a bijection (by, say, appending it with the
statement, Let Gigglywiggly be the bijection thus constructed).  I
could certainly sympathize if someone wanted to object that this isn't
really a definition, but I wouldn't be confident in averring one way
or the other.
-- 
Run mathematicians, RUN!!!  I'm coming for you.  It may take a few
months, but I'll get [computer verification of my proof] and then your
lives will be ended as you previously knew it. -- JSH meets PVS
====
 > 
 >To define a set it is not necessarily true that you need a
 >membership test.  On the other hand it should be clear that either
 >an element is in the set or is not, and that should *not* be
 >dependent on what is already put in the set or not.
 > 
 > I was going to agree with this comment, but I got to thinking about it
 > a bit.
 > 
 > I suppose whether or not I agree depends on what we count as a
 > definition.  The Cantor-Bernstein(-Schroeder?) theorem constructs a
 > bijection in which subsequent choices depend on previous choices.
That is something different, I think.  (But I also think the Axiom of
Choice is leering behind ;-).)  You may need subsequent choices, and
I think it is valid *when you do not need backtracking*.  That is,
at every point you have to chose, your choice will not invalidate
all other possible choices at some future point.  With James' deinition
we may arrive at a position where we are stuck.  Do we now have the
Object ring?  With other earlier choices we might have come at another
point.  With the theorem you cite you need only a bijection.
-- 
dik t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, 
+31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~dik/
====
|I suppose whether or not I agree depends on what we count as a
|definition.  The Cantor-Bernstein(-Schroeder?) theorem constructs a
|bijection in which subsequent choices depend on previous choices.  It
|is perhaps just a semantic quibble whether the proof of the theorem
|could be said to *define* a bijection (by, say, appending it with the
|statement, Let Gigglywiggly be the bijection thus constructed).  I
|could certainly sympathize if someone wanted to object that this isn't
|really a definition, but I wouldn't be confident in averring one way
|or the other.
I don't think Cantor-Bernstein would normally be considered at all
problematic. I don't know what you mean by subsequent choices
depend on previous choices.
We assume that there exist one-to-one functions f:X->Y and g:Y->X.
Assume X and Y are disjoint. We then define a bijection between X
and Y. Start by considering the transitive closure of the relation on
the union of X and Y which is the union of f and g, i.e., contains
(x,f(x)) for every x and (g(y),y) for every y.
The equivalence classes under that relation are only of a few different
general types. For instance, one might have an element x which is
not in the image of y, and the sequence
   x, f(x), g(f(x)), f(g(f(x))),...
would be one equivalence class. Or we might have a cycle, or an
infinite chain
  ..., x_-1, y_-1, x_0, y_0, x_1, y_1, ...
where each x_{i+1}=g(y_i) and y_i=f(x_i). In each case, we define a
one-to-one correspondence on the equivalence class which matches
the elements in Y with the elements in X. For the first example above,
it makes sense to let x<->f(x), g(f(x))<->f(g(f(x))) and so on. There's
a little bit of arbitrariness, in that in infinite chains or cycles, one 
might
equally well want to associate each element of X with the element of
Y coming after it, or vice-versa. But this is just one arbitrary choice
we make once and for all. Having decided, once for all, the one-to-one
correspondence is entirely explicit.
Note in particular that Cantor-Bernstein is a theorem of ZF, not needing
the axiom of choice.
Keith Ramsay
P.S. When I was making sure I correctly remembered which theorem
was known as Cantor-Bernstein, I got a hit for Cantor Bernstein at
the site www.bethhillel.com. Dr. Bernstein serves as their cantor.
Cantor Bernstein was formerly a full-time professional musician....
====
> |I suppose whether or not I agree depends on what we count as a
> |definition.  The Cantor-Bernstein(-Schroeder?) theorem constructs a
> |bijection in which subsequent choices depend on previous choices.  It
> |is perhaps just a semantic quibble whether the proof of the theorem
> |could be said to *define* a bijection (by, say, appending it with the
> |statement, Let Gigglywiggly be the bijection thus constructed).  I
> |could certainly sympathize if someone wanted to object that this isn't
> |really a definition, but I wouldn't be confident in averring one way
> |or the other.
 I don't think Cantor-Bernstein would normally be considered at all
> problematic. I don't know what you mean by subsequent choices
> depend on previous choices.
 We assume that there exist one-to-one functions f:X->Y and g:Y->X.
> Assume X and Y are disjoint. We then define a bijection between X
> and Y. Start by considering the transitive closure of the relation on
> the union of X and Y which is the union of f and g, i.e., contains
> (x,f(x)) for every x and (g(y),y) for every y.
 The equivalence classes under that relation are only of a few different
> general types. For instance, one might have an element x which is
> not in the image of y, and the sequence
    x, f(x), g(f(x)), f(g(f(x))),...
 would be one equivalence class. Or we might have a cycle, or an
> infinite chain
   ..., x_-1, y_-1, x_0, y_0, x_1, y_1, ...
 where each x_{i+1}=g(y_i) and y_i=f(x_i). In each case, we define a
> one-to-one correspondence on the equivalence class which matches
> the elements in Y with the elements in X. For the first example above,
> it makes sense to let x<->f(x), g(f(x))<->f(g(f(x))) and so on. There's
> a little bit of arbitrariness, in that in infinite chains or cycles, one 
might
> equally well want to associate each element of X with the element of
> Y coming after it, or vice-versa. But this is just one arbitrary choice
> we make once and for all. Having decided, once for all, the one-to-one
> correspondence is entirely explicit.
 Note in particular that Cantor-Bernstein is a theorem of ZF, not needing
> the axiom of choice.
the correction.
Well, maybe I should've used the well-ordering theorem as an example,
but it seems maybe less plausible that someone would claim that the
proof of that theorem defines a well-ordering.
noggin.
-- 
I AM serious about this being a short route to a Ph.d for some of
you, but just remember, I'm the guy who proved Fermat's Last Theorem
in just a bit over 6 years [...]  My standards are kind of high.
                  --James Harris, founding a new mathematical school
====
   [.snip.]
>There actually *is* a definition, yes.
You mean, there are ways of interpreting what is written so as to make
it something which is a definition. 
>  But what it entails is unclear.
>And my thinking is that to clarify that is just as difficult, if not
>more difficult, than proving FLT.
Well, does the set of subrings R of C [I assume they contain 1, hence
Z] which satisfy R intersect Q is equal to Z satisfy Zorn's Lemma?
Note that the second condition given is vacuous, since two integers
a,b are coprime (in the sense of having no common nonunit divisors in
the integers) if and only if there exist integers r and s such that
ra+sb=1, so any common factor of a and b in any ring will necessarily
be a unit.
Let S = {R contained in C:  Z contained in R, and R intersect Q equals Z}.
The set is trivially nonempty. Order it by inclusion of rings. If C is
a chain in S, then the union of C is a subring of C contained in C,
and if there is an element of the union which lies in Q intersect the
union, then it lies in Q intersect one of the rings in C, hence lies
in Z. 
So S has maximal elements. However, we have already seen that S does
not have a ->maximum<- element, as you noted in your reply; since the
definition does not have a referent. There is no such (uniquely
determined) ring. There are many subrings R of C which satisfy both R
intersect Q is equal to Z and are maximal with respect to inclusion. 
-- 
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
 <87vfnvwy7a.fsf@phiwumbda.org>
 <1rh8vvsdekd46h57vq7iro8a5f9485k7kl@4ax.com>
====
> Hmm. Since a set is determined by its elements there's a definite
> sense in which defining a set _is_ the same as specifying what
> the elements are. In a sense - this doesn't say anything about
> a feasible test for membership...
 So it's not so clear to me whether you have a point or not.
> Would be much more compelling if you _had_ an example
> in mind where a set is defined in a way that does not in
> some sense give a test for membership.
Yes, it would be more compelling if I had an example.  I expect that a
cleverer lad than I am could toss off a fixed point construction
fairly easily in which determining whether a particular guy is an
element of the, say, greatest fixed point is not an easy task.  If the
construction also requires an application of the axiom of choice, one
could imagine that the task isn't really feasible in any reasonable
sense.
Too bad I'm not a cleverer lad than I am.  
-- 
Jesse F. Hughes
What you call reasonable is suspect since you've proven yourself to
be an enemy of mathematics. -- James S. Harris defends the cause.
====
> On Thu, 01 Jan 2004 15:16:57 +0100, jesse@phiwumbda.org (Jesse F.
> 
> I don't think this works as a definition.  The Object Ring is a 
set
>> of numbers, OK, that's a start.  The definition is finished when, by
>> using it one can know whether or not a particular number is a member 
of
>> the Object Ring.  So, one wants a sentence something like this: the
>> object ring is the set of all numbers z such that ...(something about
>> z).  The 'something' might be, either z is an algebraic integer, 
or
>> else it ... (some other condition).

>Both you and Dik mention that a definition ought to give a rule for
>determining whether a particular number is in the set or not.  I don't
>see why this is so.
I'm not defending James's definition here, but I don't see why a
>definition should necessarily yield principles for determining
>membership.  I don't have any examples at hand, but in principle, if I
>can prove that there is a unique set satisfying some particular
>property, then that property is suitable for a definiens[1], whether
>or not there is an easily applicable rule determining membership.
Attack James's definition where one should: It's not at all clear
>whether there is a unique set satisfying his property.  Don't make up
>rules about what definitions must satisfy (like feasible membership
>tests).
> 
> Hmm. Since a set is determined by its elements there's a definite
> sense in which defining a set _is_ the same as specifying what
> the elements are. In a sense - this doesn't say anything about
> a feasible test for membership...
> 
> So it's not so clear to me whether you have a point or not.
> Would be much more compelling if you _had_ an example
> in mind where a set is defined in a way that does not in
> some sense give a test for membership.
Let Z be the ring of integers. Consider the following example.
The token ring is a commutative ring that includes numbers of Z
where any two members commute.
Then, I can't determine whether 30 and 16 are members of the
token ring.
This may not be an entirely satisfactory example.
Recall Harris' definition:
The Object Ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime.
I assume that Harris is taking numbers to mean complex numbers.
The word all makes the definition tricky. I am going to interpret
it to mean a maximal ring satisfying the stated conditions, where
there could be several such maximal rings within the complex numbers C.
Note that existence is not needed for a definition to be ok.
I don't require that such an object ring actually exists.
The condition for x to be a member of the object ring does not
depend upon properties of x, but rather on properties of the
object ring set itself. That is, the definition is choosing
certain subrings of the ring of complex numbers to have the
name of object ring.
But, even considering all that, I guess you could give the
following properties for determining whether x is a member
or not. Let x be a complex number. Find a subring S of C
such x is in S, Z intersect units of S is {1, -1}, there
are no non-unit member of S that is a factor of any
two integers that coprime, and there does not exist
a subring T of C that properly contains S, where Z
intersect units of T is {1, -1}, and there are no
non-unit member of T that is a factor of any two
integers that are coprime. If no such subring S can
be found, then x is not a member of *an* object ring.
Of course all elements might be members of an object ring.
Thus, the question is not really whether an element is
a member of an object ring. The question is what are
the object rings themselves.
In summary, the definition of object ring characterizes
certain subrings of C. It does not characterize directly
certain elements of C. Thus, it differs from the
definition of algebraic integer, which does characterize
certain elements of C. Hence, it is not required that
there be a membership test for an element of C to be
an object ring element.
-- Bill Hale
====
On Thu, 01 Jan 2004 12:20:35 -0600, hale@tulane.edu (William Hale)
> On Thu, 01 Jan 2004 15:16:57 +0100, jesse@phiwumbda.org (Jesse F.
>> 
> I don't think this works as a definition.  The Object Ring is a 
set
> of numbers, OK, that's a start.  The definition is finished when, by
> using it one can know whether or not a particular number is a member 
of
> the Object Ring.  So, one wants a sentence something like this: the
> object ring is the set of all numbers z such that ...(something about
> z).  The 'something' might be, either z is an algebraic integer, 
or
> else it ... (some other condition).
>>Both you and Dik mention that a definition ought to give a rule for
>>determining whether a particular number is in the set or not.  I don't
>>see why this is so.
>>I'm not defending James's definition here, but I don't see why a
>>definition should necessarily yield principles for determining
>>membership.  I don't have any examples at hand, but in principle, if I
>>can prove that there is a unique set satisfying some particular
>>property, then that property is suitable for a definiens[1], whether
>>or not there is an easily applicable rule determining membership.
>>Attack James's definition where one should: It's not at all clear
>>whether there is a unique set satisfying his property.  Don't make up
>>rules about what definitions must satisfy (like feasible membership
>>tests).
>> 
>> Hmm. Since a set is determined by its elements there's a definite
>> sense in which defining a set _is_ the same as specifying what
>> the elements are. In a sense - this doesn't say anything about
>> a feasible test for membership...
>> 
>> So it's not so clear to me whether you have a point or not.
>> Would be much more compelling if you _had_ an example
>> in mind where a set is defined in a way that does not in
>> some sense give a test for membership.
Let Z be the ring of integers. Consider the following example.
The token ring is a commutative ring that includes numbers of Z
>where any two members commute.
Then, I can't determine whether 30 and 16 are members of the
>token ring.
This may not be an entirely satisfactory example.
It's certainly not an example of what Jesse was talking about,
namely a valid definition of a set that does not specify the
elements - your definition of the token ring is no definition
at all, since there is more than one ring satisfying the
given condition.
>Recall Harris' definition:
The Object Ring is a commutative ring that includes all numbers such
>that -1 and 1 are the only members that are both a unit and an
>integer, where no non-unit member is a factor of any two integers that
>are coprime.
I assume that Harris is taking numbers to mean complex numbers.
The word all makes the definition tricky. I am going to interpret
>it to mean a maximal ring satisfying the stated conditions, where
>there could be several such maximal rings within the complex numbers C.
Nobody's ever claimed that it's impossible to give a definition
for the phrase Object Ring - the claim is that what he _says_
the definition is makes no sense.
>Note that existence is not needed for a definition to be ok.
>I don't require that such an object ring actually exists.
The condition for x to be a member of the object ring does not
>depend upon properties of x, but rather on properties of the
>object ring set itself. That is, the definition is choosing
>certain subrings of the ring of complex numbers to have the
>name of object ring.
But, even considering all that, I guess you could give the
>following properties for determining whether x is a member
>or not. Let x be a complex number. Find a subring S of C
>such x is in S, Z intersect units of S is {1, -1}, there
>are no non-unit member of S that is a factor of any
>two integers that coprime, and there does not exist
>a subring T of C that properly contains S, where Z
>intersect units of T is {1, -1}, and there are no
>non-unit member of T that is a factor of any two
>integers that are coprime. If no such subring S can
>be found, then x is not a member of *an* object ring.
Of course all elements might be members of an object ring.
>Thus, the question is not really whether an element is
>a member of an object ring. The question is what are
>the object rings themselves.
In summary, the definition of object ring characterizes
>certain subrings of C. It does not characterize directly
>certain elements of C. Thus, it differs from the
>definition of algebraic integer, which does characterize
>certain elements of C. Hence, it is not required that
>there be a membership test for an element of C to be
>an object ring element.
Except that you're simply _revising_ the definition in
important ways. He talks about _the_ Object Ring,
and he talks about _objects_, defining an object
of object _is_ analogous to the notion of algebraic
integer in this sense. 
You decided for some reason to change this to a 
definition of what it means for a ring to be _an_ object 
ring. So you're no longer defining a set, you're defining
a class of sets. My comments were regarding the
situation where one has defined a set.
I really don't get this stuff about what happens if
we assume he doesn't mean what he says but
means something entirely different. If when he
says there's an error in core mathematics what
he actually means is that 2 + 2 = 4 then yes, what
he means is correct...
>-- Bill Hale
************************
David C. Ullrich
====
>On Thu, 01 Jan 2004 15:16:57 +0100, jesse@phiwumbda.org (Jesse F.
 I don't think this works as a definition.  The Object Ring is a set
> of numbers, OK, that's a start.  The definition is finished when, by
> using it one can know whether or not a particular number is a member of
> the Object Ring.  So, one wants a sentence something like this: the
> object ring is the set of all numbers z such that ...(something about
> z).  The 'something' might be, either z is an algebraic integer, or
> else it ... (some other condition).
>>Both you and Dik mention that a definition ought to give a rule for
>>determining whether a particular number is in the set or not.  I don't
>>see why this is so.
>>I'm not defending James's definition here, but I don't see why a
>>definition should necessarily yield principles for determining
>>membership.  I don't have any examples at hand, but in principle, if I
>>can prove that there is a unique set satisfying some particular
>>property, then that property is suitable for a definiens[1], whether
>>or not there is an easily applicable rule determining membership.
>>Attack James's definition where one should: It's not at all clear
>>whether there is a unique set satisfying his property.  Don't make up
>>rules about what definitions must satisfy (like feasible membership
>>tests).
Hmm. Since a set is determined by its elements there's a definite
>sense in which defining a set _is_ the same as specifying what
>the elements are. In a sense - this doesn't say anything about
>a feasible test for membership...
So it's not so clear to me whether you have a point or not.
>Would be much more compelling if you _had_ an example
>in mind where a set is defined in a way that does not in
>some sense give a test for membership.
>
Having a test for membership is something of a red herring - there is no
test for membership of the field of algebraic numbers, for example.
It seems to me that there are two types of definitions in common
usage. The first is of the type An algebraic number is a complex
number which satisfies a polynomial equation over the integers and
is completely self-sufficient and unambiguous. The second type, like 
The Fitting subgroup of a group G is defined to be the largest normal
nilpotent subgroup of G requires us to prove that there is such a thing.
Provided that we can do that, we do not need to have any means of deciding
whether or not an element of G lies in the Fitting subgroup.
James' definition of an object ring (or whatever) is of the second type,
and could conceivably make sense if he could prove that there was a
unique maximal subring of the complex numbers (?) that had the required
properties. Not much chance of that though, is there?
Derek Holt.
====
On Thu, 1 Jan 2004 17:36:31 +0000 (UTC),
>>On Thu, 01 Jan 2004 15:16:57 +0100, jesse@phiwumbda.org (Jesse F.
> I don't think this works as a definition.  The Object Ring is a 
set
>> of numbers, OK, that's a start.  The definition is finished when, by
>> using it one can know whether or not a particular number is a member 
of
>> the Object Ring.  So, one wants a sentence something like this: the
>> object ring is the set of all numbers z such that ...(something about
>> z).  The 'something' might be, either z is an algebraic integer, 
or
>> else it ... (some other condition).

>Both you and Dik mention that a definition ought to give a rule for
>determining whether a particular number is in the set or not.  I don't
>see why this is so.
I'm not defending James's definition here, but I don't see why a
>definition should necessarily yield principles for determining
>membership.  I don't have any examples at hand, but in principle, if I
>can prove that there is a unique set satisfying some particular
>property, then that property is suitable for a definiens[1], whether
>or not there is an easily applicable rule determining membership.
Attack James's definition where one should: It's not at all clear
>whether there is a unique set satisfying his property.  Don't make up
>rules about what definitions must satisfy (like feasible membership
>tests).
>>Hmm. Since a set is determined by its elements there's a definite
>>sense in which defining a set _is_ the same as specifying what
>>the elements are. In a sense - this doesn't say anything about
>>a feasible test for membership...
>>So it's not so clear to me whether you have a point or not.
>>Would be much more compelling if you _had_ an example
>>in mind where a set is defined in a way that does not in
>>some sense give a test for membership.
Having a test for membership is something of a red herring - there is no
>test for membership of the field of algebraic numbers, for example.
Certainly there is, in the sense in which I, and it seems to me
Arturo and Dik, meant the phrase: x is an algebraic number if and
only if it is a root of some polynomial with integer coefficients.
Yes, there are senses in which that's not a test, but there
is also a much weaker sense in which it _is_ a test, and the
point about the definition of the Object Ring is that it does
not give a test even in this weaker sense.
>It seems to me that there are two types of definitions in common
>usage. The first is of the type An algebraic number is a complex
>number which satisfies a polynomial equation over the integers and
>is completely self-sufficient and unambiguous. The second type, like 
>The Fitting subgroup of a group G is defined to be the largest normal
>nilpotent subgroup of G requires us to prove that there is such a thing.
>Provided that we can do that, we do not need to have any means of deciding
>whether or not an element of G lies in the Fitting subgroup.
The definition gives such a test: an element is in the Fitting
subgroup if and only if it is in a nilpotent subgroup which is
in no larger nilpotent subgroup.
>James' definition of an object ring (or whatever) is of the second type,
Last time I tried to read the definition it was nowhere near as
coherent as the largest normal nilpotent subgroup. Lemme
try again:
The Object Ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime.
Nope, I can't make sense of this. When you say the Fitting subgroup
is the largest normal nilpotent subgroup it's not clear to me that
there is such a thing, but it _is_ clear what the definition means;
to tell whether H is the Fitting subgroup of G one looks at all the
normal nilpotent subgroups of G and checks that H contains all
the others. I can't figure out how to tell whether a ring is the
Object Ring in the same sense.
>and could conceivably make sense if he could prove that there was a
>unique maximal subring of the complex numbers (?) that had the required
>properties. 
You're changing the definition. If he'd said the Object Ring was the
largest subring (or the unique maximal subring) such that [etc]
then I'd know at least what the definition meant. But that's not
what he said.
>Not much chance of that though, is there?
Derek Holt.
************************
David C. Ullrich
 <87vfnvwy7a.fsf@phiwumbda.org>
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 <bt1lqv$hq5$1@wisteria.csv.warwick.ac.uk>
 <ros8vvk338r3k5ncaibec11k98uu8h20lp@4ax.com>
====
> Certainly there is, in the sense in which I, and it seems to me
> Arturo and Dik, meant the phrase: x is an algebraic number if and
> only if it is a root of some polynomial with integer coefficients.
Arturo?  I think you mean Christopher Henrich.
>>It seems to me that there are two types of definitions in common
>>usage. The first is of the type An algebraic number is a complex
>>number which satisfies a polynomial equation over the integers and
>>is completely self-sufficient and unambiguous. The second type, like 
>>The Fitting subgroup of a group G is defined to be the largest normal
>>nilpotent subgroup of G requires us to prove that there is such a 
thing.
>>Provided that we can do that, we do not need to have any means of 
deciding
>>whether or not an element of G lies in the Fitting subgroup.
 The definition gives such a test: an element is in the Fitting
> subgroup if and only if it is in a nilpotent subgroup which is
> in no larger nilpotent subgroup.
Now, this, I think, is stretching matters a bit.  After all, any
definition of, say, X, comes with the principle that x is in X iff x
is in the unique set with satisfies the definiens for X.  The same
principle would trivially apply to James's definition, if there were
indeed a unique ring satisfying his requirements for the object ring.
Therefore, this cannot be the sense in which James's definition fails.
-- 
Jesse Hughes 
Radicals are interesting because they were considered 'radical' by
modern mathematics depends on. --Another JSH history lesson
====
On Fri, 02 Jan 2004 02:58:30 +0100, jesse@phiwumbda.org (Jesse F.
> Certainly there is, in the sense in which I, and it seems to me
>> Arturo and Dik, meant the phrase: x is an algebraic number if and
>> only if it is a root of some polynomial with integer coefficients.
Arturo?  I think you mean Christopher Henrich.
It seems to me that there are two types of definitions in common
>usage. The first is of the type An algebraic number is a complex
>number which satisfies a polynomial equation over the integers and
>is completely self-sufficient and unambiguous. The second type, like 
>The Fitting subgroup of a group G is defined to be the largest normal
>nilpotent subgroup of G requires us to prove that there is such a 
thing.
>Provided that we can do that, we do not need to have any means of 
deciding
>whether or not an element of G lies in the Fitting subgroup.
>> The definition gives such a test: an element is in the Fitting
>> subgroup if and only if it is in a nilpotent subgroup which is
>> in no larger nilpotent subgroup.
Now, this, I think, is stretching matters a bit.  After all, any
>definition of, say, X, comes with the principle that x is in X iff x
>is in the unique set with satisfies the definiens for X.  The same
>principle would trivially apply to James's definition, if there were
>indeed a unique ring satisfying his requirements for the object ring.
That's not the way it looks to me. Have you tried to read the
definition? Here it is:
The Object Ring is a commutative ring that includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer, where no non-unit member is a factor of any two integers that
are coprime.
I _don't_ see an intelligible condition on the ring or on the elements
of the ring there: The ring is supposed to include all numbers with
a certain property, but the stated property is not a property that
a number can have! includes all numbers such
that -1 and 1 are the only members that are both a unit and an
integer
_Is_ it true that some numbers satisfy the property 
-1 and 1 are the only members that are both a unit and an
integer? No. The phrase all numbers such that -1 and 1 are the only
members that are both a unit and an
integer simply makes no sense.
>Therefore, this cannot be the sense in which James's definition fails.
************************
David C. Ullrich
 <87vfnvwy7a.fsf@phiwumbda.org>
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====
> On Fri, 02 Jan 2004 02:58:30 +0100, jesse@phiwumbda.org (Jesse F.
 Certainly there is, in the sense in which I, and it seems to me
> Arturo and Dik, meant the phrase: x is an algebraic number if and
> only if it is a root of some polynomial with integer coefficients.
>>Arturo?  I think you mean Christopher Henrich.
>>It seems to me that there are two types of definitions in common
>>usage. The first is of the type An algebraic number is a complex
>>number which satisfies a polynomial equation over the integers and
>>is completely self-sufficient and unambiguous. The second type, like 
>>The Fitting subgroup of a group G is defined to be the largest normal
>>nilpotent subgroup of G requires us to prove that there is such a 
thing.
>>Provided that we can do that, we do not need to have any means of 
deciding
>>whether or not an element of G lies in the Fitting subgroup.
 The definition gives such a test: an element is in the Fitting
> subgroup if and only if it is in a nilpotent subgroup which is
> in no larger nilpotent subgroup.
>>Now, this, I think, is stretching matters a bit.  After all, any
>>definition of, say, X, comes with the principle that x is in X iff x
>>is in the unique set with satisfies the definiens for X.  The same
>>principle would trivially apply to James's definition, if there were
>>indeed a unique ring satisfying his requirements for the object ring.
 That's not the way it looks to me. Have you tried to read the
> definition? Here it is:
 The Object Ring is a commutative ring that includes all numbers such
> that -1 and 1 are the only members that are both a unit and an
> integer, where no non-unit member is a factor of any two integers that
> are coprime.
 I _don't_ see an intelligible condition on the ring or on the elements
> of the ring there: The ring is supposed to include all numbers with
> a certain property, but the stated property is not a property that
> a number can have! includes all numbers such
> that -1 and 1 are the only members that are both a unit and an
> integer
I agree that this definition fails to pick out a unique ring.  This is
a bad definition.  I disagree that the problem has to do with whether
or not there is a principle for elementhood.
One might as well complain that it's a bad definition because we have
no means for determining subsethood, the transitive closure, this and
that other thing.  But none of this is really to the point.  A
definition is acceptable iff there is provably a unique structure
satisfying the definiens.  
Now, you seem to interpret the membership test liberally: Any
legitimate definition comes with a principle for determining
elementhood.  I tended to interpret the complaint differently: one
needs an elementhood test for a definition to be legitimate whether or
not there is a unique structure satisfying the definiens.  The latter
interpretation is simply and plainly false.  The former interpretation
(yours?) makes the complaint seemingly valid, but rather indirect and
confusing.  The problem has nothing to do with elementhood, but a
failure to satisfy the unique existence part.
Put another way: if your interpretation is correct, then a definition
has a membership test if and only if it satisfies the unique existence
clause.  James's definition is problematic because it fails the unique
existence clause.  As corollary, it fails the membership test, but
only in a funky way.  There is no particular set X which the
definition picks out, and so there is no test for whether a given x is
in X (because there's no privileged X!). Just seems a funny way to
criticize the definition.
At no point did I try to claim that James has offered a valid
definition, but only that the elementhood test is either a very
indirect complaint or simply based on a false intuition about
definitions.
> _Is_ it true that some numbers satisfy the property 
> -1 and 1 are the only members that are both a unit and an
> integer? No. The phrase all numbers such that -1 and 1 are the only
> members that are both a unit and an
> integer simply makes no sense.
Don't let 
-- 
I've ... contacted [some of the...] highest I.Q.'s in the country...
I've even helped the FBI out a few times...  I've met at least one
governor..., a senator... and I've had some really good seats at
sports games.  My experiences are not your experiences. --JSH != you
====
On Fri, 02 Jan 2004 17:25:22 +0100, jesse@phiwumbda.org (Jesse F.
> On Fri, 02 Jan 2004 02:58:30 +0100, jesse@phiwumbda.org (Jesse F.
> Certainly there is, in the sense in which I, and it seems to me
>> Arturo and Dik, meant the phrase: x is an algebraic number if and
>> only if it is a root of some polynomial with integer coefficients.
Arturo?  I think you mean Christopher Henrich.
It seems to me that there are two types of definitions in common
>usage. The first is of the type An algebraic number is a complex
>number which satisfies a polynomial equation over the integers and
>is completely self-sufficient and unambiguous. The second type, like 
>The Fitting subgroup of a group G is defined to be the largest 
normal
>nilpotent subgroup of G requires us to prove that there is such a 
thing.
>Provided that we can do that, we do not need to have any means of 
deciding
>whether or not an element of G lies in the Fitting subgroup.
>> The definition gives such a test: an element is in the Fitting
>> subgroup if and only if it is in a nilpotent subgroup which is
>> in no larger nilpotent subgroup.
Now, this, I think, is stretching matters a bit.  After all, any
>definition of, say, X, comes with the principle that x is in X iff x
>is in the unique set with satisfies the definiens for X.  The same
>principle would trivially apply to James's definition, if there were
>indeed a unique ring satisfying his requirements for the object ring.
>> That's not the way it looks to me. Have you tried to read the
>> definition? Here it is:
>> The Object Ring is a commutative ring that includes all numbers such
>> that -1 and 1 are the only members that are both a unit and an
>> integer, where no non-unit member is a factor of any two integers that
>> are coprime.
>> I _don't_ see an intelligible condition on the ring or on the elements
>> of the ring there: The ring is supposed to include all numbers with
>> a certain property, but the stated property is not a property that
>> a number can have! includes all numbers such
>> that -1 and 1 are the only members that are both a unit and an
>> integer
I agree that this definition fails to pick out a unique ring.  This is
>a bad definition.  I disagree that the problem has to do with whether
>or not there is a principle for elementhood.
One might as well complain that it's a bad definition because we have
>no means for determining subsethood, the transitive closure, this and
>that other thing.  But none of this is really to the point.  A
>definition is acceptable iff there is provably a unique structure
>satisfying the definiens.  
Now, you seem to interpret the membership test liberally: Any
>legitimate definition comes with a principle for determining
>elementhood.  I tended to interpret the complaint differently: one
>needs an elementhood test for a definition to be legitimate whether or
>not there is a unique structure satisfying the definiens.  The latter
>interpretation is simply and plainly false.  The former interpretation
>(yours?) makes the complaint seemingly valid, but rather indirect and
>confusing.  The problem has nothing to do with elementhood, but a
>failure to satisfy the unique existence part.
Put another way: if your interpretation is correct, then a definition
>has a membership test if and only if it satisfies the unique existence
>clause.  James's definition is problematic because it fails the unique
>existence clause.  As corollary, it fails the membership test, but
>only in a funky way.  There is no particular set X which the
>definition picks out, and so there is no test for whether a given x is
>in X (because there's no privileged X!). Just seems a funny way to
>criticize the definition.
At no point did I try to claim that James has offered a valid
>definition, but only that the elementhood test is either a very
>indirect complaint or simply based on a false intuition about
>definitions.
> _Is_ it true that some numbers satisfy the property 
>> -1 and 1 are the only members that are both a unit and an
>> integer? No. The phrase all numbers such that -1 and 1 are the only
>> members that are both a unit and an
>> integer simply makes no sense.
Don't let 
Ok, I won't. 
************************
David C. Ullrich
 <87vfnvwy7a.fsf@phiwumbda.org>
 <1rh8vvsdekd46h57vq7iro8a5f9485k7kl@4ax.com>
 <bt1lqv$hq5$1@wisteria.csv.warwick.ac.uk>
 <ros8vvk338r3k5ncaibec11k98uu8h20lp@4ax.com>
 <877k0bt8l5.fsf@phiwumbda.org>
 <3tqavv0kuedgeiiuaesnpm4aeup1a9b5d9@4ax.com>
 <87r7yixqq5.fsf@phiwumbda.org>
 <8e8bvvo338smf0mggqsc160dnrqetg9qvj@4ax.com>
====
>>Don't let 
 Ok, I won't. 
Give me a break.  I'm logged into my machine in the Netherlands while
I sit in Oklahoma City.  Sometimes, the editor lags a bit and I
overlook errors.
-- 
Jesse F. Hughes
And hey, if you're moping and miserable because mathematics tests you,
then maybe, if you think you're a mathematician, you might want to try
a different field. -- Another James S. Harris self-diagnosis.
====
On Fri, 02 Jan 2004 18:26:04 +0100, jesse@phiwumbda.org (Jesse F.
Don't let 
>> Ok, I won't. 
Give me a break.  
Was just trying to be agreeable...
>I'm logged into my machine in the Netherlands while
>I sit in Oklahoma City.  Sometimes, the editor lags a bit and I
>overlook errors.
Huh. (Probably if I asked what machine you logged into
while you were in the Netherlands you'd ask for another
break, eh? Sorry...)
************************
David C. Ullrich
====
> Certainly there is, in the sense in which I, and it seems to me
>> Arturo and Dik, meant the phrase: x is an algebraic number if and
>> only if it is a root of some polynomial with integer coefficients.
Arturo?  I think you mean Christopher Henrich.
Well, I said it often enough back when.
>It seems to me that there are two types of definitions in common
>usage. The first is of the type An algebraic number is a complex
>number which satisfies a polynomial equation over the integers and
>is completely self-sufficient and unambiguous. The second type, like 
>The Fitting subgroup of a group G is defined to be the largest normal
>nilpotent subgroup of G requires us to prove that there is such a 
thing.
>Provided that we can do that, we do not need to have any means of 
deciding
>whether or not an element of G lies in the Fitting subgroup.
>> The definition gives such a test: an element is in the Fitting
>> subgroup if and only if it is in a nilpotent subgroup which is
>> in no larger nilpotent subgroup.
Now, this, I think, is stretching matters a bit.  After all, any
>definition of, say, X, comes with the principle that x is in X iff x
>is in the unique set with satisfies the definiens for X.  The same
>principle would trivially apply to James's definition, if there were
>indeed a unique ring satisfying his requirements for the object ring.
>Therefore, this cannot be the sense in which James's definition fails.
I think this is just a problem of people perhaps not choosing the best
way of expressing themselves. Dik and others' complaint is not
necessarily that there is no black box we can put a complex number
into and decide if it is or is not in the object ring. The complaint
is really that the statement given is not sufficient to determine what
is meant. The ring is not defined to be the 'largest', or 'maximal',
with a property. It is simply stated that it is the ring in which
two conditions are met. 
If we defined the Fitting subgroup as the normal subgroup of G which
is nilpotent, then clearly we have not provided a coherent
definition. If we try to define the Uberabelian Subgroup of G as the
subgroup of G in which any two elements commute, we would have the
same sort of problem.
-- 
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
 <ros8vvk338r3k5ncaibec11k98uu8h20lp@4ax.com>
 <877k0bt8l5.fsf@phiwumbda.org> <bt2jv3$ov4$1@agate.berkeley.edu>
====
> Certainly there is, in the sense in which I, and it seems to me
> Arturo and Dik, meant the phrase: x is an algebraic number if and
> only if it is a root of some polynomial with integer coefficients.
>>Arturo?  I think you mean Christopher Henrich.
 Well, I said it often enough back when.
Okay.  I didn't recall that.
>It seems to me that there are two types of definitions in common
>>usage. The first is of the type An algebraic number is a complex
>>number which satisfies a polynomial equation over the integers and
>>is completely self-sufficient and unambiguous. The second type, like 
>>The Fitting subgroup of a group G is defined to be the largest normal
>>nilpotent subgroup of G requires us to prove that there is such a 
thing.
>>Provided that we can do that, we do not need to have any means of 
deciding
>>whether or not an element of G lies in the Fitting subgroup.
 The definition gives such a test: an element is in the Fitting
> subgroup if and only if it is in a nilpotent subgroup which is
> in no larger nilpotent subgroup.
>>Now, this, I think, is stretching matters a bit.  After all, any
>>definition of, say, X, comes with the principle that x is in X iff x
>>is in the unique set with satisfies the definiens for X.  The same
>>principle would trivially apply to James's definition, if there were
>>indeed a unique ring satisfying his requirements for the object ring.
>>Therefore, this cannot be the sense in which James's definition fails.
 I think this is just a problem of people perhaps not choosing the best
> way of expressing themselves. Dik and others' complaint is not
> necessarily that there is no black box we can put a complex number
> into and decide if it is or is not in the object ring. The complaint
> is really that the statement given is not sufficient to determine what
> is meant. The ring is not defined to be the 'largest', or 'maximal',
> with a property. It is simply stated that it is the ring in which
> two conditions are met. 
If that's what Dik and others mean, then I agree with the complaint
regarding James's definition and also the characterization that this
complaint isn't being clearly expressed.
It seems to me that there are two distinct issues.
(1) Whether or not James's definition actually characterizes a unique
structure.
(2) Whether or not his definition yields a means of determining which
complex numbers are elements of that structure.
Obviously, if he fails the first (as he has), then the second isn't
really applicable.  But, if he succeeds in the first, then the second
isn't particularly relevant in evaluating whether he has given a
proper definition -- at least not in  the way I read the second.  He's
given an adequate definition if and only if he (provably) satisfies
(1), near as I can figure.
This is why I objected to (2) recently.
> If we defined the Fitting subgroup as the normal subgroup of G which
> is nilpotent, then clearly we have not provided a coherent
> definition. If we try to define the Uberabelian Subgroup of G as the
> subgroup of G in which any two elements commute, we would have the
> same sort of problem.
Yes, of course I agree with this.
-- 
No feeling sympathy for mathematicians who start marching with signs
like 'Will work for food' in the future...  I will not show mercy
going forward.  I was trained as a soldier in the United States Army
after all... We play to win. --James Harris, feel his wrath!
====
   [.snip.]
>I am uneasy with the apparent assumption that only 1 and -1 are both
>units and integers in this ring.  sqrt(2)+1 and sqrt(2)-1 are algebraic
>numbers that are units in the ring of algebraic integers.  (Note that
>their product is 1.)  Bringing more numbers into the ring cannot
>destroy that property.  But perhaps I am misreading the above
>statement.
Maybe this will help: Assuming we are working with a subring R of the
complex numbers which contains the integers, the property alluded to
is equivalent to the statement that the intersection of R with Q is
equal to Z:
PROP. Let R be a subring of the complex numbers, and let U(R) be the
(multiplicative) group of units of R. If R contains the integers, then
the following are equivalent:
  (1) If u in U(R) is an integer, then u=1 or u=-1.
  (2) R intersect Q is equal to Z.
Proof. (2)->(1). Let u in U(R) intersect Z. Then 1/u is in R intersect
                 Q, hence lies in Z. Thus u is a unit in Z, and
                 therefore u=1 or u=-1.
(1)->(2) Let p/q be an element of R intersect Q, p and q integers. We
         may assume gcd(p,q)=1, and q>0. Since gcd(p,q)=1, there exist
         integers r and s such that rp+sq = 1. Since p/q lies in R, so
         does r*(p/q) + s; and this is also in Q. We have
         r*(p/q) + s = [(rp)/q] + [sq/q] = (rp+sq)/q = 1/q.
         Therefore, the integer q is a unit in R, since 1/q also lies
         in R. By (1), this implies that q=1 or q=-1. Since q>0, this
         means that q=1, so p/q = p is an integer. Therefore, R
         intersect Q is contained in the integers. Since the integers
         are all in R, this implies that R intersect Q is equal to Z,
         as claimed.
QED
The ring of all algebraic integers has this property, as does any
subring that contains Z. Other subrings of C also have the property:
Z[a] does, for any transcendental number a. There are also certain
subrings of the algebraic numbers which do not consist only of
algebraic integers, but have the property. Someone had posted a nice
characterisation of some of these rings, but I could not find it
through google. But one possible construction would be to pick a
quadratic extension K of Q, and a rational prime p which splits into
two distinct prime ideals, (p) = PQ. Then take the ring of integers A
of K, and invert all elements in P-Q; this ring is not contained in
the algebraic integers, but intersects Q at Z.
On the other hand, it's already been pointed out that any
generalization of the algebraic integers which is closed under Galois
conjugates is unlikely to be of use for the purpose. See Bill
Dubuque's interesting discussion about it:
http://google.com/groups?selm=y8zllty36hr.fsf%40nestle.ai.mit.edu
-- 
======================================================================
It's not denial. I'm just very selective about
 what I accept as reality.
    --- Calvin (Calvin and Hobbes)
======================================================================
Arturo Magidin
magidin@math.berkeley.edu
====
Turing's test design is flawed.
It is not double blind: the tester knows some are machines and some
are not.
C++ Simulator of a Universal Turing Machine can be downloaded at :
} * http://alexvn.freeservers.com/s1/utm.html
} * http://sourceforge.net/projects/turing-machine/

}     The program simulates a Universal Turing Machine (UTM).
    The UTM used in the Simulator is three-tape Turing Machine:
}     * Tape#0 contains transition table and initial instantaneous 
description
}       of a Particular Turing Machine (TM);
}     * Tape#1 and Tape#2 are working UTM-tapes.
    The UTM can simulate the behavior of a Multitape TM.
    The package consists of two executable files :
}     * t2u - compiler TM-to-UTM
}       which translates description and input of TM to UTM-language;
}       t2u generates several output files, one of them is used as input of 
the utm.
}     * utm - the Simulator itself.
    Detailed log file is generated.
}     Resources used (input size, output size, UTM-space, UTM-time) are 
computed as well.

}     Testsuites. Two Turing Machines (TM-1 and TM-2) are used to create 
inputs for UTM.
}     Each of them is an addition program which adds two numbers:
}     * TM-1 is one-tape TM,
}     * TM-2 is two-tape TM.
====
> Turing's test design is flawed.
 It is not double blind: the tester knows some are machines and some
> are not.
>
how?
wasn't it just a pushrod doing morse code?
Herc
====