mm-45 === 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 issueswe have been arguing about. If mitch already owns thisbook then he can be counted on to quote parts of them outof context in mistaken attacks on other people's positions.I can only hope to be able to check this book out whenthe library re-opens after Xmas break; then I should be ableto quote right back, in rebuttal. But even feeling that oneneeds to quote authorities is ridiculous. My main reasonfor making this threat is the hope that mitch can be deterredin advance from excerpting these papers in dishonest and misleadingways.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 Li?ge).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: Jos? 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 H?jek (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 formallyexpressed 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 appropriatemathematical 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 representationof predicate term negation as a unary connective arebound to be imperfect in Avron's sense. Indeed, theonly negation among the unary connectives that emergesas perfect from both the syntactic and semantic point ofview is the Boolean negation of classical logic.Now for a stupid math person like me, Avron's terms describing implication as internal and external are completelymeaningless. But, I sure do know why I expect a criterion that identifies the strength of Boolean negation. Mostly, it isbecause I understand it with respect to geometric reflection.For what it is worth, thanks. === This is a variation of the same proof I give inCardinality of Computable Numbers.I define two Turing machines.The first machine (TM1) has these instructions:1) Write a 12) Move right one positionRepeatAssume we give this machine a tape that hasan infinite string of 0's. It would seem thatTM1 will output an infinite string of 1's.Instructions for TM2:1) Scan right until a 0 is found2) Scan right until a second 0 is found3) Backup and write a 1 on the previous 0RepeatAssume we give TM2 a tape that containsan infinite string of 0's.Even if we assume that TM2 performs aninfinite number of operations, the tapeproduced by TM2 will contain an initialsegment with a finite number of 1'sfollowed by a 0.TM2 is incapable of writing an infinitenumber of 1's. If TM2 can not writean infinite number of 1's, we can notassume 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 alphabetisand 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 thenumber 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 thespecificquadruples 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 willcontinueprinting 1's forever. I don't see why you would claim otherwise. Ifyoumake such a shocking claim, perhaps you should provide us with somesort of proof or argument as to why you believe it to be true, so thatwecan 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 cannotwrite an infinite number of 1's either. TM1 never halts and neveraccomplishes 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 despitetheprofoundly fundamental flaws in your reasoning.Let me suggest a remedy:You believe that there is an *actual* infinity, a place that you cangetto by repeatedly adding 1's enough times. You believe an infinitetaskcan be completed. This is what your intuition tells you. Intuitionaboutinfinity can be very misleading. That is why it is important to makeyour thoughts mathematically rigorous. Until you do, it is pointlessarguing with mathematicians about the mathematics of infinity. But ifyoudo try to make your point of view rigorous, you will see that yourcurrentbeliefs are inconsistent.In mathematics, there is no *actual* infinity (in a sense). Infinite processes are nevercompleted. Infinity is only *potential*. Some would say, wait aminute,Cantor gave us the actually infinite. The infinite ordinals areperfectly rigorous objects. But this doesn't tell the whole story. Modern settheory *postulates* the existence of omega and then proceeds tomanipulateit like every other mathematical object. Omega is never*constructed*.This is fine. ZF is quite a rigorous and fascinating theory. Alsonotethat to speak about omega and the other transfinite numbers, once hasto liftoneself beyond the natural numbers. Within the ordinary arithmetic ofthe natural numbers (I am speaking about the standard model of thePeanopostulates), 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 twoago,I've encountered at least half a dozen people arguing exactly the samepoint of view as you, but in different contexts. Now we're in thecontextof Turing Machines. That's very humorous to me. I didn't know peoplewith your beliefs existed, and now I find that there are at leastseveralof you. Have you heard of this guy Phil who used to post absurdthings likethe 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. Imyself still don't quite get ordinal numbers, although I've gotthe cardinals down pretty well now. :) What I don't understand is how some people who don't get infinityseem to compulsively post *wrong* statements to the Internet, ratherthan trying to understand *right* ones. And how a guy like Russellcan seem to have such a reasonable grasp of what a Turing machineis, without having even a basic conception of the properties of theinteger 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 completethat thought: All natural numbers *do* have finitely many digits!But Phil made a leap from that true statement to the false statementthat *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 neverhalt.> 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. Byconstruction, at any given time-step, a Turing machine has modifiedonly finitely many cells, but there is no upper bound on the number ofcells a Turing machine can modify. (There is a glaringly obviousanalogy 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>RepeatAssume 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:Computing machines.If an a-machine prints two kinds of symbols, of which the first kind (calledfigures) consists entirely of 0 and 1 (the others being called symbols ofthe second kind), then the machine will be called a computing machine. Ifthe machine is supplied with a blank tape and set in motion, starting fromthe correct initial m-configuration, the subsequence of the symbols printedby it which are of the first kind will be called the sequence computed bythe machine. The real number whose expression as a binary decimal isobtained by prefacing this sequence by a decimal point is called the numbercomputed 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-configurationwill be said to describe the complete configuration at that stage. Thechanges of the machine and tape between successive complete configurationswill 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 saidto be circle-free.A machine will be circular if it reaches a configuration from which there isno possible move, or if it goes on moving, and possibly printing symbols ofthe second kind, but cannot print any more symbols of the first kind. Thesignificance 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-freemachine. A number is computable if it differs by an integer from the numbercomputed by a circle-free machine.We shall avoid confusion by speaking more often of computable sequences thanof computable numbers.According to this definition, any TM that halts is circular and does NOTproduce 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 andsubstitute 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>RepeatAssume 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 sequencethat 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 whythe diagonal argument doesn't work with computable numbers.The problem can be converted into determining whetherevery TM is circular or not.Turing proves this is impossibleTM2 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 manycells.Russell- 2 many 2 count === > 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, andin traditional programming jargon, the only way you can see aprogram's output is to wait for it to halt, and then look atwhat it's produced.> The word halt does not appear in Turing's paper. I'm willing to bet that a majority of participants in thisdiscussion have not read Turing's paper in the last 20 or 30years. I certainly have never read it. I get my knowledgesecond-hand, from people who can explain ideas more clearly (onehopes) than the original geniuses who came up with them. (Let'shear 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 talkingabout, here, specifically.> This is the definition of a computable number given by Turing: 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 printedby 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 theadvantage of intuitiveness, but chronological makes more sensemathematically to me. Please clarify this point: briefly, what doesTuring 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 haveany idea why Turing chose these particular words to describe the twokinds 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.> 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 ofthem, we can conclusively say that for instance your TM1 computesthe sequence 11111..., which implies that all numbers differing fromthe 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 tryto 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 sequenceof 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 wordprefacing 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 digitsin pi, or a recipe for granite cheesecake.> .111...1110 (base 2) This is not correct notation. It reminds me very strongly ofPhil's ramblings, and I really do suggest you take a look atGoogle Groups for sci.math, and search on rational numberscountable, 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 TuringMachine X is circular, for some value of X, it doesn't necessarilyfollow that the computable numbers are uncountable. For that,you'd need to actually give a reference to Turing's proof, sothat we could look at it and see whether it proves what you thinkit does. It may very well prove what you think it does. Before youposted this quote, I think most participants assumed you weretalking 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 onTuring's definition of m-configuration. The question of whetherTM2 computes a number depends entirely on Turing's definitions ofsequence 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 botheredto write out its state transitions: .10110111011110111110111111011111110111111110111111111011111111110... That number, which is approximately 0.71673, is computable, butcertainly not rational! Also computable: pi and e, among manyothers.-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 hadaddressed 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 describingwere 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: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 symbolsof>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, startingfrom>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 computedby>the machine. The real number whose expression as a binary decimal is>obtained by prefacing this sequence by a decimal point is called thenumber>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 thebeginning (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 aboutGodel's Entscheidungsproblem than in creating thefoundations 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 writingsymbols of the first kind on every other square.the 0s and 1's. He states that these other symbols are removedat 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 scannedsquare,>the complete sequence of all symbols on the tape, and them-configuration>will be said to describe the complete configuration at that stage. The>changes of the machine and tape between successive completeconfigurations>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 issaid>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 TMcan'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 whichthere is>no possible move, or if it goes on moving, and possibly printing symbolsof>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 acircle-free>machine. A number is computable if it differs by an integer from thenumber>computed by a circle-free machine.We shall avoid confusion by speaking more often of computable sequencesthan>of computable numbers.>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 beoverwritten. 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>RepeatUsing 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 severalthreads 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 numbersis certainly not as strange as the idea that there aremore real numbers than natural numbers.Several people have suggested that this proof saysthere is a finite number of natural numbers.This is incorrect. This proof shows that no setcan 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 becausethere 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 finitelymany>cells.>No it doesn't.At least, no one can prove that it does.It is simple to show that the number of 1'swritten by TM1 is some multiple of the numberof 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'sfollowed 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 isa 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 consideringsituations that can only hold _after_ the run of the TM,not _during_ the run, such as: an infinite number of 1sbeing written by the first TM. At any time _during_ therun, 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 considerwhat happens after a run, for a TM run which does nothalt. It doesn't _have_ an after; that's what doesnot halt _means_, and that's why your description ofthe behavior of TM2 makes no sense at all. You aredescribing the situation after a finite number of stepsas if it were the situation after the run, but again,there _is_ no after, so your description is not merelyincorrect, it is _meaningless_, just like an argumentbased on characteristics of members of the empty set.Please don't post _at all_ again about infinite behaviorsuntil you can understand this simple objection to yourmethods at the most profound level. Intuitive argumentsdon't _work_ for infinities, that's why mathematicallysound arguments are the only appropriate tools for discussinginfinite 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 becx where c is a constant.You are saying that cx = infinity wherec 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 unitsay TM2 is at position log X>and TM1 is at position X.even if X->oo, TM2 is always < TM1in 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 makethose 2 statements in the same context. I'd put cx->oo.You asserted TM2 has an infinite run length, so you're allowinginfinity into your experiment. All your methods are constructable,you have to consider a multiple level processor queue now.Hercold OS lecture floods back to memory === > No TM can write an infinite string of symbols.say X is the time unitsay TM2 is at position log X>and TM1 is at position X.even if X->oo, TM2 is always < TM1in 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 countshow 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 representsan infinite number of 1's.Russell- 2 many 2 count === Also available at http://math.ucr.edu/home/baez/week200.htmlThis Week's Finds in Mathematical Physics - Week 200John 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 explainn-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 itmaterializes - but so far I've just got a tentative outline:http://math.ucr.edu/home/baez/hda.htmlUnfortunately, 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 backwhenever 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 calledFunctorial 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 mathematiciansand philosophers - especially Alberto Peruzzi, a philosopher at theUniversity 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 incredibleart and architecture: the Ponte Vecchio, the Piazza del Duomo, the SantaCroce 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 theprocess, 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 toolsto really make progress in understanding these notions. After all, thiswas 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 motionof 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 andtruth values, in such a way that all the axioms get mapped to true. Since theories involve playing around with symbols according to fixedrules, the study of theories is often called syntax. Since themeaning of a theory is revealed when you look at its models, thestudy of models is called semantics. The details vary a lot depending on what you want to do, and physicists rarely bother to formulate theirtheories 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 othercategories as well. And to do this, he wanted a new way of describingtheories, 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 thereare 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 likefor 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 likefor 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 categoriestheories expressed in this logic can have models in. This point maysound esoteric, but the underlying principle should be familiar. Whichis 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 adrill press around, so now the electric drill sounds best. But anotherthing: 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. Specializedtools 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 LAWg^{-1}g = 1 LEFT INVERSE LAW gg^{-1} = 1 RIGHT INVERSE LAWSuch 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 morphismf: 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 categorywith finite products whose only goal in life is to contain a groupobject. To build this category, first we put in an object G Since our category has finite products this automatically meansit 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 morphismm: G x G -> GWe also put in a unary operation called inverse:inv: G -> Gand a nullary operation called the unit:i: 1 -> GAnd then we say a bunch of diagrams commute, which express allthe 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 indeedit can! A model of theory of groups in some category X with finiteproducts is just a product-preserving functorF: Th(Grp) -> XBy the way things are set up, this gives us an objectF(G)in C, together with morphismsF(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 elementfor F(G)... all making a bunch of diagrams commute, that expressthe 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) = GrpThat'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 theircategory 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 itis 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 semanticsand a precise duality between the two columns!I would love to dig deeper in this direction - I've really justscratched the surface so far, and I'm afraid the experts will bedisappointed... 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 defaultsetting, 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 doingit! Lawvere started doing this in 1967 - he called it synthetic differential geometry. Basically, he cooked up some axioms on a toposthat let you do calculus and differential geometry with infinitesimals. The most famous topos like this is the topos of schemes - algebraicgeometers 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 themain 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 bereally tiny, but still point somewhere. In other words, the headcan'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 technicaljargon, 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 sometopos with an infinitesimal arrow object, T. I'll call the objectsof this topos smooth spaces and the morphisms smooth maps. Howdoes 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 -> Xthat is, a way of drawing an infinitesimal arrow in X. In general, themaps from any object A of a topos to any other object B form an objectcalled 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 justa 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 callg^T: X^T -> Y^TMoreover, it's pathetically easy to check the chain rule:(gh)^T = g^T h^TAnd 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 derivativeswe start with T x T, which looks like an infinitesimal square. Thenwe mod out by the mapS_{T,T}: T x T -> T x Tthat 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 nameT^2/2!and if we now use some axioms about T, it turns out that a smooth mapf: T^2/2! -> Xpicks out what's called a second-order jet in X. This is a conceptfamiliar 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 perfectlyanalogous way, and the visual resemblance to Taylor's theorem is by nomeans an accident... but let me stick to second derivatives, since I'mtrying 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 bundleX^{T^2/2!}Using some axioms about T, we can show there is a smooth map T^2/2! -> Twhich throws out the second-order infinitesimal data and justkeeps the first-order part. This gives a smooth mapp_X: X^{T^2/2!} -> X^Tfrom the 2nd-order jet bundle to the tangent bundle. Intuitivelyyou can think of this as sending any position-velocity-accelerationtriple, 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: thatis, a smooth mapf: X -> Ythat 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 motionf: R -> Xis precisely a trajectory in X that follows the law, meaning that the acceleration of the trajectory is the desired function of positionand velocity. This example is a setup for the classical mechanicsby replacing R by a higher-dimensional space.I'm sure many of you have the same impression that I had when seeingthis stuff, namely that it's a bit quixotic for a high-powered mathematicianto be reformulating the foundations of classical mechanics here at the turnof the 21st century, instead of working on something cutting-edge likestring theory. Even if Lawvere's approach is better, one can't help but wonder if it gives truly *new* insights, or just a clearer formulationof 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 everyoneelse 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'tneed to accept it to profit from it. In math, what really mattersis 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 aboutthis 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 hereand 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 syntacticallypresenting a category with finite products. It also serves as an introduction to topoi... umm, or at least toposes. I used to find itfearsomely difficult and dry. Now I don't, which is sort of scary.A really beautiful more advanced treatment of algebraic theories andalso 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 alreadyin 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 offunctorial semantics of algebraic theories, in Reports of the MidwestCategory Seminar, eds. Jean Benabou et al, Springer Lecture Notes inMathematics No. 61, Springer, Berlin 1968, pp. 41-61.Then came his work on doctrines, which I vaguely alluded to a whileback: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 inLawvere 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 differentialgeometry: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 andat http://www.math.ucla.edu/~asl/bsl/0902/0902-006.psand this gives a good overview of his ideas, though not easy to read!Finally, Colin McLarty - whom I was delighted to meet in Florence - hasa nice quick introduction to synthetic differential geometry inhis 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 forMathematics, this is the one reference that's really good forbeginners! Okay... now that everyone is gone except the people who are absolutelynuts 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 afew 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 importantmonads - but I've never really said what they're good for! I need tocome clean and explain this now, since a doctrine is a categorifiedversion 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 wedo, 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 areobjects of C equipped with extra structure. Then we get a pair of adjoint functors:R: D -> CL: C -> DThe 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, ifC = SetandD = Grpthen we can take the underlying set of any group, and thefree group on any set.We get a monad on C by lettingT = LR: C -> CThen, we can use facts about adjoint functors to get natural transformations called multiplicationm: TT => Tand the uniti: 1_C => TUsing more facts about adjoint functors, we can check that these satisfy associativity and the left and right unit laws. I didall this in week92 so I won't do it again here. The upshot isthat 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 formonads. 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 pairof adjoint functors as above, the main way we get T-algebras isfrom 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 issome monad on the category of sets. And we can describe abeliangroups 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 onthe 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 ofbusiness. 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 categoryof 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 nowbecause I'm about to categorify monads and get doctrines.Okay: so, monads are good for describing objects equipped with extrastructure 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 interestingkinds of categories equipped with extra structure and properties, andeach of them gives a different kind of *logic*: the logic that worksinside this kind of category! The more structure and properties ourcategory has, the more powerful logic we can use inside it. This iswhat gives the hierarchy of expressive power I was talking about.So, it pays to have a good general way to describe categories equippedwith 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 categoriesequipped 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 thatit'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 talkabout the category of *small* categories - but because what's reallyimportant is the 2-CATEGORY of all categories. It's best to thinkof Cat as a 2-category. But this suggests that we should work witha 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 becomeridiculous. I'd rather talk about weak 3-monads than pseudopseudomonads,wouldn't you? Furthermore, if you look up pseudomonad in thedictionary 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 youraverage weak monoidal 2-category, so it may not be immediately obvioushow 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 theconcept 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 oneof these. For more details, try these papers:22) R. Blackwell, G. M. Kelly, and A. J. Power, Two-dimensional monadtheory, 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.html23) 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.gzAnyway, 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 computerscience, and I recommend Crole's book cited above for more about howit's related to cartesian closed categories. A good introduction totopos 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 TheoryCompendium, Oxford U. Press, Oxford. Volume 1, comprising Part A:Toposes as Categories, and Part B: 2-categorical Aspects of ToposTheory, 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. Mypoint 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-worldapplications, 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 beobtained athttp://math.ucr.edu/home/baez/For a table of contents of all the issues of This Week's Finds, tryhttp://math.ucr.edu/home/baez/twf.htmlA simple jumping-off point to the old issues is available athttp://math.ucr.edu/home/baez/twfshort.htmlIf you just want the latest issue, go tohttp://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 tobuild a new playground for everyone to enjoy - and finance yourexpense 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 homogeneousbut not isotropic. It doesn't solve the problem, but it does spreadit out. Who says every real cow has to be spherical, homogeneous, andisotropic? Sometimes simplicity leads to impossibility. Somestructures are emergent, requiring an unavoidable threshhold ofcomplexity. Stuff should be simple, but not too simple.[snip]-- Uncle Alhttp://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> 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.. [emoticon doffs its hat in honor of work done and submitsa fervent thank you]./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 votesand the matter was settled by the House.2)1900 was not a leap year but had a presidential electionMckinley, Bryan, Woolley and Bebs ran for office.McKinley wonAnd of course, the first US presidential election:3)1789 was not a leap year but had a presidential electionWashington, 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.htmlGardyloo. -- Uncle Alhttp://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 therehave been presidential elections in 1789, 1800 & 1900, none of whichwere leap years. === Joan Baez:I'm an aspiring dice setter in the casino game of craps, so I'minterested in physics, but I am largely uneducated in math andphysics. So, as I read your post, I searched for some hint thatsomething you're talking about might be useful.Having read your post in its entirety, and having found nothingwhatsoever that I can use to any advantage, I at least wanted to posta 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 randomizationof dice depends on a near-equality of possible vectors at some pointand that such randomization can be avoided by having an overridingvector 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 amcurious 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 aboutexisting theories. One reason may be that few *physicists* know topostheory; 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 thatnobody has formulated new theories of physics using topoi, despite some work on topos theory and physics by Chris Isham and FotiniMarkopoulou (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 symmetricmonoidal categories are better than topoi as a context for quantumphysics.(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 fromthe limitation of the ring of algebraic integers, and thinking back ondiscussions, I think there's been a lot of confusion on just what Imean.Context helps so I'll mention that the idea of different types ofintegers 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 integercoefficients.The problem then is that mathematicians thought they were done, but myresult shows they are not. Understanding how that's possible isn'treally difficult, and an easy way to see it, is to consider gaussianintegers and algebraic integers again.For instance 2 is a gaussian integer, as well as just an integer. Itis also an algebraic integer. However, consider the followingequation:x^2 = 2which is outside of the ring of gaussian integers.Now sqrt(2) is well-known *today* so I think that's a good example forhow 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 algebraicintegers.With my example here x^2 = 2, the decomposition is of 2 into equalfactors.While 2 is an integer, and a gaussian integer, that decompositionleads to a result that's neither, though it's an algebraic integer.I use a polynomial decomposition to show the limitation of the ring ofalgebraic integers.Once I had my result showing that the ring of algebraic integers, likethat of gaussian integers, and of integers before them was still toosmall, 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 suchthat -1 and 1 are the only members that are both a unit and aninteger, where no non-unit member is a factor of any two integers thatare coprime.That definition isolates the key property of the numbers in question,and includes integer, gaussian integers, algebraic integers, andbeyond.Now mathematicians as a group apparently were unaware that the ring ofalgebraic integers was so limited, and proceeded for quite some timeassuming that they'd found the most inclusive ring, which is theerror.So, in case you're wondering, no, I don't think the definition foralgebraic integers needs to be changed any more than the definitionfor gaussian integers needs to be changed. What's needed is arecognition 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 tosay about mathematics, because i find that the language you use isalmost always much too ambiguous to be understood clearly; andbecause, even more importantly, you seem to have the wrong idea aboutwhat it means when people have trouble understanding you. so oftenyou seem to take it as evidence that you've demonstrated yoursuperiority over the people who can't understand you, when instead youshould be considering the possibility that you've done a bad job ofexplaining 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 reallymean, but i wonder whether what you're really trying to do is todefine what an object is, rather than what the object ring is.before discussing this any further, though, i'd like to at leasttemporarily change your terminology from object to h-number, sincefor various reasons i don't think object is a good choice of namehere.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 usescomplex 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 abovedefinitions of h-number, then this raises an obvious question: dothe h-numbers form a sub-ring of the larger ring?so, can you prove that the h-numbers form a sub-ring of the largerring? (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-numbersor the h-number ring and people would probably understand what youwere 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 tofigure 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. Theerror 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 algebraicintegers. It is a subset of a larger ring -- a fact that has been knownsince it was first defined. Any definition of any entity which includessome elements and excludes others is limited in the sense you use theword. So what? You can point to the quantity 1/2 to show the limitation ofthe 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 -- andthe 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 itdoesn't help to delete out the relevant mathematics, and then call ittrivial 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 inthose rings the only unit integers are -1 and 1.My research shows that there must be another more inclusive ringbeyond algebraic integers with the same property i.e. that -1 and 1are the only integers in that ring that are units.Now then, if you C. Bond are simply too emotional to objectivelydiscuss the mathematics which backs up my claims, then you're probablyjust 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' isriddled 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 itis 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 atlength and the final score is:James Harris = 0, Mathematicians = 100And 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 projectingyou errors, this time onto mathematicians in general.Exactly what evidence do you have that mathematicians thoughtthe 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 mathematicsafter 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 algebraicintegers...>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'tbe reason for people to argue with me further. As the discoverer of amore inclusive ring, I get to name it, and I have, so then discussionscan move from antagonistic to considerations of that ring and itsproperties.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 integersare not the most inclusive ring. The arguments are about otherstatements>As the discoverer of a>more inclusive ring, Wow. Choose anything that's not an algebraic integer. Adjoin itto 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'venever done that.(One can see that the definition you give is incoherent becauseof all the discussions about what you might mean by it - oneperson says you mean this, another person says you meansomething else, others say the definition is simply meaningless.If the definition you gave were coherent those discussionswould 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. UllrichDavid MoranChief MeteorologistOklahoma Storm Team === In sci.math, James Harris<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 HarrisI'm going to have to go through your mathblogs in any event, to tryto trace through the impossibly complex contortions of this so-calledproof, and the threads arguing such. So lessee...You give as an exampleP(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) + 7^3This equality verifies fine, although at some point I'mgoing to have to dig up where you got the '7' from, asthe equation came from somewhere else in a grander proof.But that doesn't matter at present.You then stateP(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 cubicR(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'sare not explicitly shown as functions of x, but that's nothorribly important; I've also inserted GP/Pari-compatiblenotation 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 verifiedit in another post and won't reprise it here unless absolutelynecessary.Now Q(x) = P(x)/49 also has integer coefficients, and it turns outone can in fact divide, generatingQ(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 suchthings asQ'(x) = (5*a_1(x) + 7) * (5/7*a_2(x) + 1) * (5/7*a_3(x) + 1)orQ(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 bealgebraic integers, at least as far as I can understand your blog.It is my contention that one cannot conclude suchwithout at least an idea of the value of x, as x is afree variable. Certainly for x = 0, 5/7*a_1(x) and5/7*a_2(x) are in fact algebraic integers (namely, 0).However, it turns out Dik Winter has written up in a priorpost ( ) an extremely elegant methodhe found for determining the roots of an arbitrary cubicwith 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 overloadsthe 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))^2r(x) = -108 * (-49*(2401*x^3 - 147*x^2 + 3*x)) - 8*(3*(-1 + 49*x))^3k1(x) = cbrt(r(x) + sqrt(q(x)^3 + r(x)^2))/2k2(x) = cbrt(r(x) - sqrt(q(x)^3 + r(x)^2))/2w = (-1 + sqrt(-3))/2, w^2 = (-1 - sqrt(-3))/2, w^3 = 1where cbrt(z) is the cube root of z nearest the positive real axisin the complex plane. (Not that it matters all that much, aslong as conjugate(cbrt(z)) = cbrt(conjugate(z)).)The roots are thereforea_1(x) = (-(3*(-1 + 49*x)) + w*k1(x) + w^2*k2(x))/3a_2(x) = (-(3*(-1 + 49*x)) + w^2*k1(x) + w^1*k2(x))/3a_3(x) = (-(3*(-1 + 49*x)) + k1(x) + k2(x))/3It is obvious that I now have an explicit solution fora_1(x) et al, written in a rather ugly but workableform, and can compute them by simply inserting x andgrinding it out.For checking purposes, setting x = 0 yieldsq(0) = -36r(0) = 216k1(0) = cbrt(216 + sqrt((-36)^3 + 216^2))/2 = 3k2(0) = cbrt(216 - sqrt((-36)^3 + 216^2))/2 = 3a_1(0) = (3 + w*3 + w^2*3)/3 = 0a_2(0) = (3 + w^2*3 + w*3)/3 = 0a_3(0) = (3 + 3 + 3)/3 = 3which shows I didn't make any gross computational errors.If I set x = 1/49 I get:q(1/49) = 0r(1/49) = 108k1(1/49) = cbrt(108 + sqrt(108^2))/2 = 3k2(1/49) = cbrt(108 - sqrt(108^2))/2 = 0a_1(0) = (0 + w*3 + w^2*0)/3 = wa_2(0) = (0 + w^2*3 + w*0)/3 = w^2a_3(0) = (0 + 3 + 0)/3 = 1which is entirely consistent, as R(a,1/49) = a^3 - 1.Unfortunately, neither 5*a_1(0)/7 nor 5*a_2(0)/7 isan algebraic integer. For 5/7 this is obvious.For 5/7 * w and 5/7 * w^2 it turns out that all threeare roots of the equationu^3 - 5^3/7^3 = 0Since 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'llhave to work hard at it -- in fact, there's an indefinitenumber of counterexamples. If one sets x=1, for example,one gets:R(a,1) = a^3 + 144*a^2 - 110593This polynomial is irreducible. The explicit rootsare 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 isnot an algebraic integer. To do that requires the usualsubstitution:R(7/5*b,1)*5^3/7^3 = b^3 + 720/7*b^2 - 282125/7Oops, doesn't work.R(7/5*b,2)*5^3/7^3 = b^3 + 1455/7*b^2 - 2328250/7also doesn't work, so x=2 is out.R(7/5*b,-1)*5^3/7^3 = b^3 - 750/7*b^2 + 318875/7Nope; 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*xI can prove that, if x is in the rational numbers, andthis equation is irreducible (one might get lucky forcertain rational x; I can't say without a lot more work), then, forthe 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 grindingthe coefficient -1/27*z^3 - 125/343. This can never be aninteger for any integer z; therefore b_1(), b_2(), and b_3()cannot be integer for rational x except for certain specialcases 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 excludedat times; they're excluded most of the time.I will now conclude with a diatribe on divisibility. It is truethat P(x) can be divided by 49. It is not true that this means much.The polynomialx^4 - 4*x^3 + 6*x^2 - 4*x - 1is also divisible by 49, in the algebraic number field, yieldingof course the rather silly polynomialx^4/49 - 4/49*x^3 + 6/49*x^2 - 4/49*x - 1So what do we have at the end of the day? Not a whole lot,from what I can see.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === 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 provesthere's more beyond algebraic integers in terms of rings with theproperty that -1 and 1 are the only unit integers!!!So you have it reversed.My research involves the step beyond algebraic integers, like beforealgebraic integers were a step beyond gaussian integers, whilegaussian integers were a step beyond integers.Understand?James Harris === In sci.math, James Harrison 1 Jan 2004 08:52:53 -0800<3c65f87.0401010852.5aa5d45b@posting.google.com>:> 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 HarrisAh, OK. Mind you, everyone knows there is a next stepbeyond algebraic integers -- namely, algebraic numbers.Unless you're trying to show there's a set of numberswith units -1 and +1 and algebraic-integer-like propertiesthat constitute a ring, with the algebraic integers as aproper subset.Using standard multiplication, however, you will run intoa problem, as *every* unit of the algebraic integers willbe a unit of your set as well. This includes numberssuch 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.netIt's still legal to go .sigless. === > In sci.math, James Harris> <3c65f87.0401010852.5aa5d45b@posting.google.com>:> 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 areunits in the ring.James Harris === In sci.math, James Harrison 1 Jan 2004 18:21:10 -0800<3c65f87.0401011821.10845d56@posting.google.com>:> In sci.math, James Harris> on 1 Jan 2004 08:52:53 -0800> <3c65f87.0401010852.5aa5d45b@posting.google.com>:> 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 algebraicintegers, *not* step beyond algebraic numbers (which leads usto 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.netIt'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 thecomplex numbers, most of them not fields. You can localize at anynumber of places to obtain a ring that still satisfies the conditiongiven: the intersection with Q is equal to Z. Or you can throw in anyof a number of different transcendentals. I do not really see a nextstep, 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 Magidinmagidin@math.berkeley.edu === In sci.math, Arturo Magidinon Thu, 1 Jan 2004 23:13:27 +0000 (UTC):> [.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 thering of algebraic integers.However, any ring containing the algebraic integers willby necessity contain its units as well. An object ring Owith units [-1, +1] containing A simply doesn't work.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === >In sci.math, Arturo Magidin>:> [.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 ALSOintegers be 1 and -1. That is, that the intersection of the ring withQ 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 isthat 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 Magidinmagidin@math.berkeley.edu === In sci.math, Arturo Magidinon Fri, 2 Jan 2004 05:12:38 +0000 (UTC):>In sci.math, Arturo Magidin>on Thu, 1 Jan 2004 23:13:27 +0000 (UTC)>:> [.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 onlyunits which are integers are -1 and +1.Color me confused, but from the looks of it he might be tryingto construct A for some x(where a_1(x) etc. is the root of his second polynomial). Presumablythis 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 looksan awful lot like a special case of something, but I can'tremember what now.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === >In sci.math, Arturo Magidin> [.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 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 nonintegralrationals; for example, since a_1, a_2, and a_3 are the threeconjugate, 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 = 0a1 + a2 + a3 = -3(-1+49x).So if you have a_1/7 and a_2/7, then you have-3(-1+49x)/7 - a3/7and other elements, which will usually lead you to nonintegerrationals. If you include several values of x, things get nastier. See forexample an elementary analsysi of this problem with an earlierattempt (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 Magidinmagidin@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 timethat 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 aparticular 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, +31205924131home: 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 isnot a unit, then for any pair of coprime integers it can not be a factorof 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 completeset of numbers that are in the ring. To be more precise. The ring ofalgebraic integers satisfy the definition. Given an arbitrary algebraicnumber not in that ring, can we add it without contradiction? Within thering of algebraic numbers it is a unit, but will it be so in the new ringwhen we adjoin it to that ring?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: 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, gaussianintegers, and algebraic integers are ALL distinguished by the factthat -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 algebraicintegers 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 includesthe 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 thatis also not a field. Also already known a long time. There are aninfinite number of rings that are not fields between the algebraicintegers and the algebraic numbers. So what you state is neither newnor 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 ringof 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, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > Knowledge is a good thing Dik Winter. James HarrisThen 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 believehappen, because so many of the things you say turn out to be false.Give us some _evidence_ in support of your assertion that until youcame along mathematicians assumed that the algebraic integerswere 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 ofdetail every time some poster makes a false statement, as it happens alot. 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'venever explained, not even once.(Which of course is no surprise, since what you're saying about whatmathematicians 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 setof numbers, OK, that's a start. The definition is finished when, byusing it one can know whether or not a particular number is a member ofthe Object Ring. So, one wants a sentence something like this: theobject ring is the set of all numbers z such that ...(something aboutz). The 'something' might be, either z is an algebraic integer, orelse 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 bothunits and integers in this ring. sqrt(2)+1 and sqrt(2)-1 are algebraicnumbers that are units in the ring of algebraic integers. (Note thattheir product is 1.) Bringing more numbers into the ring cannotdestroy that property. But perhaps I am misreading the abovestatement.> 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 === 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 ringslike integers, gaussian integers, and algebraic integers from otherrings like rings that are fields is the fact that the only integerunits in the ring are -1 and 1.I also have where no non-unit member is a factor of any two integersthat are coprime, as that's also interesting in its own right, andit's another differentiating property.Now then, it may seem like a subtle definition, but think about it fora 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 mentionfrom rings that are fields. The ring you get when you add 1/2 to thering 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 somethingelse? 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, +31205924131home: 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 tooverrings. The property given is vacuous (i.e., always holds). If you assume thatcoprimeness if being defined with respect to the ring in question, R,then by definition, any factor of two elements coprime in R willnecessarily be a unit, in particular for integers.If you assume that coprime is in some intermediate ring, then inconjunction with R intersect Q is equal to Z, you get that theproperty is also vacuous.PROP. Let S be a subring of C which intersects Q in Z, and let x and ybe two integers. If x and y are coprime in the sense of (*) in anysubring of S, then they are coprime in Z; therefore, there exist r ands 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 arational prime p which divides x and y in Z, and therefore in S. Sincex and y are assumed coprime in S, it follows that p must be a unit inS. But that implies that 1/p lies in S intersect Q, which isimpossible. Therefore, x and y are coprime in Z in the sense of(*). It is well known that condition (*) in Z implies the existence ofr and s. If R is ANY ring containing the integers, and u is an elementof 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 arecoprime 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 Magidinmagidin@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 fordetermining whether a particular number is in the set or not. I don'tsee why this is so.I'm not defending James's definition here, but I don't see why adefinition should necessarily yield principles for determiningmembership. I don't have any examples at hand, but in principle, if Ican prove that there is a unique set satisfying some particularproperty, then that property is suitable for a definiens[1], whetheror not there is an easily applicable rule determining membership.Attack James's definition where one should: It's not at all clearwhether there is a unique set satisfying his property. Don't make uprules about what definitions must satisfy (like feasible membershiptests).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 acorntoward 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 definitesense in which defining a set _is_ the same as specifying whatthe elements are. In a sense - this doesn't say anything abouta 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 examplein mind where a set is defined in a way that does not insome 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 beentalking about.) Say I define S = {1} if Goldbach's conjecture istrue and S = {2} if Goldbach's conjecture is false. There is nofeasible test for membership in S. I nonetheless _have_ specifiedthe members of S, and given a test for membership: 1 is a member if and only if every even number > 2 is the sum oftwo 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 noproof that a unique structure satisfies the definition. Elementhoodtests (feasible or in principle or whatever) have nothing to do withit.I interpreted the elementhood complaint in terms of feasible tests,just because I can't figure out what the complaint is supposed to meanotherwise. -- So, at this time, I'd like to assure you that I am not interested inI'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 thatyou need a membership test. On the other hand it should be clear thateither an element is in the set or is not, and that should *not* bedependent on what is already put in the set or not. Also someconsistency 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 setif and only if TM-2 is not in the set. (It is the case that you canput TM-1 in the set if and only if TM-2 is in the set, but that is noproblem.)Compare James' ring (which has more structure than a set in itself).that in a number of cases two conjugate complex numbers can not gotogether in the ring, but that one of them should go in it. There isno way to show which one should go in, nor is there a way to show thatyour choice conflicts with other choices you make, or not, until youmay have to make a third choice that creates a conflicting situation.But even with his current requirements (that a number of algebraicintegers divided by 7 go into the ring) the definition can not be shownto 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 notmore difficult, than proving FLT.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: 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 noalgorithmic membership test. There certainly is a membershiptest 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 testis the best word for this (come to think of it membershipcriterion would be much better) there is no membership testin 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> === > 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 ita bit.I suppose whether or not I agree depends on what we count as adefinition. The Cantor-Bernstein(-Schroeder?) theorem constructs abijection in which subsequent choices depend on previous choices. Itis perhaps just a semantic quibble whether the proof of the theoremcould be said to *define* a bijection (by, say, appending it with thestatement, Let Gigglywiggly be the bijection thus constructed). Icould certainly sympathize if someone wanted to object that this isn'treally a definition, but I wouldn't be confident in averring one wayor the other.-- Run mathematicians, RUN!!! I'm coming for you. It may take a fewmonths, but I'll get [computer verification of my proof] and then yourlives 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 ofChoice is leering behind ;-).) You may need subsequent choices, andI think it is valid *when you do not need backtracking*. That is,at every point you have to chose, your choice will not invalidateall other possible choices at some future point. With James' deinitionwe may arrive at a position where we are stuck. Do we now have theObject ring? With other earlier choices we might have come at anotherpoint. With the theorem you cite you need only a bijection.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: 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 allproblematic. I don't know what you mean by subsequent choicesdepend 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 Xand Y. Start by considering the transitive closure of the relation onthe 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 differentgeneral types. For instance, one might have an element x which isnot 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 aninfinite 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 aone-to-one correspondence on the equivalence class which matchesthe 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'sa little bit of arbitrariness, in that in infinite chains or cycles, one mightequally well want to associate each element of X with the element ofY coming after it, or vice-versa. But this is just one arbitrary choicewe make once and for all. Having decided, once for all, the one-to-onecorrespondence is entirely explicit.Note in particular that Cantor-Bernstein is a theorem of ZF, not needingthe axiom of choice.Keith RamsayP.S. When I was making sure I correctly remembered which theoremwas known as Cantor-Bernstein, I got a hit for Cantor Bernstein atthe 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 theproof of that theorem defines a well-ordering.noggin.-- I AM serious about this being a short route to a Ph.d for some ofyou, but just remember, I'm the guy who proved Fermat's Last Theoremin 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 makeit 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, henceZ] which satisfy R intersect Q is equal to Z satisfy Zorn's Lemma?Note that the second condition given is vacuous, since two integersa,b are coprime (in the sense of having no common nonunit divisors inthe integers) if and only if there exist integers r and s such thatra+sb=1, so any common factor of a and b in any ring will necessarilybe 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 isa 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 theunion, then it lies in Q intersect one of the rings in C, hence liesin Z. So S has maximal elements. However, we have already seen that S doesnot have a ->maximum<- element, as you noted in your reply; since thedefinition does not have a referent. There is no such (uniquelydetermined) ring. There are many subrings R of C which satisfy both Rintersect 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 Magidinmagidin@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 acleverer lad than I am could toss off a fixed point constructionfairly easily in which determining whether a particular guy is anelement of the, say, greatest fixed point is not an easy task. If theconstruction also requires an application of the axiom of choice, onecould imagine that the task isn't really feasible in any reasonablesense.Too bad I'm not a cleverer lad than I am. -- Jesse F. HughesWhat you call reasonable is suspect since you've proven yourself tobe 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 Zwhere any two members commute.Then, I can't determine whether 30 and 16 are members of thetoken ring.This may not be an entirely satisfactory example.Recall Harris' definition:The Object Ring is a commutative ring that includes all numbers suchthat -1 and 1 are the only members that are both a unit and aninteger, where no non-unit member is a factor of any two integers thatare coprime.I assume that Harris is taking numbers to mean complex numbers.The word all makes the definition tricky. I am going to interpretit to mean a maximal ring satisfying the stated conditions, wherethere 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 notdepend upon properties of x, but rather on properties of theobject ring set itself. That is, the definition is choosingcertain subrings of the ring of complex numbers to have thename of object ring.But, even considering all that, I guess you could give thefollowing properties for determining whether x is a memberor not. Let x be a complex number. Find a subring S of Csuch x is in S, Z intersect units of S is {1, -1}, thereare no non-unit member of S that is a factor of anytwo integers that coprime, and there does not exista subring T of C that properly contains S, where Zintersect units of T is {1, -1}, and there are nonon-unit member of T that is a factor of any twointegers that are coprime. If no such subring S canbe 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 isa member of an object ring. The question is what arethe object rings themselves.In summary, the definition of object ring characterizescertain subrings of C. It does not characterize directlycertain elements of C. Thus, it differs from thedefinition of algebraic integer, which does characterizecertain elements of C. Hence, it is not required thatthere be a membership test for an element of C to bean 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 theelements - your definition of the token ring is no definitionat all, since there is more than one ring satisfying thegiven 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 definitionfor 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 inimportant ways. He talks about _the_ Object Ring,and he talks about _objects_, defining an objectof object _is_ analogous to the notion of algebraicinteger 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 defininga class of sets. My comments were regarding thesituation where one has defined a set.I really don't get this stuff about what happens ifwe assume he doesn't mean what he says butmeans something entirely different. If when hesays there's an error in core mathematics whathe actually means is that 2 + 2 = 4 then yes, whathe 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 notest for membership of the field of algebraic numbers, for example.It seems to me that there are two types of definitions in commonusage. The first is of the type An algebraic number is a complexnumber which satisfies a polynomial equation over the integers andis completely self-sufficient and unambiguous. The second type, like The Fitting subgroup of a group G is defined to be the largest normalnilpotent 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 decidingwhether 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 aunique maximal subring of the complex numbers (?) that had the requiredproperties. 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 meArturo and Dik, meant the phrase: x is an algebraic number if andonly if it is a root of some polynomial with integer coefficients.Yes, there are senses in which that's not a test, but thereis also a much weaker sense in which it _is_ a test, and thepoint about the definition of the Object Ring is that it doesnot 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 Fittingsubgroup if and only if it is in a nilpotent subgroup which isin 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 ascoherent as the largest normal nilpotent subgroup. Lemmetry again:The Object Ring is a commutative ring that includes all numbers suchthat -1 and 1 are the only members that are both a unit and aninteger, where no non-unit member is a factor of any two integers thatare coprime.Nope, I can't make sense of this. When you say the Fitting subgroupis the largest normal nilpotent subgroup it's not clear to me thatthere 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 thenormal nilpotent subgroups of G and checks that H contains allthe others. I can't figure out how to tell whether a ring is theObject 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 thelargest subring (or the unique maximal subring) such that [etc]then I'd know at least what the definition meant. But that's notwhat he said.>Not much chance of that though, is there?Derek Holt.************************David C. Ullrich <87vfnvwy7a.fsf@phiwumbda.org> <1rh8vvsdekd46h57vq7iro8a5f9485k7kl@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, anydefinition of, say, X, comes with the principle that x is in X iff xis in the unique set with satisfies the definiens for X. The sameprinciple would trivially apply to James's definition, if there wereindeed 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' bymodern 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 thedefinition? Here it is:The Object Ring is a commutative ring that includes all numbers suchthat -1 and 1 are the only members that are both a unit and aninteger, where no non-unit member is a factor of any two integers thatare coprime.I _don't_ see an intelligible condition on the ring or on the elementsof the ring there: The ring is supposed to include all numbers witha certain property, but the stated property is not a property thata number can have! includes all numbers suchthat -1 and 1 are the only members that are both a unit and aninteger_Is_ it true that some numbers satisfy the property -1 and 1 are the only members that are both a unit and aninteger? No. The phrase all numbers such that -1 and 1 are the onlymembers that are both a unit and aninteger simply makes no sense.>Therefore, this cannot be the sense in which James's definition fails.************************David C. Ullrich <87vfnvwy7a.fsf@phiwumbda.org> <1rh8vvsdekd46h57vq7iro8a5f9485k7kl@4ax.com> <877k0bt8l5.fsf@phiwumbda.org> <3tqavv0kuedgeiiuaesnpm4aeup1a9b5d9@4ax.com> === > 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> integerI agree that this definition fails to pick out a unique ring. This isa bad definition. I disagree that the problem has to do with whetheror not there is a principle for elementhood.One might as well complain that it's a bad definition because we haveno means for determining subsethood, the transitive closure, this andthat other thing. But none of this is really to the point. Adefinition is acceptable iff there is provably a unique structuresatisfying the definiens. Now, you seem to interpret the membership test liberally: Anylegitimate definition comes with a principle for determiningelementhood. I tended to interpret the complaint differently: oneneeds an elementhood test for a definition to be legitimate whether ornot there is a unique structure satisfying the definiens. The latterinterpretation is simply and plainly false. The former interpretation(yours?) makes the complaint seemingly valid, but rather indirect andconfusing. The problem has nothing to do with elementhood, but afailure to satisfy the unique existence part.Put another way: if your interpretation is correct, then a definitionhas a membership test if and only if it satisfies the unique existenceclause. James's definition is problematic because it fails the uniqueexistence clause. As corollary, it fails the membership test, butonly in a funky way. There is no particular set X which thedefinition picks out, and so there is no test for whether a given x isin X (because there's no privileged X!). Just seems a funny way tocriticize the definition.At no point did I try to claim that James has offered a validdefinition, but only that the elementhood test is either a veryindirect complaint or simply based on a false intuition aboutdefinitions.> _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 onegovernor..., a senator... and I've had some really good seats atsports 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> integerI 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> <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 whileI sit in Oklahoma City. Sometimes, the editor lags a bit and Ioverlook errors.-- Jesse F. HughesAnd hey, if you're moping and miserable because mathematics tests you,then maybe, if you think you're a mathematician, you might want to trya 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 intowhile you were in the Netherlands you'd ask for anotherbreak, 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 bestway of expressing themselves. Dik and others' complaint is notnecessarily that there is no black box we can put a complex numberinto and decide if it is or is not in the object ring. The complaintis really that the statement given is not sufficient to determine whatis 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 whichtwo conditions are met. If we defined the Fitting subgroup as the normal subgroup of G whichis nilpotent, then clearly we have not provided a coherentdefinition. If we try to define the Uberabelian Subgroup of G as thesubgroup of G in which any two elements commute, we would have thesame sort of problem.-- === ===============================================It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === =============================================== Arturo Magidinmagidin@math.berkeley.edu <877k0bt8l5.fsf@phiwumbda.org> === > 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 complaintregarding James's definition and also the characterization that thiscomplaint 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 uniquestructure.(2) Whether or not his definition yields a means of determining whichcomplex numbers are elements of that structure.Obviously, if he fails the first (as he has), then the second isn'treally applicable. But, if he succeeds in the first, then the secondisn't particularly relevant in evaluating whether he has given aproper definition -- at least not in the way I read the second. He'sgiven 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 signslike 'Will work for food' in the future... I will not show mercygoing forward. I was trained as a soldier in the United States Armyafter 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 thecomplex numbers which contains the integers, the property alluded tois equivalent to the statement that the intersection of R with Q isequal 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, thenthe 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.QEDThe ring of all algebraic integers has this property, as does anysubring that contains Z. Other subrings of C also have the property:Z[a] does, for any transcendental number a. There are also certainsubrings of the algebraic numbers which do not consist only ofalgebraic integers, but have the property. Someone had posted a nicecharacterisation of some of these rings, but I could not find itthrough google. But one possible construction would be to pick aquadratic extension K of Q, and a rational prime p which splits intotwo distinct prime ideals, (p) = PQ. Then take the ring of integers Aof K, and invert all elements in P-Q; this ring is not contained inthe algebraic integers, but intersects Q at Z.On the other hand, it's already been pointed out that anygeneralization of the algebraic integers which is closed under Galoisconjugates is unlikely to be of use for the purpose. See BillDubuque'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 Magidinmagidin@math.berkeley.edu === Turing's test design is flawed.It is not double blind: the tester knows some are machines and someare 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 ===