mm-1189 === Subject: Re: No Set Contains Every Computable Natural You cant prove that an infinite set exists. > You have to assume it exists. > This is why ZFC has the Axiom of Infinity. In order to prove omega is a set you need the axiom of infinity. However to prove it is infinite you only need its definition and the definition of an ordinal. The axiom of infinity actually says nothing about the existence of infinite sets, mearly that a certain type of set is constructable. > Now consider Cantors diagonal proof. > The input is an infinite list of real numbers. > Obviously, this list is not finite. > The algorithm requires that we compute an > infinite number of digits. Again, not finite. > The output is a real number. > This number probably doesnt have a finite number of digits. > Cantors proof completely fails to be an algorithm. I dislike Cantors diagonal proof, since it requires you contruct something you can prove unconstructable. However a proof is not required to be an algorithm. There are better ways to prove the same result. > There is no way that Cantors proof that the reals are > uncountable can be computed by a TM in a finite > number of steps. In other words, a TM can easily handle FOL. > If we redefine algorithm to include TMs that can > perform an infinite number of operations, we have > to deal with the many paradoxes that result. So dont bother. Does a TM Ôunderstand what 7 is? If not why does it have to able to understand what the set of reals is? All it has to know is what it can do with it. I read somewhere that the Bourabki(sp?) definiton of 1 needs around 4*10^15 terms. My (approximately TM equivelent) computer can;t even address that many terms, yet it will still tell me the result of 1+1. > If we accept Cantors proof then we must also accept > these other proofs as well. If you confuse the map with the territory of course you have to accept the consequences. -- If you can read this youve gone too far. Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: No Set Contains Every Computable Natural > I dislike Cantors diagonal proof, since it requires you contruct something you can prove unconstructable. However a proof is not required to be an algorithm. If you think that, then you dont understand the proof. The object that is constructed does indeed exist. Possibly you are objecting to the reductio ad absurdum form of the proof. Cantor did not formulate the proof that way. > There are better ways to prove the same result. Would you care to show us your improvement on the following? Theorem (Cantor). X < 2^X for each cardinal X. Proof. There is an obvious injection from X into 2^X. Let f: X -> 2^X be given. We are to show that f is not a surjection. Let S = { x in X : not(x in f(x)) }. Then S is a member of 2^X and is clearly not in the range of f, Q.E.D. Corollary. |N| < |2^R|. Proof. There is a natural bijection between 2^N, the power set of the naturals, and C, the Cantor set, which is therefore an uncountable subset of the reals. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: No Set Contains Every Computable Natural something you can prove unconstructable. However a proof is not >required to be an algorithm. > If you think that, then you dont understand the proof. The object > that is constructed does indeed exist. > Possibly you are objecting to the reductio ad absurdum form of the > proof. Cantor did not formulate the proof that way. Ok, Ive seen many different forms of the proof but I may be getting the names wrong. What I was referring to was the creating a list of all the reals written base n form. Then making a number where the first digit is different to the first digit of the first number on the list, the second digit is different to the second digit of the second number etc... I may be wrong in calling this Cantor Diaganol proof. Please inform me if I am, finals are soming up soon. >There are better ways to prove the same result. > Would you care to show us your improvement on the following? The proof you gave is what I was referring to as the better proof. -- If you can read this youve gone too far. Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: No Set Contains Every Computable Natural >>I dislike Cantors diagonal proof, since it requires you contruct >>something you can prove unconstructable. However a proof is not >>required to be an algorithm. >> If you think that, then you dont understand the proof. The object >> that is constructed does indeed exist. >> Possibly you are objecting to the reductio ad absurdum form of the >> proof. Cantor did not formulate the proof that way. > Ok, Ive seen many different forms of the proof but I may be getting the > names wrong. What I was referring to was the creating a list of all the > reals written base n form. Then making a number where the first digit is > different to the first digit of the first number on the list, the second > digit is different to the second digit of the second number etc... I may > be wrong in calling this Cantor Diaganol proof. Please inform me if I > am, finals are soming up soon. That is often called the Cantor diagonal proof, but notice that it is not necessary to start by creating a list of all the reals. That is what I meant by not using a reductio ad absurdum approach. It is sufficient to consider a mapping f: N -> R and then show that such a mapping cannot be a surjection. >>There are better ways to prove the same result. >> Would you care to show us your improvement on the following? > === Subject: Re: No Set Contains Every Computable Natural not necessary to start by creating a list of all the reals. That is > what I meant by not using a reductio ad absurdum approach. It is > sufficient to consider a mapping f: N -> R and then show that such a > mapping cannot be a surjection. Yes and this can be done without recourse to Cantors diagonal proof. Its not the result I have difficulty with, but the diagonal proof itself. I fully accept that the result is true from the mapping argument, but not that DP is a proof that |R| > |N|. > That version of the proof can also be recast as a proof by > contradiction. Is your objection to the concept of using base-n > numbers, or is it the concept of an indirect proof that you dislike? It seems that you would have to show that the number being generated in the DP grows faster than the list of numbers you are generating it from. I have no problems with base n numbers, and I dont know what you mean by indirect proof. -- If you can read this youve gone too far. Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: No Set Contains Every Computable Natural >> Huh? That *is* Cantors diagonal proof (one of them, anyway). The >> set { x in X : not(x in f(x)) } is the diagonal in the proof. >> Likewise, the set of all Turing machines m that do not halt with input >> m is adiagonal in Turings proof of the unsolvability of the halting >> problem. > Oh. I wasnt aware of that. Every thing Ive read on the subject seemed > to imply that the two were different proofs of the same fact. The one > being presented as Cantors Diagonal Proof and the other as Cantors > Theorem. For some reason it never clicked that they were the same thing. > Something still doesnt seem quite right but Im sure Ill figure it > /me goes to find a better textbook. Cantor proved the uncountability of the reals in more than one way. The decimal-digit version of the proof is an immediate consequence of the argument for power sets. Cantors original proof of the uncountability of the reals was not a diagonal argument. It used a sequence of nested intervals. It has frequently been discussed in this newsgroup. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: No Set Contains Every Computable Natural set { x in X : not(x in f(x)) } is the diagonal in the proof. > Likewise, the set of all Turing machines m that do not halt with input > m is adiagonal in Turings proof of the unsolvability of the halting > problem. Oh. I wasnt aware of that. Every thing Ive read on the subject seemed to imply that the two were different proofs of the same fact. The one being presented as Cantors Diagonal Proof and the other as Cantors Theorem. For some reason it never clicked that they were the same thing. Something still doesnt seem quite right but Im sure Ill figure it /me goes to find a better textbook. -- If you can read this youve gone too far. Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: No Set Contains Every Computable Natural >> That is often called the Cantor diagonal proof, but notice that it is >> not necessary to start by creating a list of all the reals. That is >> what I meant by not using a reductio ad absurdum approach. It is >> sufficient to consider a mapping f: N -> R and then show that such a >> mapping cannot be a surjection. > Yes and this can be done without recourse to Cantors diagonal proof. Huh? That *is* Cantors diagonal proof (one of them, anyway). The set { x in X : not(x in f(x)) } is the diagonal in the proof. Likewise, the set of all Turing machines m that do not halt with input m is a diagonal in Turings proof of the unsolvability of the halting problem. > Its not the result I have difficulty with, but the diagonal proof > itself. I fully accept that the result is true from the mapping > argument, but not that DP is a proof that |R| > |N|. So you like some diagonals better than others, apparently. >> That version of the proof can also be recast as a proof by >> contradiction. Is your objection to the concept of using base-n >> numbers, or is it the concept of an indirect proof that you dislike? > It seems that you would have to show that the number being generated in > the DP grows faster than the list of numbers you are generating it from. > I have no problems with base n numbers, and I dont know what you mean > by indirect proof. Indirect proof means proof by contradiction, or a reductio ad absurdum proof. The version of the proof that you mentioned begins by asserting that you have a complete list of real numbers, and then proceeds to derive a contradiction. That approach is not needed. It is sufficient to start with a list (actually a mapping f: N -> R), and then show that the mapping is not a surjection. Thats what I mean by a direct proof. There is no number being generated in the proof. There is a number that is generated, fully and completely in a single step, just as the diagonal set S is generated fully and completely by the formula S = { x in X : not(x in f(x))}. The number given by d = sum_n a_n * 10^n, where a_n = 1, if the n-th digit of f(n) is not 1, a_n = 2, otherwise, is a constant. It does not grow. The number d differs from each number in the given list in at least one decimal place, and it is constructed in a way that guarantees that dual representation (the 0.999... = 1 phenomenon) is not an issue, since there are no 9s or 0s after the decimal point. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: No Set Contains Every Computable Natural Discussion, linux) >>I dislike Cantors diagonal proof, since it requires you contruct >>something you can prove unconstructable. However a proof is not >>required to be an algorithm. >> If you think that, then you dont understand the proof. The object >> that is constructed does indeed exist. >> Possibly you are objecting to the reductio ad absurdum form of the >> proof. Cantor did not formulate the proof that way. > Ok, Ive seen many different forms of the proof but I may be getting the > names wrong. What I was referring to was the creating a list of all the > reals written base n form. Then making a number where the first digit is > different to the first digit of the first number on the list, the second > digit is different to the second digit of the second number etc... I may > be wrong in calling this Cantor Diaganol proof. Please inform me if I > am, finals are soming up soon. Is that number really unconstructable, given the function N -> [0,1] which was the starting assumption? In what sense is it unconstructable? Looks like an algorithm to me. -- Jesse F. Hughes The sole cause of all human misery is the inability of people to sit quietly in their rooms. -- Blaise Pascal === Subject: Re: No Set Contains Every Computable Natural <87oerravz2.fsf@phiwumbda.orgOk, Ive seen many different forms of the proof but I may be getting >the names wrong. What I was referring to was the creating a list of >all the reals written base n form. Then making a number where the >first digit is different to the first digit of the first number on >the list, the second digit is different to the second digit of the >second number etc... I may be wrong in calling this Cantor Diaganol >proof. Please inform me if I am, finals are soming up soon. > Is that number really unconstructable, given the function N -> [0,1] > which was the starting assumption? > In what sense is it unconstructable? Looks like an algorithm to me. Unconstructable was not the right word. The trouble I have is I find it very hard to pinpoint what seems wrong about it to me. Its just every time I look at it or think about it, something just doesnt seem quite right about it. Obviously it cant be constructed in finite time, but can be easily defined. Ill try to define my problem in more detail. At step n of the algorithm you have an n digit number, x, which cannot be the initial part one of the first n elements of the list. However it seems obvious that there exist at least aleph_0 elements in the list which do have x as their initial parts. This will be true for any and all n. Hence any possible x must be a member of the list. Or does this show that the list has a size strictly aleph_0 greater than itself and hence a contradiction? -- If you can read this youve gone too far. Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: No Set Contains Every Computable Natural > Ill try to define my problem in more detail. At step n of the algorithm > you have an n digit number, x, which cannot be the initial part one of > the first n elements of the list. However it seems obvious that there > exist at least aleph_0 elements in the list which do have x as their > initial parts. This will be true for any and all n. Hence any possible x > must be a member of the list. Or does this show that the list has a size > strictly aleph_0 greater than itself and hence a contradiction? Consider the following list of rational numbers between 0 and 1: 1/2 1/3 2/3 1/4 3/4 1/5 2/5 3/5 4/5 1/6 5/6 and so on. The numbers 6/11, 7/13, and 8/15 (among others) start with .5 . Would you argue that all three of those are the first element of the list? Or, would you argue that 1/6, 1/7, and 1/8, and sqrt(2)/10 are all the tenth element of the list (since they all start with .1)? -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: No Set Contains Every Computable Natural >>Ok, Ive seen many different forms of the proof but I may be getting >>the names wrong. What I was referring to was the creating a list of >>all the reals written base n form. Then making a number where the >>first digit is different to the first digit of the first number on >>the list, the second digit is different to the second digit of the >>second number etc... I may be wrong in calling this Cantor Diaganol >>proof. Please inform me if I am, finals are soming up soon. >> Is that number really unconstructable, given the function N -> [0,1] >> which was the starting assumption? >> In what sense is it unconstructable? Looks like an algorithm to me. > Unconstructable was not the right word. The trouble I have is I find it > very hard to pinpoint what seems wrong about it to me. Its just every > time I look at it or think about it, something just doesnt seem quite > right about it. Obviously it cant be constructed in finite time, but > can be easily defined. > Ill try to define my problem in more detail. At step n of the algorithm > you have an n digit number, x, which cannot be the initial part one of > the first n elements of the list. However it seems obvious that there > exist at least aleph_0 elements in the list which do have x as their > initial parts. This will be true for any and all n. Hence any possible x > must be a member of the list. Or does this show that the list has a size > strictly aleph_0 greater than itself and hence a contradiction? What do you mean by step n of the algorithm? I dont see anything in the argument that is concerned with terminating decimals or with finite subsets of the given list at all. If you dont think it is possible to perform operations on infinite sets, then you also must think the real numbers do not exist, in which case the question of whether they are countable simply does not arise. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: No Set Contains Every Computable Natural <87oerravz2.fsf@phiwumbda.org> Discussion, linux) >>The algorithm for generating a real number that is (supposedly) not in >>the list of all reals, showing that no such list exists. The nth step >>would be setting the nth digit of x to a digit which is not equal the >>nth digit of the nth number in the list. I know its not technically >>stated as an algorithm, but treating it as one which runs in infinite >>time saves on verbage (and as far as Im aware isnt invalid). > It also avoids all those embarrasing questions about the validity > of the algorithm. >> But the proof makes no reference to any partial results of this process. >> The proof mentions only the completed number which is the final result. >> You are getting yourself bogged down in irrelevant details by worrying >> about what happens if you stop the process at some finite stage instead >> of carrying it through to completion. All that matters is that the >> number thus produced is not in the given list (not the list of all real >> numbers, because that assumption is not needed). > Why get bogged down in irrelevant details? > I can compute a natural number that is not on the tape. > That is all that matters. Russell, I know this is hard to believe, but Stephen and Dave found something to discuss that did not include your completely moronic assertions. > If you accept that real numbers exist and are complete, > then you have to agree there exists a real number whose > binary representation is a solution to the halting problem. Duh. -- Jesse Hughes Besides, discoverers are too proud to kiss butt. Indiana Jones would never kiss some academics ass to get published, and neither will I. --James Harris === Subject: Re: No Set Contains Every Computable Natural >The algorithm for generating a real number that is (supposedly) not in >the list of all reals, showing that no such list exists. The nth step >would be setting the nth digit of x to a digit which is not equal the >nth digit of the nth number in the list. I know its not technically >stated as an algorithm, but treating it as one which runs in infinite >time saves on verbage (and as far as Im aware isnt invalid). It also avoids all those embarrasing questions about the validity of the algorithm. > But the proof makes no reference to any partial results of this process. > The proof mentions only the completed number which is the final result. > You are getting yourself bogged down in irrelevant details by worrying > about what happens if you stop the process at some finite stage instead > of carrying it through to completion. All that matters is that the > number thus produced is not in the given list (not the list of all real > numbers, because that assumption is not needed). Why get bogged down in irrelevant details? I can compute a natural number that is not on the tape. That is all that matters. People keep complaining that the input is infinite. I have no idea how they can make this argument with a straight face. My proof shows that the input tape is finite. They keep claiming they know the input is infinite. This is like me claiming Cantors proof is wrong because I know the reals are countable. I point out that Cantors proof requires an infinite input. They say that some algorithms can have infinite input. (Some of them even claim they can prove the reals are uncountable in a finite number of steps. The proof must be too long to fit in a posting, because none of them have actually presented such a proof.) What is their criteria for algorithms that accept infinite input? It seems like proofs they like are acceptable, but proofs they dont like are wrong. Russell - 2 many 2 count ps If you accept that real numbers exist and are complete, then you have to agree there exists a real number whose binary representation is a solution to the halting problem. === Subject: Re: No Set Contains Every Computable Natural >> Ill try to define my problem in more detail. At step n of the >> algorithm you have an n digit number, x, which cannot be the initial >> part one of the first n elements of the list. However it seems >> obvious that there exist at least aleph_0 elements in the list which >> do have x as their initial parts. This will be true for any and all >> n. Hence any possible x must be a member of the list. Or does this >> show that the list has a size strictly aleph_0 greater than itself >> and hence a contradiction? What do you mean by step n of the algorithm? I dont see anything >in the argument that is concerned with terminating decimals or with >finite subsets of the given list at all. > The algorithm for generating a real number that is (supposedly) not in > the list of all reals, showing that no such list exists. The nth step > would be setting the nth digit of x to a digit which is not equal the > nth digit of the nth number in the list. I know its not technically > stated as an algorithm, but treating it as one which runs in infinite > time saves on verbage (and as far as Im aware isnt invalid). It hardly saves on verbage, since it generates verbage. Since it doesnt matter what the ones or twos or plus do, its the *zeroes* than generate the verbage, O Set Theory wankers of infinite blunder. > Lets see if I can explain myself better with a table: > N | R (base 2| possible x > --+----------+----------- > 1 | 0.010... | 0.1 > 2 | 0.110... | 0.10 > 3 | 0.100... | 0.101 > . . . > . . . > . . . > There surely are aleph_0 members of the R column that begin with /any/ > given value in the x column, and at least one of those must be x00000... > = x. Hence all x are in the R column. Youve explained nothing. Even worse youve unexplained everything. Since Cantor invented the diagonal proof for the express purpose of getting rid of mathematicians and their idiot dots. === Subject: Re: No Set Contains Every Computable Natural > Is x not a 0 error approximation to x? X itself is not in that column. > No, but its not the same situation. No where is it claimed the list > should contain 1/3. Why would such a claim be any less tenable than your claim about x? -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: No Set Contains Every Computable Natural <87n07fo9qr.fsf@phiwumbda.org> <87k72jnxtp.fsf@phiwumbda.org> <87ptcau3xj.fsf@phiwumbda.org> <87oerravz2.fsf@phiwumbda.org> <1g9kek3.dw2e85h6xqf6N%panoptes@iquest.netThe algorithm for generating a real number that is (supposedly) not >in the list of all reals, showing that no such list exists. The nth >step would be setting the nth digit of x to a digit which is not >equal the nth digit of the nth number in the list. > The nth step is IDENTIFYING the nth digit of x as a digit which is not > equal to the nth digit of the nth number in the list. Ah, I dont think Ive seen it stated in those terms. >There surely are aleph_0 members of the R column that begin with >/any/ given value in the x column, and at least one of those must be >x00000...= x. Hence all x are in the R column. > Hence all approximations to x are probably in the R column. There is > only one x. Is x not a 0 error approximation to x? > Consider the following list: > 1/4 > 5/16 = 1/4 + 1/16 > 21/64 = 5/16 + 1/64 > 85/256 = 21/64 + 1/256 > ... > Note that each element of the list, if written in lowest terms, has a > denominator which is a power of 4. Would you assert that 1/3 is on > the list? No, but its not the same situation. No where is it claimed the list should contain 1/3. -- If you can read this youve gone too far. Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: No Set Contains Every Computable Natural >>Ill try to define my problem in more detail. At step n of the >>algorithm you have an n digit number, x, which cannot be the initial >>part one of the first n elements of the list. However it seems >>obvious that there exist at least aleph_0 elements in the list which >>do have x as their initial parts. This will be true for any and all >>n. Hence any possible x must be a member of the list. Or does this >>show that the list has a size strictly aleph_0 greater than itself >>and hence a contradiction? >> What do you mean by step n of the algorithm? I dont see anything >> in the argument that is concerned with terminating decimals or with >> finite subsets of the given list at all. > The algorithm for generating a real number that is (supposedly) not in > the list of all reals, showing that no such list exists. The nth step > would be setting the nth digit of x to a digit which is not equal the > nth digit of the nth number in the list. I know its not technically > stated as an algorithm, but treating it as one which runs in infinite > time saves on verbage (and as far as Im aware isnt invalid). But the proof makes no reference to any partial results of this process. The proof mentions only the completed number which is the final result. You are getting yourself bogged down in irrelevant details by worrying about what happens if you stop the process at some finite stage instead of carrying it through to completion. All that matters is that the number thus produced is not in the given list (not the list of all real numbers, because that assumption is not needed). It doesnt matter how many numbers in the list are partial matches for the number defined by the diagonal argument; all that matters is that no number can be a complete match. Each f(n) differs at least in the n-th digit. > Lets see if I can explain myself better with a table: > N | R (base 2| possible x > --+----------+----------- > 1 | 0.010... | 0.1 > 2 | 0.110... | 0.10 > 3 | 0.100... | 0.101 > . . . > . . . > . . . > There surely are aleph_0 members of the R column that begin with /any/ > given value in the x column, and at least one of those must be x00000... >= x. Hence all x are in the R column. Your possible x column has nothing to do with whats actually in the proof. No one claimed you couldnt have numbers in the list that partially match the generated number. Apparently it does not bother you that the same sort of pseudo-argument could be applied to the power-set version of the diagonal proof, where you could take definitions of the diagonal on finite subsets of the given set X and then wave your hands and argue about partial matches on those sets. -- Dave Seaman Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: No Set Contains Every Computable Natural > The algorithm for generating a real number that is (supposedly) not in > the list of all reals, showing that no such list exists. The nth step > would be setting the nth digit of x to a digit which is not equal the > nth digit of the nth number in the list. The nth step is IDENTIFYING the nth digit of x as a digit which is not equal to the nth digit of the nth number in the list. > I know its not technically > stated as an algorithm, but treating it as one which runs in infinite > time saves on verbage (and as far as Im aware isnt invalid). > Lets see if I can explain myself better with a table: > N | R (base 2| possible x | | approximation to x > --+----------+----------- > 1 | 0.010... | 0.1 > 2 | 0.110... | 0.10 > 3 | 0.100... | 0.101 > . . . > . . . > . . . > There surely are aleph_0 members of the R column that begin with /any/ > given value in the x column, and at least one of those must be x00000... > = x. Hence all x are in the R column. Hence all approximations to x are probably in the R column. There is only one x. Consider the following list: 1/4 5/16 = 1/4 + 1/16 21/64 = 5/16 + 1/64 85/256 = 21/64 + 1/256 ... Note that each element of the list, if written in lowest terms, has a denominator which is a power of 4. Would you assert that 1/3 is on the list? -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: No Set Contains Every Computable Natural <87oerravz2.fsf@phiwumbda.org> algorithm you have an n digit number, x, which cannot be the initial >part one of the first n elements of the list. However it seems >obvious that there exist at least aleph_0 elements in the list which >do have x as their initial parts. This will be true for any and all >n. Hence any possible x must be a member of the list. Or does this >show that the list has a size strictly aleph_0 greater than itself >and hence a contradiction? > What do you mean by step n of the algorithm? I dont see anything > in the argument that is concerned with terminating decimals or with > finite subsets of the given list at all. The algorithm for generating a real number that is (supposedly) not in the list of all reals, showing that no such list exists. The nth step would be setting the nth digit of x to a digit which is not equal the nth digit of the nth number in the list. I know its not technically stated as an algorithm, but treating it as one which runs in infinite time saves on verbage (and as far as Im aware isnt invalid). Lets see if I can explain myself better with a table: N | R (base 2| possible x --+----------+----------- 1 | 0.010... | 0.1 2 | 0.110... | 0.10 3 | 0.100... | 0.101 . . . . . . . . . There surely are aleph_0 members of the R column that begin with /any/ given value in the x column, and at least one of those must be x00000... = x. Hence all x are in the R column. -- If you can read this youve gone too far. Stephen Jones (Zombywuf) D5A4 5342 E7BD E524 710A E44F E997 4422 7FEC E44E === Subject: Re: No Set Contains Every Computable Natural >> He wont *if* we apply the condition that his input tape has finitely >> many ones, but there is nothing about that convention that is forced >> on us. Why not allow that he has a tape with all of N? After all, >> that assumption is not his problem. His problem is only in reasoning >> about whats on the tape after an infinite number of steps. > Yes, of course I agree that a TM can have anything written on its tape >to start. If thats the case then it would seem to me that it is very easy for a TM to solve the halting problem (although procuring an appropriate input tape might be a little tricky). What am I missing? Alan -- Defendit numerus === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p Discussion, linux) > He wont *if* we apply the condition that his input tape has finitely > many ones, but there is nothing about that convention that is forced > on us. Why not allow that he has a tape with all of N? After all, > that assumption is not his problem. His problem is only in reasoning > about whats on the tape after an infinite number of steps. >> Yes, of course I agree that a TM can have anything written on its tape >>to start. > If thats the case then it would seem to me that it is very easy for > a TM to solve the halting problem (although procuring an appropriate > input tape might be a little tricky). What am I missing? What does it mean to solve the halting problem? As far as I know it means that, given a tape consisting of two natural numbers m and n, the machine halts and outputs 1 iff machine m halts on input n and halts and outputs 0 else. I dont see how allowing arbitrary input tapes (including an infinite tape, say, the characteristic function of the halting problem) helps a bit. No tapes with an infinite number of 1s are relevant for whether a particular machine solves the halting problem. -- Jesse Hughes Certainly he who can digest a second or third ßuxion need not, methinks, be squeamish about any point in divinity. George Berkeley, 1734 === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org What does it mean to solve the halting problem? As far as I know it > means that, given a tape consisting of two natural numbers m and n, > the machine halts and outputs 1 iff machine m halts on input n and > halts and outputs 0 else. I dont see how allowing arbitrary input > tapes (including an infinite tape, say, the characteristic function of > the halting problem) helps a bit. > No tapes with an infinite number of 1s are relevant for whether a > particular machine solves the halting problem. Let RM(x) = Russells machine (a Turing machine) with *any* input x Let x = the string giving: cell i equals 1 iff i(th) TM halts on empty input. Now, you might recall from your first course in complexity theory that (*) Given m and n, TM m halts on input n is reducible to (**) Given m, TM m halts on empty input You might also recall the power of Turing machines to be able to call other machines (simulation.) To solve (**), create a machine M that takes an input m. Then simulate RM(x) for the above x, move to tape cell m and get the result. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> <87ptc6df8r.fsf@phiwumbda.org> <873c92coz3.fsf@phiwumbda.org> <87r7wmb1uu.fsf@phiwumbda.org> The anolog is again that of you saying but if we introduce transfinite > numbers, *then* there is a natural number greater than any other natural > number. No, the unsolvability of the halting problem under discussion > is a standard result of standard definitions. If you change those > definitions, such as introducing TMs with a non-standard kind of tape, > then youre *not* referring to the same unsolvability result. Right, as I said, this was what I was trying to explain to the original poster. But he wanted to interpret Turing Machines to mean much more than they are defined to do (like doing an infinite amount of work in a finite amount of time) and I was merely showing that it Ôsolves the halting problem if he allows that. That can either be interpreted as that extension being invalid or interpreted as the new model of computation is stronger than the old one. In either way, he shouldnt have been using the term Turing Machines. J === Subject: Re: No Set Contains Every Computable Natural >I see you still havent understood the objection to this kind >of misuse of terminology. Specifically, the halting problem >_for_finite_inputs_ is unsolvable. Since it is *defined* as >being _for_finite_inputs_, it is a misuse of terminology to >say that it is solved by a TM with inputs that are not finite. > And I am simply saying that the halting problem (for finite inputs) is > easily (in fact, trivially) solvable when a (finite input) turing machine > has the power to invoke an oracle tape that solves the halting problem. Thats just another version of the same kind of misuse of terminology. The anolog is again that of you saying but if we introduce transfinite numbers, *then* there is a natural number greater than any other natural number. No, the unsolvability of the halting problem under discussion is a standard result of standard definitions. If you change those definitions, such as introducing TMs with a non-standard kind of tape, then youre *not* referring to the same unsolvability result. This isnt a merely semantical issue, because you incorrectly claim that certain nonstandard TMs solve the halting problem and thereby show an inconsistency inherent in the use of tapes with infinite input. The only inconsistency that has so far been shown is in your misuse of terminology, not in the use of tapes with infinite input. >Its quite frustrating that you continue to ignore this point. > Im not ignoring the point... I tried to use the fact that turing > machines are defined with Ôfinite input but Russell kept insisting that > his magical machines had infinite input, and so I was showing him that > such machines would be solving the kinds of problems people have already > shown to be undecidable. But you havent shown that. --r.e.s. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> <87ptc6df8r.fsf@phiwumbda.org> <873c92coz3.fsf@phiwumbda.org> <87r7wmb1uu.fsf@phiwumbda.org> of misuse of terminology. Specifically, the halting problem > _for_finite_inputs_ is unsolvable. Since it is *defined* as > being _for_finite_inputs_, it is a misuse of terminology to > say that it is solved by a TM with inputs that are not finite. And I am simply saying that the halting problem (for finite inputs) is easily (in fact, trivially) solvable when a (finite input) turing machine has the power to invoke an oracle tape that solves the halting problem. > Its quite frustrating that you continue to ignore this point. Im not ignoring the point... I tried to use the fact that turing machines are defined with Ôfinite input but Russell kept insisting that his magical machines had infinite input, and so I was showing him that such machines would be solving the kinds of problems people have already shown to be undecidable. > Its simply a misuse of the terminology, as Im quite sure you know, > so why keep it up? Again, this argument has all been under the premise that these infinite tape machine can behave like normal TMs. I hope you understand this now. J === Subject: Re: No Set Contains Every Computable Natural >> As you note that it does, above. >As I note that it does above *when* the input string is finite. > Which is what Ive been using ALL ALONG. >> Again, n is not an infinite tape, Im talking about the normal, known, >> undecidable problem (with finite n). Im also operating under the premise >> that these magical machines that can handle the infinite tapes can >> actually process the whole (infinite) tape like how Russell was originally >> doing. >Thats fine, because even if we adopt the convention that allows an >output after omega steps, the halting problem is still undecidable. > Well, not if that machine is equipped with an orcale tape solving the > halting problem (that is, the halting problem for finite inputs.) I see you still havent understood the objection to this kind of misuse of terminology. Specifically, the halting problem _for_finite_inputs_ is unsolvable. Since it is *defined* as being _for_finite_inputs_, it is a misuse of terminology to say that it is solved by a TM with inputs that are not finite. Its quite frustrating that you continue to ignore this point. To give a very crude analogy, its as though we were to agree that there is no natural number larger than all other natural numbers -- and then you were to say but if we introduce transfinite numbers, *then* theres a natural number greater than all other natural numbers. Its simply a misuse of the terminology, as Im quite sure you know, so why keep it up? --r.e.s. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> <87ptc6df8r.fsf@phiwumbda.org> <873c92coz3.fsf@phiwumbda.org> <87r7wmb1uu.fsf@phiwumbda.org As you note that it does, above. > As I note that it does above *when* the input string is finite. Which is what Ive been using ALL ALONG. > Again, n is not an infinite tape, Im talking about the normal, known, >undecidable problem (with finite n). Im also operating under the premise >that these magical machines that can handle the infinite tapes can >actually process the whole (infinite) tape like how Russell was originally >doing. > Thats fine, because even if we adopt the convention that allows an > output after omega steps, the halting problem is still undecidable. Well, not if that machine is equipped with an orcale tape solving the halting problem (that is, the halting problem for finite inputs.) J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> <87ptc6df8r.fsf@phiwumbda.org> <873c92coz3.fsf@phiwumbda.org> <87r7wmb1uu.fsf@phiwumbda.org> Discussion, linux) > Thats fine, because even if we adopt the convention that allows an > output after omega steps, the halting problem is still undecidable. Yes and no. The classical halting problem is undecidable, since by definition, it requires that the machine which decides it does so in a finite time. But, as Russell pointed out, there is a sense in which its decidable if we allow for the decision to take infinitely many steps. This is not contradictory, because this is a different notion of decidability. The usual diagonal argument cannot be applied here, since the machine which thwarts the halting problem decider would have to wait for an infinite computation *before* outputting anything. -- Jesse F. Hughes C is for Cookie. Thats good enough for me. Cookie Monsters === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> <87ptc6df8r.fsf@phiwumbda.org> <873c92coz3.fsf@phiwumbda.org> Discussion, linux) >> Ôn is just some input string. Ôm is a natural number serving as an >>index to some ordering of the Turing Machines. >> If string means finite sequence of symbols (i.e., finite number of >> non-blanks), then (*) reduces to (**) trivially. If string means >> a possibly infinite tape, then it (*) doesnt reduce to (**). > No, Im talking about the standard problems (of having finite input) > which anyone would see in their first course in computability theory. >> No it doesnt. You are begging the question here by assuming that (*) >> reduces to (**). > As you note that it does, above. As I note that it does above *when* the input string is finite. >> Look, how do you prove that, for every m and n, there is a machine k >> such that T_k halts on the empty tape (which Ill denote ) iff T_m >> halts on tape n and T_k() = T_m(n)? I dont know how youd prove >> tape n on the tape, then goes back to the starting position and runs >> m. >> If this is our proof, then it doesnt work when n is an infinite tape, >> does it? We cant write down and infinite sequence and *then* do >> something else. > Again, n is not an infinite tape, Im talking about the normal, known, > undecidable problem (with finite n). Im also operating under the premise > that these magical machines that can handle the infinite tapes can > actually process the whole (infinite) tape like how Russell was originally > doing. Thats fine, because even if we adopt the convention that allows an output after omega steps, the halting problem is still undecidable. >> R.e.s. is correct. Alan Morgan was mistaken. Allowing an arbitrary >> infinite tape does not allow one to solve the halting problem. > Again, if a machine takes input m on tape one and is initialized with > the magical tape on tape two, then it can determine whether Machine m > halts by looking at the second tape. Yes, absolutely. And in this case, it does not decide the halting problem, because the halting problem does not allow two tapes, the second of which has an infinite input consisting of the characteristic function for the halting problem. [...] -- Ive been thinking about my problems with getting any kind of admission that my math arguments showing the core error in mathematics are correct, so Ive gone to marketing books. -- James S. Harris, on when mathematics isnt enough === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> <87ptc6df8r.fsf@phiwumbda.org> <873c92coz3.fsf@phiwumbda.org Ôn is just some input string. Ôm is a natural number serving as an >index to some ordering of the Turing Machines. > If string means finite sequence of symbols (i.e., finite number of > non-blanks), then (*) reduces to (**) trivially. If string means > a possibly infinite tape, then it (*) doesnt reduce to (**). No, Im talking about the standard problems (of having finite input) which anyone would see in their first course in computability theory. > No it doesnt. You are begging the question here by assuming that (*) > reduces to (**). As you note that it does, above. > Look, how do you prove that, for every m and n, there is a machine k > such that T_k halts on the empty tape (which Ill denote ) iff T_m > halts on tape n and T_k() = T_m(n)? I dont know how youd prove > tape n on the tape, then goes back to the starting position and runs > m. > If this is our proof, then it doesnt work when n is an infinite tape, > does it? We cant write down and infinite sequence and *then* do > something else. Again, n is not an infinite tape, Im talking about the normal, known, undecidable problem (with finite n). Im also operating under the premise that these magical machines that can handle the infinite tapes can actually process the whole (infinite) tape like how Russell was originally doing. > R.e.s. is correct. Alan Morgan was mistaken. Allowing an arbitrary > infinite tape does not allow one to solve the halting problem. Again, if a machine takes input m on tape one and is initialized with the magical tape on tape two, then it can determine whether Machine m halts by looking at the second tape. > Machines are typically given by just the program, not the program and > tape. You never said what the program for RM is, but never mind. > Your argument is mistaken. Yes, youre absolutely correct... I was not keeping the tape and machine distinct. My intended use of RM was any machine that can have an arbitrary infinite tape and process that tape in whichever way it pleases. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> <87ptc6df8r.fsf@phiwumbda.org> Discussion, linux) >> Let RM(x) = Russells machine (a Turing machine) with *any* input x >> Let x = the string giving: cell i equals 1 iff i(th) TM halts on empty >>input. >> Now, you might recall from your first course in complexity theory that >> (*) Given m and n, TM m halts on input n >> is reducible to >> (**) Given m, TM m halts on empty input >> You might also recall the power of Turing machines to be able to call >>other machines (simulation.) >> What is n in your (*)? Is n a natural number? If so, this fact has >> nothing to do with simulating machine m when the input is infinite. > Ôn is just some input string. Ôm is a natural number serving as an > index to some ordering of the Turing Machines. If string means finite sequence of symbols (i.e., finite number of non-blanks), then (*) reduces to (**) trivially. If string means a possibly infinite tape, then it (*) doesnt reduce to (**). >> Is n supposed to be an arbitrary tape? If so, the fact is almost >> certainly false in the current context (in which we allow infinitely >> many ones on the input tape). In fact, it clearly *is* false, since >> it would allow one to solve the halting problem and the halting >> problem is not solvable even in the current context[1]. > Thats my point... allowing such a tape would allow for a solution to > the halting problem. No it doesnt. You are begging the question here by assuming that (*) reduces to (**). Look, how do you prove that, for every m and n, there is a machine k such that T_k halts on the empty tape (which Ill denote ) iff T_m halts on tape n and T_k() = T_m(n)? I dont know how youd prove tape n on the tape, then goes back to the starting position and runs m. If this is our proof, then it doesnt work when n is an infinite tape, does it? We cant write down and infinite sequence and *then* do something else. > Another poster (Alan Morgan) pointed out that allowing an arbitrary > and infinite input tape would solve the halting proble, and then a > different poster (r.e.s) said he didnt see how the two were > related. R.e.s. is correct. Alan Morgan was mistaken. Allowing an arbitrary infinite tape does not allow one to solve the halting problem. >> To solve (**), create a machine M that takes an input m. Then simulate >>RM(x) for the above x, move to tape cell m and get the result. >> I dont know what you mean here. Im not familiar with Russells >> machine. I dont know what solving (**) means. > Russells machine was described above, and you even copied it into your > reponse. It is a TM with the Ômagical starting tape. Machines are typically given by just the program, not the program and tape. You never said what the program for RM is, but never mind. Your argument is mistaken. > solving (**) means to solve the problem above marked as (**) as > opposed to solving the problem marked (*), for example. >> I assume you mean to give a sketch of how to solve the halting problem >> by simulating a given machine with an infinite tape. This cannot be >> done, of course. In fact, if this is what you meant to do, then you >> could have concluded that (*) is not reducible to (**) when n is allowed >> to contain infinitely many ones. > Yes, that was my intention. Both problem (*) and (**) are undecidable... > other posters were using (*) as Ôthe halting problem, but I didnt care > to specify an input, so I preferred to argue on (**). Then its a different problem. (*) is *not* reducible to (**) in the case that the input tape is infinite. The fact that its not reducible is provable by your argument. If it *were* reducible, then the halting problem would be decidable. The halting problem is provably undecidable. Therefore it is not reducible. If you want to claim that infinite tapes lead to an inconsistent setting, then you bloody well better argue why (*) reduces to (**) in the current context. -- Jesse F. Hughes A factor is simply something that multiplies against another factor to produce a Ôproduct. -- James Harris offers a definition. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> <87ptc6df8r.fsf@phiwumbda.org Let RM(x) = Russells machine (a Turing machine) with *any* input x > Let x = the string giving: cell i equals 1 iff i(th) TM halts on empty >input. > Now, you might recall from your first course in complexity theory that > (*) Given m and n, TM m halts on input n > is reducible to > (**) Given m, TM m halts on empty input > You might also recall the power of Turing machines to be able to call >other machines (simulation.) > What is n in your (*)? Is n a natural number? If so, this fact has > nothing to do with simulating machine m when the input is infinite. Ôn is just some input string. Ôm is a natural number serving as an index to some ordering of the Turing Machines. > Is n supposed to be an arbitrary tape? If so, the fact is almost > certainly false in the current context (in which we allow infinitely > many ones on the input tape). In fact, it clearly *is* false, since > it would allow one to solve the halting problem and the halting > problem is not solvable even in the current context[1]. Thats my point... allowing such a tape would allow for a solution to the halting problem. Another poster (Alan Morgan) pointed out that allowing an arbitrary and infinite input tape would solve the halting proble, and then a different poster (r.e.s) said he didnt see how the two were related. > To solve (**), create a machine M that takes an input m. Then simulate >RM(x) for the above x, move to tape cell m and get the result. > I dont know what you mean here. Im not familiar with Russells > machine. I dont know what solving (**) means. Russells machine was described above, and you even copied it into your reponse. It is a TM with the Ômagical starting tape. solving (**) means to solve the problem above marked as (**) as opposed to solving the problem marked (*), for example. > I assume you mean to give a sketch of how to solve the halting problem > by simulating a given machine with an infinite tape. This cannot be > done, of course. In fact, if this is what you meant to do, then you > could have concluded that (*) is not reducible to (**) when n is allowed > to contain infinitely many ones. Yes, that was my intention. Both problem (*) and (**) are undecidable... other posters were using (*) as Ôthe halting problem, but I didnt care to specify an input, so I preferred to argue on (**). J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87u11jawar.fsf@phiwumbda.org> Discussion, linux) >> What does it mean to solve the halting problem? As far as I know it >> means that, given a tape consisting of two natural numbers m and n, >> the machine halts and outputs 1 iff machine m halts on input n and >> halts and outputs 0 else. I dont see how allowing arbitrary input >> tapes (including an infinite tape, say, the characteristic function of >> the halting problem) helps a bit. >> No tapes with an infinite number of 1s are relevant for whether a >> particular machine solves the halting problem. > Let RM(x) = Russells machine (a Turing machine) with *any* input x > Let x = the string giving: cell i equals 1 iff i(th) TM halts on empty > input. > Now, you might recall from your first course in complexity theory that > (*) Given m and n, TM m halts on input n > is reducible to > (**) Given m, TM m halts on empty input > You might also recall the power of Turing machines to be able to call > other machines (simulation.) What is n in your (*)? Is n a natural number? If so, this fact has nothing to do with simulating machine m when the input is infinite. Is n supposed to be an arbitrary tape? If so, the fact is almost certainly false in the current context (in which we allow infinitely many ones on the input tape). In fact, it clearly *is* false, since it would allow one to solve the halting problem and the halting problem is not solvable even in the current context[1]. > To solve (**), create a machine M that takes an input m. Then simulate > RM(x) for the above x, move to tape cell m and get the result. I dont know what you mean here. Im not familiar with Russells machine. I dont know what solving (**) means. I assume you mean to give a sketch of how to solve the halting problem by simulating a given machine with an infinite tape. This cannot be done, of course. In fact, if this is what you meant to do, then you could have concluded that (*) is not reducible to (**) when n is allowed to contain infinitely many ones. Of course, Im not an expert on Turing computability. It is always possible I have an error here or there. But I dont think so this time. Footnotes: [1] Not even if we define the phrase T halts with output x after an infinite number of steps in the way we mentioned. -- Jesse F. Hughes Im better than you, and you know it. -- James Harris === Subject: Re: No Set Contains Every Computable Natural >What does it mean to solve the halting problem? As far as I know it >means that, given a tape consisting of two natural numbers m and n, >the machine halts and outputs 1 iff machine m halts on input n and >halts and outputs 0 else. I dont see how allowing arbitrary input >tapes (including an infinite tape, say, the characteristic function of >the halting problem) helps a bit. >No tapes with an infinite number of 1s are relevant for whether a >particular machine solves the halting problem. > Let RM(x) = Russells machine (a Turing machine) with *any* input x > Let x = the string giving: cell i equals 1 iff i(th) TM halts on empty > input. > Now, you might recall from your first course in complexity theory that > (*) Given m and n, TM m halts on input n > is reducible to > (**) Given m, TM m halts on empty input ^^^^^^^ > You might also recall the power of Turing machines to be able to call > other machines (simulation.) > To solve (**), create a machine M that takes an input m. Then > simulate RM(x) for the above x, move to tape cell m and get the result. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Youve assumed an impossibility in your final sentence, because the TM thats given m (only) cant simulate a TM thats given an infinite input. We know it cant do this because to do so would solve the halting problem for such TMs, which we know is impossible. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p > He wont *if* we apply the condition that his input tape has finitely >> many ones, but there is nothing about that convention that is forced >> on us. Why not allow that he has a tape with all of N? After all, >> that assumption is not his problem. His problem is only in reasoning >> about whats on the tape after an infinite number of steps. > Yes, of course I agree that a TM can have anything written on its tape >to start. > If thats the case then it would seem to me that it is very easy for > a TM to solve the halting problem (although procuring an appropriate > input tape might be a little tricky). What am I missing? Another excellent point against starting with whatever tape you like. My use of anything is too broad, and maybe someone would say any sequence of values that is actually computable can be an initial tape... but then that goes back to requiring a finitely-lengthed input. J === Subject: Re: No Set Contains Every Computable Natural > He wont *if* we apply the condition that his input tape has finitely > many ones, but there is nothing about that convention that is forced > on us. Why not allow that he has a tape with all of N? After all, > that assumption is not his problem. His problem is only in reasoning > about whats on the tape after an infinite number of steps. >> Yes, of course I agree that a TM can have anything written on its tape >>to start. >If thats the case then it would seem to me that it is very easy for >a TM to solve the halting problem (although procuring an appropriate >input tape might be a little tricky). What am I missing? > Another excellent point against starting with whatever tape you like. > My use of anything is too broad, and maybe someone would say any > sequence of values that is actually computable can be an initial tape... > but then that goes back to requiring a finitely-lengthed input. I dont see how Alans point is applicable here. The behaviour of infinite-input TMs is irrelevant to the finite-input halting problem -- by its definition that halting problem would be solved only by a finite-input TM. Similarly, although a TM could simply halt immediately on input of a sequence thats non-computable by finite-input TMs (thus computing it), that would be irrelevant to the fact that the sequence remains non-computable by any finite-input TM. --r.e.s. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p My use of anything is too broad, and maybe someone would say any >sequence of values that is actually computable can be an initial tape... >but then that goes back to requiring a finitely-lengthed input. > I dont see how Alans point is applicable here. The behaviour > of infinite-input TMs is irrelevant to the finite-input halting > problem -- by its definition that halting problem would be > solved only by a finite-input TM. Its not just the act of allowing an infinite input, but allowing *any* infinite input would lead to solvability of any problem, like the halting problem. Let a TM have initial tape with cell T_i = 1 if and only if the i(th) Turing Machine halts on an empty input. Then this TM can decide the question: Given k, does the k(th) Turing Machine halt on empty input? (This problem is undeciable.) J === Subject: Re: No Set Contains Every Computable Natural >> Another excellent point against starting with whatever tape you like. >> My use of anything is too broad, and maybe someone would say any >> sequence of values that is actually computable can be an initial tape... >> but then that goes back to requiring a finitely-lengthed input. >I dont see how Alans point is applicable here. The behaviour >of infinite-input TMs is irrelevant to the finite-input halting >problem -- by its definition that halting problem would be >solved only by a finite-input TM. > Its not just the act of allowing an infinite input, but allowing *any* > infinite input would lead to solvability of any problem, like the halting > problem. Let a TM have initial tape with cell T_i = 1 if and only if the > i(th) Turing Machine halts on an empty input. Then this TM can decide the > question: Given k, does the k(th) Turing Machine halt on empty input? ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > (This problem is undeciable.) No, that would *not* solve the halting problem that you just stated. That problem requires the TM to be given k only, i.e. finite input. The behavior of TMs with infinite input isnt relevant to the problem. --r.e.s. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ > (This problem is undeciable.) > No, that would *not* solve the halting problem that you just stated. > That problem requires the TM to be given k only, i.e. finite input. > The behavior of TMs with infinite input isnt relevant to the problem. The only behaviour of TMs with infinite and arbitrary input required here is that they exist... The TM that takes finite input k can simulate the magical machine and retrieve the answer. If you dont like simulation of magical machines but are still accepting the premise of their existence (which, personally, I dont) then consider a two-tape machine, one tape is the tape containing the information that tape cell i = 1 iff i(th) TM halts on empty input and the other tape is reserved for the input. This is just an argument against the existence of machines with any arbitrary starting configuration on their tape. J === Subject: Re: No Set Contains Every Computable Natural >> There are no discernible operations in Cantors proof. It is not an >> algorithm. > Cantor is not restricted to computable methods. Hes not doing > computability theory. >> And even if we *were* to view it algorithmically, we have no >> imposed convergence conditions (like completes in finitely many steps). > At any time n, there is a blank on the tape. At time omega, there are > no blanks. > There is no position omega, you silly boy. There is no time omega, either. > But, after he completes *every* step for n < omega, there are no > blanks. I know you dont believe me. You just said there is no position omega. How can a TM complete *every* step for n < omega if we both agree there is no position omega? > Youre incapable of > understanding this point, evidently. But we started with a tape that > has squares enumerated 0,1,2,3,.... It does not have a square labeled > omega. It simply doesnt. So you cant have a blank in the square > marked omega. And a TM cant perform omega steps. There will always be a blank on the tape. That is the point of the proof. It proves the non-existence of omega. Russell - Zeno was right. Motion is impossible. === Subject: Re: No Set Contains Every Computable Natural > How can it construct a counterexample without an algorithm? How can Detroit construct a car without a buggy whip holder? > Its not a proof unless it does. Its not a vehicle unless it does. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vßykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> <87ishxjs4d.fsf@phiwumbda.org> <87n079e239.fsf@phiwumbda.org> <878yitdkf4.fsf@phiwumbda.org> Discussion, linux) >> There are no discernible operations in Cantors proof. It is not an >> algorithm. > How can it construct a counterexample without an algorithm? > Its not a proof unless it does. Russell, go look up the word proof. Come back when you find where it says that algorithms are a necessary feature of existence proofs. >> Maybe one of the local constructivists will disagree, I >> dont know. But as far as the classical interpretation of Cantors >> proof goes, there is not a procedure, but the specification of a real >> number. > An uncomputable real number. > How do I know this number is not in my list if I cant compute it? I never said that the real number was uncomputable. It is relatively computable, namely relative to the function f:N -> R given in the hypothesis. Note, however, that computability of reals means something different than computability of naturals. No sense in discussing computability of reals if you dont understand computability of naturals. And there is no need for a number to be computable in order to show that it is different than every number on the list. Cantor is not restricted to computable methods. Hes not doing computability theory. >> And even if we *were* to view it algorithmically, we have no >> imposed convergence conditions (like completes in finitely many steps). > Or any proof it exists. Any sequence of digits defines a real number. Cantors proof yields a sequence of digits. Therefore, it defines a real number. > There are no > operations that I can see. There is a proof of the existence of a > real number not on the list. > But we cant compute this real number. > Very convenient. Who said we cant? We can, *given* the function f:N -> R. If the function f is uncomputable, then of course, we cant really compute the number, but none of this is relevant. >>Then I prove there exists a natural number not on the input tape. >> No, you dont. Russell, Im not arguing that your proof fails because >> it converges only in omega and no fewer steps. >> Your proof fails >> because after omega steps, there is not a single natural number on >> your tape. > We know there was a non-zero number of blanks on the initial tape. > My TM cant overwrite the last blank on the tape. > It is incapable of doing so. > There will be a blank on the tape. Wrong, wrong, wrong. At any time n, there is a blank on the tape. At time omega, there are no blanks. >> Ive told you this. Ive given the proof that your tape >> contains no blanks (right of the start state). By the n+1st iteration >> of your loop, the nth state is not blank. Therefore, after omega >> steps there are no blanks. > Then there is a blank on my tape at position omega. There *is* no ing position omega on the tape, you twit. There is no position omega, you silly boy. > You keep talking about omega steps like it really means something. > I have shown why a TM always performs a finite number of > operations. There is no such thing as omega steps. > There will still be blanks on the tape after any number of steps. But, after he completes *every* step for n < omega, there are no blanks. I know you dont believe me. Youre incapable of understanding this point, evidently. But we started with a tape that has squares enumerated 0,1,2,3,.... It does not have a square labeled omega. It simply doesnt. So you cant have a blank in the square marked omega. Im done for now. Youre just an idiot and the conversation goes round and round. Listen, you have fun with your delusions, Ôkay? -- Ive been thinking about my problems with getting any kind of admission that my math arguments showing the core error in mathematics are correct, so Ive gone to marketing books. -- James S. Harris, on when mathematics isnt enough === Subject: Re: No Set Contains Every Computable Natural > [earlier in the thread:] >The number of operations in Cantors proof is not finite. > [and now:] >The diagonal construction is not an algorithm? >The construction of a set in P(N) not in N is not an algorithm? >What is it? Wishful thinking? > I think you might benefit by reading (or re-reading) the post > by Dave Seaman 2-22 1:24Pm in this same thread, concerning > both direct and indirect proofs. The diagonal set can be > regarded as generated, fully and completely in a single step. An infinite set in a single step! That is an impressive TM. Can I get one? Russell - Zeno was right. Motion is impossible. === Subject: Re: No Set Contains Every Computable Natural > There are no discernible operations in Cantors proof. It is not an > algorithm. How can it construct a counterexample without an algorithm? Its not a proof unless it does. > Maybe one of the local constructivists will disagree, I > dont know. But as far as the classical interpretation of Cantors > proof goes, there is not a procedure, but the specification of a real > number. An uncomputable real number. How do I know this number is not in my list if I cant compute it? > And even if we *were* to view it algorithmically, we have no > imposed convergence conditions (like completes in finitely many steps). Or any proof it exists. >> There are no >> operations that I can see. There is a proof of the existence of a >> real number not on the list. But we cant compute this real number. Very convenient. >Then I prove there exists a natural number not on the input tape. > No, you dont. Russell, Im not arguing that your proof fails because > it converges only in omega and no fewer steps. > Your proof fails > because after omega steps, there is not a single natural number on > your tape. We know there was a non-zero number of blanks on the initial tape. My TM cant overwrite the last blank on the tape. It is incapable of doing so. There will be a blank on the tape. > Ive told you this. Ive given the proof that your tape > contains no blanks (right of the start state). By the n+1st iteration > of your loop, the nth state is not blank. Therefore, after omega > steps there are no blanks. Then there is a blank on my tape at position omega. You keep talking about omega steps like it really means something. I have shown why a TM always performs a finite number of operations. There is no such thing as omega steps. There will still be blanks on the tape after any number of steps. > Want to take omega steps to construct your tape? Fine. Go ahead. Cantor uses omega steps to construct his missing real. Unless you are still claiming this real cant be constructed. > Take your time, we have all day here. You still utterly fail to > create a natural number not on the list. Cantor utterly fails to create a real number not on the list. You have already admitted that a hyper-TM can solve the halting problem. Cantors algorithm requires a hyper-TM. Why do you accept some hyper-TM proofs and not others? Russell - Zeno was right. Motion is impossible. === Subject: Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vßykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> <87ishxjs4d.fsf@phiwumbda.org> <87n079e239.fsf@phiwumbda.org> Discussion, linux) >> There is plenty mushy, beginning with the fact that the number of >> operations is utterly without meaning in Cantors proof. > Not finite has a meaning. > The number of operations in Cantors proof is not finite. > (Except in the trivial case where the original list is finite.) There are no discernible operations in Cantors proof. It is not an algorithm. Maybe one of the local constructivists will disagree, I dont know. But as far as the classical interpretation of Cantors proof goes, there is not a procedure, but the specification of a real number. And even if we *were* to view it algorithmically, we have no imposed convergence conditions (like completes in finitely many steps). >> There are no >> operations that I can see. There is a proof of the existence of a >> real number not on the list. > Then I prove there exists a natural number not on the input tape. No, you dont. Russell, Im not arguing that your proof fails because it converges only in omega and no fewer steps. Your proof fails because after omega steps, there is not a single natural number on your tape. Ive told you this. Ive given the proof that your tape contains no blanks (right of the start state). By the n+1st iteration of your loop, the nth state is not blank. Therefore, after omega steps there are no blanks. Want to take omega steps to construct your tape? Fine. Go ahead. Take your time, we have all day here. You still utterly fail to create a natural number not on the list. Why are you confused on this point? Because, after I dont know how many years, you are still incapable of going five minutes without commuting quantifiers. You have a proof that, for each step n, there is an x which is blank at n and you believe this is sufficient to show that there is an x such that for all n, x is blank at n. This is just the same error over and over and over. >> When I say that the TM converged in the omegath step, I mean that >> (1) he didnt halt in a finite step and >> (2) he produced a tape after an infinite number of steps. > The output on the tape converges after a finite number of steps. > We might have to wait for an infinite number of steps to be > sure the output has converged. This point is irrelevant but correct, more or less. However, Ive chosen to define the term convergence here as above, so your observation is a bit confusing with the definition Im using. >>Halt is an instruction. Any TM that performs a halt instruction must >>do so in a finite number of steps. >> OK. (Though, in most conventions, halt is not an instruction.) >> The rest of your post is uncontroversial, as long as we adopt your >> idea of the halting problem for this setting. > OK > Lets agree that our hyper-TM can perform an infinite number of > operations in finite time, but it can NOT halt on step omega. > In all other respects, the hyper-TM behaves like a normal TM. > A hyper-TM can solve the halting problem, even for hyper-TMs. No, it cant, but Im not going down this road with you. There is no sense in making the picture more complicated at this point. Lets wait until you grasp basic Turing machines, or perhaps those which converge in omega steps, before we go galloping through the hierarchy of transfinite ordinals. -- Just because youre ... in a Ph.d program it does not mean that youre up to the challenge of being a real mathematician. Only those who have a purity of mind and dedication to the truth as the highest ideal have a chance. --James Harris, as Sir Galahad the Pure. === Subject: Re: No Set Contains Every Computable Natural [earlier in the thread:] > The number of operations in Cantors proof is not finite. [and now:] > The diagonal construction is not an algorithm? > The construction of a set in P(N) not in N is not an algorithm? > What is it? Wishful thinking? I think you might benefit by reading (or re-reading) the post by Dave Seaman 2-22 1:24Pm in this same thread, concerning both direct and indirect proofs. The diagonal set can be regarded as generated, fully and completely in a single step. --r.e.s. === Subject: Re: No Set Contains Every Computable Natural >> This is lame and sloppy thinking. I dont really care what you >> accept, but I wont get into silly arguments that Cantors proof >> doesnt count if TMs have to halt within a finite number of steps. >> Try to be precise and correct in your reasoning. Dont use mushy and >> ignorant analogies. >There is nothing mushy about pointing out that Cantors proof >requires an infinite number of operations. > There is plenty mushy, beginning with the fact that the number of > operations is utterly without meaning in Cantors proof. Not finite has a meaning. The number of operations in Cantors proof is not finite. (Except in the trivial case where the original list is finite.) > There are no > operations that I can see. There is a proof of the existence of a > real number not on the list. Then I prove there exists a natural number not on the input tape. >> Whether the output was written before step omega is inconsequential. >> The machine didnt converge until step omega. >> No, you cant solve the halting problem for machines with infinite >> computations with our convention. Let m be any machine which, on >> input x, converges in the omegath step and not before. >There is no such machine. >We are assuming that a TM can perform an infinite number of >operations in a finite amount of time. We specifically are NOT >assuming that a TM can perform an infinite number of operations >and then halt. > When I say that the TM converged in the omegath step, I mean that > (1) he didnt halt in a finite step and > (2) he produced a tape after an infinite number of steps. The output on the tape converges after a finite number of steps. We might have to wait for an infinite number of steps to be sure the output has converged. > What do you want the halting problem decider to say in that case? > Seems sensible to me to say that our decider ought to output 1 (the > halt output), but if you think he ought to output 0, then thats > fine. But in that case, my comment (2) doesnt apply. I think we agree that the definition of halt means the TM halts after a finite number of steps. Even if we allow a TM to halt after an infinite number of steps, I dont see how we can say the TM halts if it does so on the omegath step. >Halt is an instruction. Any TM that performs a halt instruction must >do so in a finite number of steps. > OK. (Though, in most conventions, halt is not an instruction.) > The rest of your post is uncontroversial, as long as we adopt your > idea of the halting problem for this setting. OK Lets agree that our hyper-TM can perform an infinite number of operations in finite time, but it can NOT halt on step omega. In all other respects, the hyper-TM behaves like a normal TM. A hyper-TM can solve the halting problem, even for hyper-TMs. Russell - Zeno was right. Motion is impossible. === Subject: Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vßykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> <87ishxjs4d.fsf@phiwumbda.org> <14-dnWhDt8HHB6fdRVn-uQ@comcast.com> Discussion, linux) >> Cantor didnt give an algorithm. Aside from that, um, yeah, good >> point. > The diagonal construction is not an algorithm? > The construction of a set in P(N) not in N is not an algorithm? > What is it? Wishful thinking? An existence proof. -- Jesse F. Hughes And a journal can beg me for the right to publish it [...] because Id rather see it in People magazine [...] --James Harris on his simple proof of Fermats last theorem === Subject: Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vßykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> <87ishxjs4d.fsf@phiwumbda.org> Discussion, linux) >> This is lame and sloppy thinking. I dont really care what you >> accept, but I wont get into silly arguments that Cantors proof >> doesnt count if TMs have to halt within a finite number of steps. >> Try to be precise and correct in your reasoning. Dont use mushy and >> ignorant analogies. > There is nothing mushy about pointing out that Cantors proof > requires an infinite number of operations. There is plenty mushy, beginning with the fact that the number of operations is utterly without meaning in Cantors proof. There are no operations that I can see. There is a proof of the existence of a real number not on the list. >> Whether the output was written before step omega is inconsequential. >> The machine didnt converge until step omega. >> No, you cant solve the halting problem for machines with infinite >> computations with our convention. Let m be any machine which, on >> input x, converges in the omegath step and not before. > There is no such machine. > We are assuming that a TM can perform an infinite number of > operations in a finite amount of time. We specifically are NOT > assuming that a TM can perform an infinite number of operations > and then halt. When I say that the TM converged in the omegath step, I mean that (1) he didnt halt in a finite step and (2) he produced a tape after an infinite number of steps. What do you want the halting problem decider to say in that case? Seems sensible to me to say that our decider ought to output 1 (the halt output), but if you think he ought to output 0, then thats fine. But in that case, my comment (2) doesnt apply. > Halt is an instruction. Any TM that performs a halt instruction must > do so in a finite number of steps. OK. (Though, in most conventions, halt is not an instruction.) The rest of your post is uncontroversial, as long as we adopt your idea of the halting problem for this setting. -- Jesse F. Hughes And Im one of my own biggest skeptics as I had *YEARS* of wrong ideas, and attempts that failed. Worse, for some of them it took *MONTHS* before I figured out where I screwed up. -- James Harris === Subject: Re: No Set Contains Every Computable Natural >I am saying my proof that no set contains every natural number >requires less machinary than Cantors proof that the reals are >uncountable. >You can claim I am inventing a new definition for algorithm, >but my proof is closer to an valid algorithm than Cantors. >Cantors algorithm is well accepted. > Cantor didnt give an algorithm. Aside from that, um, yeah, good > point. The diagonal construction is not an algorithm? The construction of a set in P(N) not in N is not an algorithm? What is it? Wishful thinking? Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural >> >>Cantors proof that the reals are uncountable is well accepted. >>This proof requires that an infinite number of operations be >>completed in a finite amount of time. If we accept Cantors >>proof, dont we also have to accept that the halting problem >>is solvable? >> (1) Because Cantor wasnt discussing computability? Because notions >> of computability are supposed to approximate actual computations >> (unlimited memory notwithstanding)? >Cantors diagonal number is not computable by a TM that halts >after a finite number of steps. Why should I accept that an >uncomputable number proves the reals are uncountable. >I can question any proof that requires an infinite number >of operations if we define algorithm to mean solvable >by a TM in a finite number of steps. > This is lame and sloppy thinking. I dont really care what you > accept, but I wont get into silly arguments that Cantors proof > doesnt count if TMs have to halt within a finite number of steps. > Try to be precise and correct in your reasoning. Dont use mushy and > ignorant analogies. There is nothing mushy about pointing out that Cantors proof requires an infinite number of operations. >> (2) Because even if we allow your machine to output its solution after >> omega many steps, it has failed to deliver correct information about >> which other machines converge in the omegath step, but not before? >The output occurs before step omega. And yes, we can decide if >these infinite TMs halt. The same proof works for them. We can >decide if an infinite TM halts after a finite number of steps. > Whether the output was written before step omega is inconsequential. > The machine didnt converge until step omega. > No, you cant solve the halting problem for machines with infinite > computations with our convention. Let m be any machine which, on > input x, converges in the omegath step and not before. There is no such machine. We are assuming that a TM can perform an infinite number of operations in a finite amount of time. We specifically are NOT assuming that a TM can perform an infinite number of operations and then halt. Halt is an instruction. Any TM that performs a halt instruction must do so in a finite number of steps. > Then, our > machine m does not halt in a finite amount of time, our machine never Proving that TM(m) did not halt in a finite number of steps. > Our machine can only write in finite steps, according > to our convention. It cannot write in the omegath step. It can not halt on the omegath step, either. > No machine > can. Therefore, it fails to correct its output after the machine m > halts. Our decider correctly responds that TM(m) did not halt in a finite number of steps. The set of all halting Turing machines is recursively enumerable even if we limit ourselves to normal (finite steps) TMs. Russell - the universe is one dimensional === Subject: Re: No Set Contains Every Computable Natural <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vßykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> <874qthldb1.fsf@phiwumbda.org> Discussion, linux) >>Cantors proof that the reals are uncountable is well accepted. >>This proof requires that an infinite number of operations be >>completed in a finite amount of time. If we accept Cantors >>proof, dont we also have to accept that the halting problem >>is solvable? >> (1) Because Cantor wasnt discussing computability? Because notions >> of computability are supposed to approximate actual computations >> (unlimited memory notwithstanding)? > Cantors diagonal number is not computable by a TM that halts > after a finite number of steps. Why should I accept that an > uncomputable number proves the reals are uncountable. > I can question any proof that requires an infinite number > of operations if we define algorithm to mean solvable > by a TM in a finite number of steps. This is lame and sloppy thinking. I dont really care what you accept, but I wont get into silly arguments that Cantors proof doesnt count if TMs have to halt within a finite number of steps. Try to be precise and correct in your reasoning. Dont use mushy and ignorant analogies. >> (2) Because even if we allow your machine to output its solution after >> omega many steps, it has failed to deliver correct information about >> which other machines converge in the omegath step, but not before? > The output occurs before step omega. And yes, we can decide if > these infinite TMs halt. The same proof works for them. We can > decide if an infinite TM halts after a finite number of steps. Whether the output was written before step omega is inconsequential. The machine didnt converge until step omega. No, you cant solve the halting problem for machines with infinite computations with our convention. Let m be any machine which, on input x, converges in the omegath step and not before. Then, our machine m does not halt in a finite amount of time, our machine never to our convention. It cannot write in the omegath step. No machine can. Therefore, it fails to correct its output after the machine m halts. But, look, Russell, you have remarkable problems with easy theories (like whether N has a largest element). I dont see that moving to ever more complicated settings is going to bring you any insight. Why dont you stick with garden variety Turing machines until you actual develop an aptitude for mathematical reasoning? >> After all, if we use this new definition of convergence, shouldnt >> that change which machines halt and which dont? > Not if we mean halts after a finite number of steps. > If we consider halt an instruction then a TM that can perform > an infinite number of operations in finite time can decide if > a halt instruction was executed. It can even do so for TMs > that perform an infinite number of operations. Wrong. Youve used up your second by the time that machine halted. Youre in the next second, but we didnt allow that. >> Your halting problem >> decider should probably be constrained to the same definition of >> convergence as the machines its deciding. > I am saying my proof that no set contains every natural number > requires less machinary than Cantors proof that the reals are > uncountable. > You can claim I am inventing a new definition for algorithm, > but my proof is closer to an valid algorithm than Cantors. > Cantors algorithm is well accepted. Cantor didnt give an algorithm. Aside from that, um, yeah, good point. For that matter, I never said you were giving a new definition for algorithm, but you are using one which isnt too common. This increases the likelihood that you will be confused by basic facts about standard TMs. Moreover, this new version allows infinite computations. I know this may come as a shock to you, Russell, but you dont really have a good reputation for reasoning about infinity. There is a teeny chance that, rather than finding extraordinary clarity and understanding with this model of computation, well, you just might become a bit bewildered as you bumble through the mess you create, tossing Cantors name willy-nilly and then jumping to proofs that the rationals are uncountable before leaping to the conclusion that the reals are countable and, oh yes, in fact all live right next door to the worlds greatest natural number. Oh, never mind. Full steam ahead. -- Jesse F. Hughes What you call reasonable is suspect since youve proven yourself to be an enemy of mathematics. -- James S. Harris defends the cause. === Subject: Re: No Set Contains Every Computable Natural > This is a bad way of claiming that the TM solves the halting problem > in a finite number of steps. It has written the correct answer in a > finite number of steps, but we do not know what the answer is until > after an infinite number of steps. > >>The only new requirement is that a TM can perform an infinite number >>of operations in a finite amount of time. Assume we have an accelerated >>TM that performs an infinite number of steps in 1 second (1st step in >1/2 >>second, 2nd in 1/4 sec, etc). The ATM solves the halting problem in >>1 second or less. >> Oh, is *that* the only new requirement? Well, a Turing machine which >> can perform an infinite number of operations in a finite time is >> certainly reasonable enough. >> Seriously, as a mathematical abstraction, its a perfectly reasonable >> one (with no evident implications for computability in the real >> world). But, as I said previously, there are no contradictions to be >> had here. >Cantors proof that the reals are uncountable is well accepted. >This proof requires that an infinite number of operations be >completed in a finite amount of time. If we accept Cantors >proof, dont we also have to accept that the halting problem >is solvable? > (1) Because Cantor wasnt discussing computability? Because notions > of computability are supposed to approximate actual computations > (unlimited memory notwithstanding)? Cantors diagonal number is not computable by a TM that halts after a finite number of steps. Why should I accept that an uncomputable number proves the reals are uncountable. I can question any proof that requires an infinite number of operations if we define algorithm to mean solvable by a TM in a finite number of steps. > (2) Because even if we allow your machine to output its solution after > omega many steps, it has failed to deliver correct information about > which other machines converge in the omegath step, but not before? The output occurs before step omega. And yes, we can decide if these infinite TMs halt. The same proof works for them. We can decide if an infinite TM halts after a finite number of steps. > After all, if we use this new definition of convergence, shouldnt > that change which machines halt and which dont? Not if we mean halts after a finite number of steps. If we consider halt an instruction then a TM that can perform an infinite number of operations in finite time can decide if a halt instruction was executed. It can even do so for TMs that perform an infinite number of operations. > Your halting problem > decider should probably be constrained to the same definition of > convergence as the machines its deciding. I am saying my proof that no set contains every natural number requires less machinary than Cantors proof that the reals are uncountable. You can claim I am inventing a new definition for algorithm, but my proof is closer to an valid algorithm than Cantors. Cantors algorithm is well accepted. Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87znbab8hp.fsf@phiwumbda.org> <87oerqb12j.fsf@phiwumbda.org> <877jydle69.fsf@phiwumbda.org> <87smh1jvlu.fsf@phiwumbda.org Yes, you can get the right behavior if you feed a machine n, m and > also the characteristic function. If we restrict to finite input, you > can get the write behavior if you feed the machine n, m and a finite > portion of the characteristic function. Like, for instance, the entry > for the pair . > In neither case have you solved the halting problem, so why should > this cast doubt on the reasonableness of allowing infinite input? Yes, youre right, the above example gives no reason for or against the use of infinite tape. Use infinite tape if you wish. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87smh2dg8v.fsf@phiwumbda.org> <87u11ib2cg.fsf@phiwumbda.org> <87ad39lebq.fsf@phiwumbda.org> Discussion, linux) > Lets leave this discussion at that. That was all I was trying to say. Ok. -- Jesse Hughes Depression hits more people than thought. --headline in Lexington, KY newspaper, as reported on NPRs Morning Edition === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87znbab8hp.fsf@phiwumbda.org> <87oerqb12j.fsf@phiwumbda.org> <877jydle69.fsf@phiwumbda.org> Discussion, linux) >> For a problem to arise, you have to design a TM which does the right >> thing on the right input. The input here is finite, so *whether or >> not* we allow the machine to take infinite inputs is completely >> irrelevant. You have to show that it does the right thing on finite >> input. > Sure, I would have to show it behaves correctly on finite input if I was > a crank trying to solve the halting problem, which I am not. Well, sorry, but I just dont get your point. Yes, you can get the right behavior if you feed a machine n, m and also the characteristic function. If we restrict to finite input, you can get the write behavior if you feed the machine n, m and a finite portion of the characteristic function. Like, for instance, the entry for the pair . In neither case have you solved the halting problem, so why should this cast doubt on the reasonableness of allowing infinite input? -- Jesse F. Hughes My baby dont allow me in the kitchen and Ive come to love her decision. -- Bad Livers === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87znbab8hp.fsf@phiwumbda.org> <87oerqb12j.fsf@phiwumbda.org> <877jydle69.fsf@phiwumbda.org For a problem to arise, you have to design a TM which does the right > thing on the right input. The input here is finite, so *whether or > not* we allow the machine to take infinite inputs is completely > irrelevant. You have to show that it does the right thing on finite > input. Sure, I would have to show it behaves correctly on finite input if I was a crank trying to solve the halting problem, which I am not. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87smh2dg8v.fsf@phiwumbda.org> <87u11ib2cg.fsf@phiwumbda.org> <87ad39lebq.fsf@phiwumbda.org Right, which is the kind of contradiction I have been arriving at all >along. Im not interested in extending the model of turing machines to a >more powerful one. Im fixing the set of Ôdecidable languages to the ones >that are normally called decidable and showing that such a magical machine >is Ôtoo powerful. > But your magical machine isnt one of the ones that weve allowed > (since it has two tapes). Thats a minor point. More importantly, > your magic machine does not solve the halting problem, so it isnt too > powerful by your standard. We now seem to be differing on what it takes to Ôsolve the halting problem. I am saying this magical machine with two tapes, first with input n and second with characteristic function, could write, perhaps on a thrid tape the contents of the n(th) cell of the characteristic function onto the third (output) tape. This would provide a yes/no answer to whether the n(th) machine halts on empty input. Having multiple tapes is not a factor in computability and it runs in a finite number of steps. Explain why this doesnt fit your criteria for solving the problem. > What if the machine has, within its trasition function, the capability >of writing out that list? Again, this is impossible for normal turing >machines, but Russell was allowing his machines to do an Ôinfinite amount >of work in finite time (his words.) > No Turing machine can write out the characteristic function of the > halting problem, not even in omega steps. Thats no easier than > solving the problem itself. Right, I understand that... if we are talking about normal, real, turing machines. But if we say the turing machine is capable of giving itself *any* characteristic function, then it can be solving any problem. > The *relevant sense* of the halting problem is this. This is what you > have to do to solve the halting problem. If you cant do this, then > you havent solved the halting problem. > You have to design a Turing machine which, when fed a single tape with > numbers m and n on it *and nothing else* (including *no other tape*), > your machine has to halt and output 1 if machine m halts on n, 0 else. Why do you insist on no other tape? Dont you believe that if there is a problem that is solved with a 7-tape machine, then there is a 1-tape machine solving that same problem? Now, of course, you will also be restricting the use of any oracles. I am not trying to solve the halting problem. I am showing that the use of an oracle tape would solve the halting problem. I fully realize that the halting problem will not be solved by any TM with finite input and finite computation. > You havent done this. Youve pointed out that its easy to design a > machine which can take two tapes, the second of which has the > characteristic function for the halting problem, and that halts and > gives the correct output. This is not showing the halting problem is > decidable. Ive never wanted to show the halting problem was decidable. Ive been trying to show that a machine with the characteristic function will easily solve the halting problem, as you write right above so I have to guess you agree that it is easy. Lets leave this discussion at that. That was all I was trying to say. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vßykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> <87smh1lu3l.fsf@phiwumbda.org> <8tydnVVJgrep-KfdRVn-sQ@comcast.com> Discussion, linux) > This is a bad way of claiming that the TM solves the halting problem > in a finite number of steps. It has written the correct answer in a > finite number of steps, but we do not know what the answer is until > after an infinite number of steps. >The only new requirement is that a TM can perform an infinite number >>of operations in a finite amount of time. Assume we have an accelerated >>TM that performs an infinite number of steps in 1 second (1st step in > 1/2 >>second, 2nd in 1/4 sec, etc). The ATM solves the halting problem in >>1 second or less. >> Oh, is *that* the only new requirement? Well, a Turing machine which >> can perform an infinite number of operations in a finite time is >> certainly reasonable enough. >> Seriously, as a mathematical abstraction, its a perfectly reasonable >> one (with no evident implications for computability in the real >> world). But, as I said previously, there are no contradictions to be >> had here. > Cantors proof that the reals are uncountable is well accepted. > This proof requires that an infinite number of operations be > completed in a finite amount of time. If we accept Cantors > proof, dont we also have to accept that the halting problem > is solvable? (1) Because Cantor wasnt discussing computability? Because notions of computability are supposed to approximate actual computations (unlimited memory notwithstanding)? (2) Because even if we allow your machine to output its solution after omega many steps, it has failed to deliver correct information about which other machines converge in the omegath step, but not before? After all, if we use this new definition of convergence, shouldnt that change which machines halt and which dont? Your halting problem decider should probably be constrained to the same definition of convergence as the machines its deciding. -- Just because youre ... in a Ph.d program it does not mean that youre up to the challenge of being a real mathematician. Only those who have a purity of mind and dedication to the truth as the highest ideal have a chance. --James Harris, as Sir Galahad the Pure. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87znbab8hp.fsf@phiwumbda.org> <87oerqb12j.fsf@phiwumbda.org> Discussion, linux) [...] >> It all depends on what counts as a problem. If, as is typical, a >> problem is a function f:N^k -> N satisfying some description (like the >> halting problem, which is the function N^2 -> N such that..., or in >> your preferred version, a function N -> N such that...), then >> what a Turing machine does to a tape with infinitely many ones is >> completely irrelevant. > What counts as a problem is any solution to any undecidable problem. I > have simply been arguing that if one is to allow arbitrary input tapes and > still keep the same computing power, then one has to be careful as to what > kind of tapes are allowed. And Ive been arguing that this is wrong. And that you havent shown that any problems arise. For a problem to arise, you have to design a TM which does the right thing on the right input. The input here is finite, so *whether or not* we allow the machine to take infinite inputs is completely irrelevant. You have to show that it does the right thing on finite input. -- Jesse F. Hughes Thats the base tautological space where by tautological space I mean a region of truth. -- James S. Harris does philosophy of mathematics. JSH is a renaissance man. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87smh2dg8v.fsf@phiwumbda.org> <87u11ib2cg.fsf@phiwumbda.org> Discussion, linux) >> That is a different notion of Turing machine. Thats a Turing machine >> with an oracle tape. With oracle tapes, the halting problem is >> decidable. > This is essentially way I was referring to any machine with an > arbitrary/infinite tape (even without describing the machine, but only > the tape) as a magical machine. Yes, now that you mentioned Oracles, > what I was trying to hit was a turing machine with an arbitrarily strong > oracle. And this was my argument that they wont exist (in our normal > world), but of course, there are theoretical models where they do. But thats *not* an argument against a Turing machine which accepts a single tape with possibly infinite input. >> Look, heres what you need to show that the halting problem is >> decidable in our setting. You need to show that there is a Turing >> machine that, when fed a tape with m and n and *no other* symbols on >> that tape and *no other* tape, the machine always halts, outputting 1 >> iff machine m halts on input n and otherwise outputting 0. This is >> the halting problem. Your two tape version is not relevant. > What is wrong with the standard TM simulates a two-tape TM > conversion? The fact that the TM that simulates the two-tape TM with an oracle for the halting problem cannot solve the halting problem, since it requires a tape with two numbers *and* a characteristic function. > And I essentially was trying to get at the fact that this TM (taking > m and n) was calling on the power of an oracle. But when he does so, he *isnt* solving the halting problem. >> Well, this is the argument against the fact that a Turing machine can >>start with any arbitrary input tape, which was my point. > I have never referenced a problem regarding the halting problem for > machines with infinite input. I have always been talking about the > standard halting problem. Not with that second tape with the characteristic function, you havent. >> tapes with infinite input. It proves that if we allow two tapes, the >> second of which may have the characteristic function of the halting >> problem, then the halting problem is decidable. > Right, which is the kind of contradiction I have been arriving at all > along. Im not interested in extending the model of turing machines to a > more powerful one. Im fixing the set of Ôdecidable languages to the ones > that are normally called decidable and showing that such a magical machine > is Ôtoo powerful. But your magical machine isnt one of the ones that weve allowed (since it has two tapes). Thats a minor point. More importantly, your magic machine does not solve the halting problem, so it isnt too powerful by your standard. >> There is no Turing machine which, when fed *just* m and n, behaves >> like your machine which is fed m, n and another tape consisting of the >> characteristic function of the halting problem. > What if the machine has, within its trasition function, the capability > of writing out that list? Again, this is impossible for normal turing > machines, but Russell was allowing his machines to do an Ôinfinite amount > of work in finite time (his words.) No Turing machine can write out the characteristic function of the halting problem, not even in omega steps. Thats no easier than solving the problem itself. [...] >> Yes, it avoids that need by adding yet another irrelevant complexity. >> Youve added an oracle to the Turing machine. It doesnt show that >> the halting problem is decidable in the relevant sense. > Your relevant sense is one you made up. Ive always been talking about > the halting problem in the normal sense, never about the halting problem > for a more powerful model of computation. The *relevant sense* of the halting problem is this. This is what you have to do to solve the halting problem. If you cant do this, then you havent solved the halting problem. You have to design a Turing machine which, when fed a single tape with numbers m and n on it *and nothing else* (including *no other tape*), your machine has to halt and output 1 if machine m halts on n, 0 else. You havent done this. Youve pointed out that its easy to design a machine which can take two tapes, the second of which has the characteristic function for the halting problem, and that halts and gives the correct output. This is not showing the halting problem is decidable. -- At the Microsoft-sponsored cocktail reception in the Galaxy Ballroom that evening, Robert Dees urges us Ôto network on behalf of the people of Iraq, -- Naomi Klein reports on Microsofts efforts to further democracy. === Subject: Re: No Set Contains Every Computable Natural >> This is a bad way of claiming that the TM solves the halting problem >> in a finite number of steps. It has written the correct answer in a >> finite number of steps, but we do not know what the answer is until >> after an infinite number of steps. >The only new requirement is that a TM can perform an infinite number >of operations in a finite amount of time. Assume we have an accelerated >TM that performs an infinite number of steps in 1 second (1st step in 1/2 >second, 2nd in 1/4 sec, etc). The ATM solves the halting problem in >1 second or less. > Oh, is *that* the only new requirement? Well, a Turing machine which > can perform an infinite number of operations in a finite time is > certainly reasonable enough. > Seriously, as a mathematical abstraction, its a perfectly reasonable > one (with no evident implications for computability in the real > world). But, as I said previously, there are no contradictions to be > had here. Cantors proof that the reals are uncountable is well accepted. This proof requires that an infinite number of operations be completed in a finite amount of time. If we accept Cantors proof, dont we also have to accept that the halting problem is solvable? Russell - solution to halting problem: Ctrl Alt Del === Subject: Re: No Set Contains Every Computable Natural >As I said in a previous post, if we allow a TM to have infinite input, >then every non-computable number is trivially computable by a TM that >does nothing but halt immediately on every input. I put the word in >scare quotes, because it doesnt have the meaning as in its finite-input >definition. Just as with the functions Minsky calls real-computable, >if we want to speak of computability with infinite inputs, the concept >needs some special terminology to avoid confusion. >Its the same situation with the phrase solving the halting problem -- >words that have been given precise finite-input definitions can be >misapplied to TMs with infinite inputs. So far, the misapplication has >been shown to produce plenty of confusion through inconsistent usage, >but no logical inconsistency has been shown. Bingo. This was precisely my problem when I raised the original objection. I suspected that I might be missing something, but wasnt sure what it was (sloppy definitions were the culprit. What a surprise). Alan -- Defendit numerus === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87d686azim.fsf@phiwumbda.org You mean that you were trying to show that, if we assume something > silly like, a Turing machine can emulate (or call) another Turing > machine with an infinite tape, then odd things happen? Youre welcome. Im glad you have finally caught on. > I dont think anyone here was under the misconception that a TM could > call another TM with an infinite tape as its input. Well, at least > not r.e.s. and me. Right, but Russell had been using machines that can take infinite input and process infinite input. And as he claims these can be done in finite time, it was implying that he was saying these machines somehow behave like normal turing machines, and I was just trying to show that these can not be kept within the realm of Ônormal turing machines because it changes the notion of what problems are decidable. >P.S. I still dont think you understood what I was trying to show in the >first place. > I certainly didnt. Still dont. Then why bother arguing when you didnt even understand the issue at hand? J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87znbab8hp.fsf@phiwumbda.org> <87oerqb12j.fsf@phiwumbda.org Again, Im not talking about infinte x. Im talking about a machine M1 >which has an infinite arbitrary tape can decide the problem: given a >Turing Machine n, does this this n(th) turing machine halt on empty input? >Youll just say this is trivial. > Yes. And Ill also say that this is not a solution to the problem as > given. Im not sure what you think the problem Ôas given is, but from your complaints, it sounds like you think someone is trying to solve the halting problem on a more general model of turing machines... Ive simply been trying to show that the more general model of turing machines can solve the standard halting problem. > In other words, given a machine which solves the Halting problem, we > can solve the Halting problem (this is an equivalence that Rogers > brießy discusses, that having an auxiliary tape is equivalent to > querying an oracle machine). Yes, I was just showing a very trivial fact that given a machine which has a tape which solves the halting proble, we can solve the halting problem. Im not sure why you threw such a huge deal about it. My point was that if we are to respect the standard notion of decidability, we should not be allowed to start a machine with any arbitrary input (because I want to stay within the frame that undecidable problems are still undecidable.) > None of this is relevant to the plausibility of allowing arbitrary > tapes as input, near as I can figure. ÔArbitrary tape meaning Ôquerying an arbitrary oracle . > If you dont need to be careful, as you write, then easily tell me that >if you are going to extend the turing machine model to allow them to have >infinite input, which input are allowed such that it does not affect what >problems are decidable. > It all depends on what counts as a problem. If, as is typical, a > problem is a function f:N^k -> N satisfying some description (like the > halting problem, which is the function N^2 -> N such that..., or in > your preferred version, a function N -> N such that...), then > what a Turing machine does to a tape with infinitely many ones is > completely irrelevant. What counts as a problem is any solution to any undecidable problem. I have simply been arguing that if one is to allow arbitrary input tapes and still keep the same computing power, then one has to be careful as to what kind of tapes are allowed. > Of course, we can also define new problems, namely problems which take > an infinite input and do something to it. Im not interested in such problems. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87smh2dg8v.fsf@phiwumbda.org> <87u11ib2cg.fsf@phiwumbda.org That is a different notion of Turing machine. Thats a Turing machine > with an oracle tape. With oracle tapes, the halting problem is > decidable. This is essentially way I was referring to any machine with an arbitrary/infinite tape (even without describing the machine, but only the tape) as a magical machine. Yes, now that you mentioned Oracles, what I was trying to hit was a turing machine with an arbitrarily strong oracle. And this was my argument that they wont exist (in our normal world), but of course, there are theoretical models where they do. > Look, heres what you need to show that the halting problem is > decidable in our setting. You need to show that there is a Turing > machine that, when fed a tape with m and n and *no other* symbols on > that tape and *no other* tape, the machine always halts, outputting 1 > iff machine m halts on input n and otherwise outputting 0. This is > the halting problem. Your two tape version is not relevant. What is wrong with the standard TM simulates a two-tape TM conversion? And I essentially was trying to get at the fact that this TM (taking m and n) was calling on the power of an oracle. > Well, this is the argument against the fact that a Turing machine can >start with any arbitrary input tape, which was my point. I have never referenced a problem regarding the halting problem for machines with infinite input. I have always been talking about the standard halting problem. > tapes with infinite input. It proves that if we allow two tapes, the > second of which may have the characteristic function of the halting > problem, then the halting problem is decidable. Right, which is the kind of contradiction I have been arriving at all along. Im not interested in extending the model of turing machines to a more powerful one. Im fixing the set of Ôdecidable languages to the ones that are normally called decidable and showing that such a magical machine is Ôtoo powerful. > There is no Turing machine which, when fed *just* m and n, behaves > like your machine which is fed m, n and another tape consisting of the > characteristic function of the halting problem. What if the machine has, within its trasition function, the capability of writing out that list? Again, this is impossible for normal turing machines, but Russell was allowing his machines to do an Ôinfinite amount of work in finite time (his words.) > Obviously, what I meant by the Ômagical TM is a TM that started with >that magical tape. > But thats not particularly interesting or troublesome, near as I can > figure. Not troublesome? As you say above, it essentially has an oracle tape, thereby making it more powerful than the standard Turing machines. > Yes, it avoids that need by adding yet another irrelevant complexity. > Youve added an oracle to the Turing machine. It doesnt show that > the halting problem is decidable in the relevant sense. Your relevant sense is one you made up. Ive always been talking about the halting problem in the normal sense, never about the halting problem for a more powerful model of computation. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> <87vßykodq.fsf@phiwumbda.org> <1tidnSHUxMqqYaTdRVn-ig@comcast.com> Discussion, linux) >> This is a bad way of claiming that the TM solves the halting problem >> in a finite number of steps. It has written the correct answer in a >> finite number of steps, but we do not know what the answer is until >> after an infinite number of steps. > The only new requirement is that a TM can perform an infinite number > of operations in a finite amount of time. Assume we have an accelerated > TM that performs an infinite number of steps in 1 second (1st step in 1/2 > second, 2nd in 1/4 sec, etc). The ATM solves the halting problem in > 1 second or less. Oh, is *that* the only new requirement? Well, a Turing machine which can perform an infinite number of operations in a finite time is certainly reasonable enough. Seriously, as a mathematical abstraction, its a perfectly reasonable one (with no evident implications for computability in the real world). But, as I said previously, there are no contradictions to be had here. Weve extended the definition to allow for certain computations that take omega-many steps. We cant run the diagonal argument here, because the machine that tries to compose with the halting problem decider would have to take *more* than omega steps. Just to forestall the acceleration of implausible conventions, yes, we *could* (I think) extend our definition to allow for machines which converge in omega + n steps, but then you lose your halting problem decider. So then we extend to allow for machines which converge in omega + omega steps, but then we lose the diagonal constructor. So, then we could.... In any case, a machine which decides the halting problem in omega many steps in not a machine which solves the halting problem. The halting problem as stated requires convergence in a finite number of steps. -- Jesse Hughes LOL. How arrogant you are. Now when you realize that I DID prove Goldbachs conjecture and that I proved Fermats Last Theorem as well, how are you going to feel then? -- James Harris === Subject: Re: No Set Contains Every Computable Natural >This is a bad way of claiming that the TM solves the halting problem >in a finite number of steps. It has written the correct answer in a >finite number of steps, but we do not know what the answer is until >after an infinite number of steps. > The only new requirement is that a TM can perform an infinite number > of operations in a finite amount of time. Assume we have an accelerated > TM that performs an infinite number of steps in 1 second (1st step in 1/2 > second, 2nd in 1/4 sec, etc). The ATM solves the halting problem in > 1 second or less. This must be well known, but this also proves that the set of all Turing machines that halt is recursively enumerable. Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural >>A Turing machine that solves the halting problem is a machine such >>that, given input m and n, halts and outputs 1 if machine m halts on n >>and halts and outputs 0 otherwise. >> >>Your machine is fed an infinite tape in order to achieve its result. >>Therefore it does not solve the halting problem. >This is simpler than I thought. >I dont even need a TM that performs an infinite number of operations. >Assume I have a UTM. >We give this UTM a finite number, n, that represents the nth TM. >We also give it a number, m, that represents the input for TM(n). >The UTM then simulates TM(n). If the simulation halts then the UTM > Clarification: Your UTM is not writing a 0 or 1. Another machine, > halts. >This UTM solves the halting problem in a finite number of steps. >(Just because the UTM finds the solution in a finite number of steps >does not mean the UTM, itself, halts after a finite number of states.) > This is a bad way of claiming that the TM solves the halting problem > in a finite number of steps. It has written the correct answer in a > finite number of steps, but we do not know what the answer is until > after an infinite number of steps. The only new requirement is that a TM can perform an infinite number of operations in a finite amount of time. Assume we have an accelerated TM that performs an infinite number of steps in 1 second (1st step in 1/2 second, 2nd in 1/4 sec, etc). The ATM solves the halting problem in 1 second or less. Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> <_YednVwym7cPfKTdRVn-ug@comcast.com> Discussion, linux) >>A Turing machine that solves the halting problem is a machine such >>that, given input m and n, halts and outputs 1 if machine m halts on n >>and halts and outputs 0 otherwise. >>Your machine is fed an infinite tape in order to achieve its result. >>Therefore it does not solve the halting problem. > This is simpler than I thought. > I dont even need a TM that performs an infinite number of operations. > Assume I have a UTM. > We give this UTM a finite number, n, that represents the nth TM. > We also give it a number, m, that represents the input for TM(n). > The UTM then simulates TM(n). If the simulation halts then the UTM Clarification: Your UTM is not writing a 0 or 1. Another machine, halts. > This UTM solves the halting problem in a finite number of steps. > (Just because the UTM finds the solution in a finite number of steps > does not mean the UTM, itself, halts after a finite number of states.) This is a bad way of claiming that the TM solves the halting problem in a finite number of steps. It has written the correct answer in a finite number of steps, but we do not know what the answer is until after an infinite number of steps. -- There are people [...] who think its socially acceptable to level accusations of mental illness in insulting exchanges to make points[...] [They] are rather sick [them]selves, and in reality, are sociopathic. --- James Harris, evidently a self-described sociopath === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> <87fzd2b0ta.fsf@phiwumbda.org> Discussion, linux) >> A Turing machine that solves the halting problem is a machine such >> that, given input m and n, halts and outputs 1 if machine m halts on n >> and halts and outputs 0 otherwise. >> Your machine is fed an infinite tape in order to achieve its result. >> Therefore it does not solve the halting problem. > You are right. > I was distracted by this infinite tape stuff. > What really matters is that we are assuming a TM can > perform an infinite number of operations. > Such a TM can solve the halting problem. I was about to write that this is false. Now I see that whether it is false or not depends on our exact definition of the halting problem. Lets first review our convention. We say that a TM converges to a tape x if, for every square x_n on the tape, there is a step t in the execution of the TM such that for all t > t, the square x_n at t contains the same bit as the square x_n at step t. With this convention in mind, contrary to my previous comments, the and emulates machine m on input n. If machine m halts, then stop, erase the 0 and write 1. Otherwise keep going. In this case, we see that after an infinite number of steps, we have a 1 on the tape iff machine m halts on n and 0 otherwise. But, of course, this isnt the classical halting problem. The classical problem requires that the problem is solved in a finite number of steps. > By definition, a TM halts if it DOESNT perform an > infinite number of operation. Assume we have a UTM > that can perform an infinite number of operations and > output a solution. This UTM can simulate any TM we give it. > We give this UTM a number, n, that represents a TM, > and a number, m, that represents the input for TM(n). > If the UTM halts after a finite number of steps then TM(n) > halts on input, m. Otherwise, TM(n) does not halt on m. > The infinite UTM can output a 1 if the simulation halts > or a 0 if the simulation runs for an infinite number of steps. > (Assuming, of course, that the UTM can distiguish > between finite and infinite.) No distinction necessary. Write a 0. If the machine halts, erase the 0 and write a 1. The diagonal argument showing that the halting problem is uncomputable doesnt apply here. If we wanted to construct the usual machine D to thwart our halting problem machine K, then machine D would have to first emulate the halting problem machine K for an infinite number of steps (omega many) and *then* decide whether to halt or not. But we have never said that we could make sense of omega + 1 steps, only omega-many. Thus, if we want to allow for infinite computations, we must be careful in formulating the halting problem. The traditional halting problem requires that the machine K halts in finite time. This problem is uncomputable. The amended problem in which we allow for output after an infinite number of steps is computable within this odd convention, but there is no contradiction to be had from it. In this respect, Jim Nastoss warnings that one must be careful are correct. But the care must be given in the meaning of an infinite computation and not with a concern over an infinite input tape. Certainly, no one should think that, because our amended halting problem is decidable in this somewhat odd sense, the uncomputability of the halting problem is doubtful. It is not. There are two different problems at hand here, and two different notions of convergence. -- Jesse Hughes Surround sound is going to be increasingly important in future offices. -- Microsoft marketing manager displays his keen insight === Subject: Re: No Set Contains Every Computable Natural >A Turing machine that solves the halting problem is a machine such >that, given input m and n, halts and outputs 1 if machine m halts on n >and halts and outputs 0 otherwise. >Your machine is fed an infinite tape in order to achieve its result. >Therefore it does not solve the halting problem. This is simpler than I thought. I dont even need a TM that performs an infinite number of operations. Assume I have a UTM. We give this UTM a finite number, n, that represents the nth TM. We also give it a number, m, that represents the input for TM(n). The UTM then simulates TM(n). If the simulation halts then the UTM This UTM solves the halting problem in a finite number of steps. (Just because the UTM finds the solution in a finite number of steps does not mean the UTM, itself, halts after a finite number of states.) Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural > A Turing machine that solves the halting problem is a machine such > that, given input m and n, halts and outputs 1 if machine m halts on n > and halts and outputs 0 otherwise. > Your machine is fed an infinite tape in order to achieve its result. > Therefore it does not solve the halting problem. You are right. I was distracted by this infinite tape stuff. What really matters is that we are assuming a TM can perform an infinite number of operations. Such a TM can solve the halting problem. By definition, a TM halts if it DOESNT perform an infinite number of operation. Assume we have a UTM that can perform an infinite number of operations and output a solution. This UTM can simulate any TM we give it. We give this UTM a number, n, that represents a TM, and a number, m, that represents the input for TM(n). If the UTM halts after a finite number of steps then TM(n) halts on input, m. Otherwise, TM(n) does not halt on m. The infinite UTM can output a 1 if the simulation halts or a 0 if the simulation runs for an infinite number of steps. (Assuming, of course, that the UTM can distiguish between finite and infinite.) Russell - 2 many 2 count === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p Discussion, linux) >> No. First of all, in trying to find an inconsistency due to the use >> of infinite inputs, youve claimed that some TM with finite input >> *can* simulate certain specific TMs with arbitrary infinite input. >> Thats just not correct, because to assume so would contradict the >> known unsolvability of the halting problem under discussion. This >> says nothing about the kind of inconsistency youre trying to show. > Right... Im arguing with the premise that if TMs with infinite input > and the ability to use the infinite input can behave like normal TMs, then > a normal (finite) TM can use them as subroutines. > And, yes, this premise contradicts the unsolvability to the halting > problem, which is what I was trying to show. You mean that you were trying to show that, if we assume something silly like, a Turing machine can emulate (or call) another Turing machine with an infinite tape, then odd things happen? >> There is no TM that, given arbitrary input n in N, decides >> Does the nth TM halt if started on a blank input tape? >> Note that the TM deciding a question is supposed to have *finite* >> input (a natural number n); consequently, no TM with infinite input >> can be said to decide this halting problem -- the deciding must, by >> definition, be done by a TM with finite input. > Again, my argument was that (under the assumption that TMs with infinite > input can behave and compute things normally reserved for finitely-input > machines) would lead to inconsistencies. I dont think anyone here was under the misconception that a TM could call another TM with an infinite tape as its input. Well, at least not r.e.s. and me. >> Solving this halting problem *means* that a TM with finite input >> decides it. Whatever else a TM with infinite input might do, it >> does not solve this problem, and hence produces no inconsistency >> on that account. > Okay, but I was operating under the premise that solve is not > restricted to solve by a TM with finite input, but instead extended to > the hypothetical machines. >> You havent yet shown an inconsistency (except in thinking that TMs >> with finite input are able to simulate those with infinite input). > The is the premise with which I used to draw contradictions. > P.S. I still dont think you understood what I was trying to show in the > first place. I certainly didnt. Still dont. -- Jesse F. Hughes Leaving things always seems to fix me, Running seems to ease my worried mind. -- Bad Livers, Honey, Ive Found a Brand New Way === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <1OqdnQsaDe1_P6TdRVn-uw@comcast.com> Discussion, linux) >> There is no TM that, given arbitrary input n in N, decides >> Does the nth TM halt if started on a blank input tape? >> Note that the TM deciding a question is supposed to have *finite* >> input (a natural number n); consequently, no TM with infinite input >> can be said to decide this halting problem -- the deciding must, by >> definition, be done by a TM with finite input. > The definition you give says NO TM can decide if the nth TM halts. > If we allow TMs to have an infinite input tape then there IS a TM > that will solve the halting problem for TMs that only accept finite > input. This is wrong. > The magic tape would consist of n, some separator like X, and the > binary representation of a real number that solves the halting problem > for the countable set of TMs that only accept finite input. > (Such a real number must exist.) A Turing machine that solves the halting problem is a machine such that, given input m and n, halts and outputs 1 if machine m halts on n and halts and outputs 0 otherwise. Your machine is fed an infinite tape in order to achieve its result. Therefore it does not solve the halting problem. > Of course, the set of all TMs doesnt exist, so the halting problem > is bogus, anyway. Yes, of course. > - I may be a crackpot, but I am getting good at it. Well, youve had practice. -- Sale or rental of this disc is ILLEGAL. If you have rented or purchased this disc, please call the MPAA at 1-800-NO-COPYS. -- The MPAA begins a new anti-piracy program, found on a DVD purchased in China === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87wu6eb7uu.fsf@phiwumbda.org> Discussion, linux) >> Heres another argument that your claim is simply wrong. It is not >> the case that, when x is an infinite input, there is a machine T >> which, when fed , simulates machine T on x. > Again, stop assuming x is infinite. My post was written before I saw any word from you that you werent really simulating here, but instead wanted an auxiliary tape. -- To assert otherwise is to say that mathematical operations and symbols mean whatever we want them to mean, varying from context to context [...] It in essence introduces post-modernism into mathematics. --Paul Lutus on absolute notation and mathematical humanism === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87znbab8hp.fsf@phiwumbda.org> Discussion, linux) >> Look, why dont you humor us poor folk that dont understand >> computability theory? Could you be so kind as to prove that, for any >> machine T_m and any (possibly infinite) tape x, there is a machine T_m >> such that T_m(x) = T_m()? Somehow, I just cant see the argument. >> Oh, I dont doubt it if we restrict to the case that x is finite. >> Its that other case that you havent given any argument for. > Again, Im not talking about infinte x. Im talking about a machine M1 > which has an infinite arbitrary tape can decide the problem: given a > Turing Machine n, does this this n(th) turing machine halt on empty input? > Youll just say this is trivial. Yes. And Ill also say that this is not a solution to the problem as given. You can see the Rogers text, Theory of Recursive Functions and Effective Computability, for a discussion of Turing machines with auxiliary tapes (sometimes called Oracle tapes). What youve argued is that, the Halting problem is decidable relative to the Halting problem, if I understand the terminology. In other words, given a machine which solves the Halting problem, we can solve the Halting problem (this is an equivalence that Rogers brießy discusses, that having an auxiliary tape is equivalent to querying an oracle machine). None of this is relevant to the plausibility of allowing arbitrary tapes as input, near as I can figure. >>Obviously, saying that a TM can only take finite input would be >>sufficient to avoid these inconsistencies, but if you want to >>start allowing infinite inputs, you have to be careful as to >>which ones are allowed and which are not, if you want to still >>keep problems like the Halting problem as undecidable. >> No you dont. > If you dont need to be careful, as you write, then easily tell me that > if you are going to extend the turing machine model to allow them to have > infinite input, which input are allowed such that it does not affect what > problems are decidable. It all depends on what counts as a problem. If, as is typical, a problem is a function f:N^k -> N satisfying some description (like the halting problem, which is the function N^2 -> N such that..., or in your preferred version, a function N -> N such that...), then what a Turing machine does to a tape with infinitely many ones is completely irrelevant. Of course, we can also define new problems, namely problems which take an infinite input and do something to it. Yes, our TMs can now solve (some) of those problems and in a crude sense they couldnt before. But they couldnt before just because these new problems didnt count as problems before. -- Jesse F. Hughes I may have invented it, but Bill [Gates] made it famous. -- David Bradley, inventor of the Ctrl-Alt-Del reboot sequence === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87smh2dg8v.fsf@phiwumbda.org> Discussion, linux) >> The only behaviour of TMs with infinite and arbitrary input required >>here is that they exist... >> The TM that takes finite input k can simulate the magical machine and >>retrieve the answer. >> Absolutely wrong. >> The universal Turing machine can simulate any other machine using >> input on the TMs tape. > Fine, fine. To avoid my poorly defined simulation, take a turing machine > that has the magical tape on tape 1 and any input n on tape 2. Then it > will solve the halting problem by looking at tape 2 then checking its > magical tape. That is a different notion of Turing machine. Thats a Turing machine with an oracle tape. With oracle tapes, the halting problem is decidable. Look, heres what you need to show that the halting problem is decidable in our setting. You need to show that there is a Turing machine that, when fed a tape with m and n and *no other* symbols on that tape and *no other* tape, the machine always halts, outputting 1 iff machine m halts on input n and otherwise outputting 0. This is the halting problem. Your two tape version is not relevant. >> The tape with the characteristic function is what is magical, not >> some machine which is fed the tape. > Well, this is the argument against the fact that a Turing machine can > start with any arbitrary input tape, which was my point. It is not an argument against allowing arbitrary tapes at all. It does not prove that the halting problem is decidable when we allow tapes with infinite input. It proves that if we allow two tapes, the second of which may have the characteristic function of the halting problem, then the halting problem is decidable. There is no Turing machine which, when fed *just* m and n, behaves like your machine which is fed m, n and another tape consisting of the characteristic function of the halting problem. >> There are no magical machines in this discussion. There is, perhaps, >> a magical tape, but so what? > Obviously, what I meant by the Ômagical TM is a TM that started with > that magical tape. But thats not particularly interesting or troublesome, near as I can figure. >>This is just an argument against the existence of machines with >>any arbitrary starting configuration on their tape. >> I think its an argument for you revisiting an introduction to Turing >> computability. Look up universal machine. See what it means. Look > I intentionally avoided the term universal turing machine for a > reason. I did, however, use the term Ôsimulation with the knowledge that > a TM has the power to run another TM, sort of like a subroutine. But using > the two-tape example above avoids my need to invoke the poor simulation > description. Yes, it avoids that need by adding yet another irrelevant complexity. Youve added an oracle to the Turing machine. It doesnt show that the halting problem is decidable in the relevant sense. >> Come back and tell us how conventions regarding allowable input to the >> Turing machine (conventions which *dont* alter the definition of the >> halting problem) lead to the undesirable solution of the halting >> problem. > My intention was the show how allowing any input (for instance, the > magical tape) would lead to Ôthe undesirable solution of the halting > problem. But you failed to do so. The undesirable solution is much worse than undesirable. It would show that our theory is inconsistent, because the halting problem is provably undecidable in our setting. But you havent shown anything of the sort, of course. -- Jesse F. Hughes Truth is common stuff, ready to your hand, but lies you have to make yourself, and you cant be sure they are any good until youve used them --- and then its too late. John Steinbeck === Subject: Re: No Set Contains Every Computable Natural > There is no TM that, given arbitrary input n in N, decides > Does the nth TM halt if started on a blank input tape? >Note that the TM deciding a question is supposed to have *finite* >input (a natural number n); consequently, no TM with infinite input >can be said to decide this halting problem -- the deciding must, by >definition, be done by a TM with finite input. > The definition you give says NO TM can decide if the nth TM halts. No, it says ... no TM that, given arbitrary input n in N, ... ^^^^^^^^^^^^^^^^^^^^^^^^^^^^ A TM with infinite input is not even in the competition. > If we allow TMs to have an infinite input tape then there IS a TM > that will solve the halting problem for TMs that only accept finite input. No, because to solve this halting problem, the TM that does the deciding must, by definition, do so with finite input, and the example you give below, like those Jim has given, employ infinite inputs. > The magic tape would consist of n, some separator like X, and the > binary representation of a real number that solves the halting problem > for the countable set of TMs that only accept finite input. As I said in a previous post, if we allow a TM to have infinite input, then every non-computable number is trivially computable by a TM that does nothing but halt immediately on every input. I put the word in scare quotes, because it doesnt have the meaning as in its finite-input definition. Just as with the functions Minsky calls real-computable, if we want to speak of computability with infinite inputs, the concept needs some special terminology to avoid confusion. Its the same situation with the phrase solving the halting problem -- words that have been given precise finite-input definitions can be misapplied to TMs with infinite inputs. So far, the misapplication has been shown to produce plenty of confusion through inconsistent usage, but no logical inconsistency has been shown. --r.e.s. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87znbab8hp.fsf@phiwumbda.org Look, why dont you humor us poor folk that dont understand > computability theory? Could you be so kind as to prove that, for any > machine T_m and any (possibly infinite) tape x, there is a machine T_m > such that T_m(x) = T_m()? Somehow, I just cant see the argument. > Oh, I dont doubt it if we restrict to the case that x is finite. > Its that other case that you havent given any argument for. Again, Im not talking about infinte x. Im talking about a machine M1 which has an infinite arbitrary tape can decide the problem: given a Turing Machine n, does this this n(th) turing machine halt on empty input? Youll just say this is trivial. >Obviously, saying that a TM can only take finite input would be sufficient >to avoid these inconsistencies, but if you want to start allowing infinite >inputs, you have to be careful as to which ones are allowed and which are >not, if you want to still keep problems like the Halting problem as >undecidable. > No you dont. If you dont need to be careful, as you write, then easily tell me that if you are going to extend the turing machine model to allow them to have infinite input, which input are allowed such that it does not affect what problems are decidable. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87wu6eb7uu.fsf@phiwumbda.org Heres another argument that your claim is simply wrong. It is not > the case that, when x is an infinite input, there is a machine T > which, when fed , simulates machine T on x. Again, stop assuming x is infinite. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p of infinite inputs, youve claimed that some TM with finite input > *can* simulate certain specific TMs with arbitrary infinite input. > Thats just not correct, because to assume so would contradict the > known unsolvability of the halting problem under discussion. This > says nothing about the kind of inconsistency youre trying to show. Right... Im arguing with the premise that if TMs with infinite input and the ability to use the infinite input can behave like normal TMs, then a normal (finite) TM can use them as subroutines. And, yes, this premise contradicts the unsolvability to the halting problem, which is what I was trying to show. > There is no TM that, given arbitrary input n in N, decides > Does the nth TM halt if started on a blank input tape? > Note that the TM deciding a question is supposed to have *finite* > input (a natural number n); consequently, no TM with infinite input > can be said to decide this halting problem -- the deciding must, by > definition, be done by a TM with finite input. Again, my argument was that (under the assumption that TMs with infinite input can behave and compute things normally reserved for finitely-input machines) would lead to inconsistencies. > Solving this halting problem *means* that a TM with finite input > decides it. Whatever else a TM with infinite input might do, it > does not solve this problem, and hence produces no inconsistency > on that account. Okay, but I was operating under the premise that solve is not restricted to solve by a TM with finite input, but instead extended to the hypothetical machines. > You havent yet shown an inconsistency (except in thinking that TMs > with finite input are able to simulate those with infinite input). The is the premise with which I used to draw contradictions. J P.S. I still dont think you understood what I was trying to show in the first place. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p Discussion, linux) Heres another argument that your claim is simply wrong. It is not the case that, when x is an infinite input, there is a machine T which, when fed , simulates machine T on x. How many input tapes are there? Uncountably many. So how many output tapes? Uncountably many (since all of them are output by the machine which does nothing). Now, can countably many machines produce uncountable output when fed just the empty input? Obviously not. They can only produce a single output per machine. Therefore, your argument that infinite tapes allow for the solving of the halting problem is simply based on an obviously false assumption. -- Jesse Hughes Like the ski resort full of girls hunting for husbands and husbands hunting for girls, the situation is not as symmetrical as it might seem. -- Alan MacKay === Subject: Re: No Set Contains Every Computable Natural > There is no TM that, given arbitrary input n in N, decides > Does the nth TM halt if started on a blank input tape? > Note that the TM deciding a question is supposed to have *finite* > input (a natural number n); consequently, no TM with infinite input > can be said to decide this halting problem -- the deciding must, by > definition, be done by a TM with finite input. The definition you give says NO TM can decide if the nth TM halts. If we allow TMs to have an infinite input tape then there IS a TM that will solve the halting problem for TMs that only accept finite input. The magic tape would consist of n, some separator like X, and the binary representation of a real number that solves the halting problem for the countable set of TMs that only accept finite input. (Such a real number must exist.) Of course, the set of all TMs doesnt exist, so the halting problem is bogus, anyway. Russell - I may be a crackpot, but I am getting good at it. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p Discussion, linux) > But, still, you can NOT allow any arbitrary input tape. Again, > if you dont like my simulation given before, then make a machine that > take as input a number n and then a separator symbol Ô# and then a string > of the characteristic function for halting turing machines. Then the > machine can decide the halting problem for any machine n on empty > input. You are mistaken. Look, why dont you humor us poor folk that dont understand computability theory? Could you be so kind as to prove that, for any machine T_m and any (possibly infinite) tape x, there is a machine T_m such that T_m(x) = T_m()? Somehow, I just cant see the argument. Oh, I dont doubt it if we restrict to the case that x is finite. Its that other case that you havent given any argument for. > My arguments are that one can not have any tape they want as > input. Yeah, but youre wrong. > Obviously, saying that a TM can only take finite input would be sufficient > to avoid these inconsistencies, but if you want to start allowing infinite > inputs, you have to be careful as to which ones are allowed and which are > not, if you want to still keep problems like the Halting problem as > undecidable. No you dont. -- It has been shown that no man can sit down to write without a very profound design. Thus to authors in general trouble is spared. A novelist, for example, need have no care of his moral. It is there -- that is to say, it is somewhere -- and the moral and the critics can take care of themselves. --E.A. Poe === Subject: Re: No Set Contains Every Computable Natural > The only behaviour of TMs with infinite and arbitrary input required > here is that they exist... > The TM that takes finite input k can simulate the magical machine and > retrieve the answer. >> No, it cant, because to do so would solve the halting problem, >> which is provably unsolvable. > This is precisely my point. No. First of all, in trying to find an inconsistency due to the use of infinite inputs, youve claimed that some TM with finite input *can* simulate certain specific TMs with arbitrary infinite input. Thats just not correct, because to assume so would contradict the known unsolvability of the halting problem under discussion. This says nothing about the kind of inconsistency youre trying to show. Secondly, you dont seem to see the reason given for there being no such inconsistency. Heres a statement of the unsolvability of the halting problem were discussing: There is no TM that, given arbitrary input n in N, decides Does the nth TM halt if started on a blank input tape? Note that the TM deciding a question is supposed to have *finite* input (a natural number n); consequently, no TM with infinite input can be said to decide this halting problem -- the deciding must, by definition, be done by a TM with finite input. > Either (1) you are trying to support my > statements (which I think is unnecessary), or (2) you believe that the > existence of a machine with an arbitrary starting tape although one can > not invoke it, or (3) you dont know what my original point was. Or (4) you havent understood the objection. >> This is just an argument against the existence of machines with >> any arbitrary starting configuration on their tape. >But they do exist, and without contradiction. I think the first >instance of such a TM that I can recall seeing is in Minskys >old textbook. He mentions them in the context of what he calls >real-computable functions, which turn out to be necessarily >continuous. (The input tape contains representations of a natural >number n and of the infinite binary expansion of the argument x of >the function, say f. The output tape contains representations of >the nth digit of f(x) and the argument x.) > But, still, you can NOT allow any arbitrary input tape. Again, > if you dont like my simulation given before, then make a machine that > take as input a number n and then a separator symbol Ô# and then a string > of the characteristic function for halting turing machines. Then the > machine can decide the halting problem for any machine n on empty input. Solving this halting problem *means* that a TM with finite input decides it. Whatever else a TM with infinite input might do, it does not solve this problem, and hence produces no inconsistency on that account. > My arguments are that one can not have any tape they want as input. > Obviously, saying that a TM can only take finite input would be sufficient > to avoid these inconsistencies, but if you want to start allowing infinite > inputs, you have to be careful as to which ones are allowed and which are > not, if you want to still keep problems like the Halting problem as > undecidable. You havent yet shown an inconsistency (except in thinking that TMs with finite input are able to simulate those with infinite input). --r.e.s. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p <87smh2dg8v.fsf@phiwumbda.org The only behaviour of TMs with infinite and arbitrary input required >here is that they exist... > The TM that takes finite input k can simulate the magical machine and >retrieve the answer. > Absolutely wrong. > The universal Turing machine can simulate any other machine using > input on the TMs tape. Fine, fine. To avoid my poorly defined simulation, take a turing machine that has the magical tape on tape 1 and any input n on tape 2. Then it will solve the halting problem by looking at tape 2 then checking its magical tape. > The tape with the characteristic function is what is magical, not > some machine which is fed the tape. Well, this is the argument against the fact that a Turing machine can start with any arbitrary input tape, which was my point. > There are no magical machines in this discussion. There is, perhaps, > a magical tape, but so what? Obviously, what I meant by the Ômagical TM is a TM that started with that magical tape. >This is just an argument against the existence of machines with >any arbitrary starting configuration on their tape. > I think its an argument for you revisiting an introduction to Turing > computability. Look up universal machine. See what it means. Look I intentionally avoided the term universal turing machine for a reason. I did, however, use the term Ôsimulation with the knowledge that a TM has the power to run another TM, sort of like a subroutine. But using the two-tape example above avoids my need to invoke the poor simulation description. > Come back and tell us how conventions regarding allowable input to the > Turing machine (conventions which *dont* alter the definition of the > halting problem) lead to the undesirable solution of the halting > problem. My intention was the show how allowing any input (for instance, the magical tape) would lead to Ôthe undesirable solution of the halting problem. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p here is that they exist... > The TM that takes finite input k can simulate the magical machine and >retrieve the answer. > No, it cant, because to do so would solve the halting problem, > which is provably unsolvable. This is precisely my point. Either (1) you are trying to support my statements (which I think is unnecessary), or (2) you believe that the existence of a machine with an arbitrary starting tape although one can not invoke it, or (3) you dont know what my original point was. > This is just an argument against the existence of machines with >any arbitrary starting configuration on their tape. > But they do exist, and without contradiction. I think the first > instance of such a TM that I can recall seeing is in Minskys > old textbook. He mentions them in the context of what he calls > real-computable functions, which turn out to be necessarily > continuous. (The input tape contains representations of a natural > number n and of the infinite binary expansion of the argument x of > the function, say f. The output tape contains representations of > the nth digit of f(x) and the argument x.) But, still, you can NOT allow any arbitrary input tape. Again, if you dont like my simulation given before, then make a machine that take as input a number n and then a separator symbol Ô# and then a string of the characteristic function for halting turing machines. Then the machine can decide the halting problem for any machine n on empty input. My arguments are that one can not have any tape they want as input. Obviously, saying that a TM can only take finite input would be sufficient to avoid these inconsistencies, but if you want to start allowing infinite inputs, you have to be careful as to which ones are allowed and which are not, if you want to still keep problems like the Halting problem as undecidable. J === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p Discussion, linux) > The only behaviour of TMs with infinite and arbitrary input required > here is that they exist... > The TM that takes finite input k can simulate the magical machine and > retrieve the answer. Absolutely wrong. The universal Turing machine can simulate any other machine using input on the TMs tape. The universal Turing machine does not magically receive any possible input available to any machine. The tape with the characteristic function is what is magical, not some machine which is fed the tape. > If you dont like simulation of magical machines but are still > accepting the premise of their existence (which, personally, I > dont) then consider a two-tape machine, one tape is the tape > containing the information that tape cell i = 1 iff i(th) TM halts > on empty input and the other tape is reserved for the input. There are no magical machines in this discussion. There is, perhaps, a magical tape, but so what? > This is just an argument against the existence of machines with > any arbitrary starting configuration on their tape. I think its an argument for you revisiting an introduction to Turing computability. Look up universal machine. See what it means. Look up halting problem. See what it means. Come back and tell us how conventions regarding allowable input to the Turing machine (conventions which *dont* alter the definition of the halting problem) lead to the undesirable solution of the halting problem. -- Jesse Hughes You see 300 of something, anything, and you go `[Man], thats a lot of stuff. -- Jim Bigler, quoted in the Pittsburgh Post-Gazette. === Subject: Re: No Set Contains Every Computable Natural >> question: Given k, does the k(th) Turing Machine halt on empty input? > ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ >> (This problem is undeciable.) >No, that would *not* solve the halting problem that you just stated. >That problem requires the TM to be given k only, i.e. finite input. >The behavior of TMs with infinite input isnt relevant to the problem. > The only behaviour of TMs with infinite and arbitrary input required > here is that they exist... > The TM that takes finite input k can simulate the magical machine and > retrieve the answer. No, it cant, because to do so would solve the halting problem, which is provably unsolvable. > If you dont like simulation of magical machines but are still accepting > the premise of their existence (which, personally, I dont) then consider > a two-tape machine, one tape is the tape containing the information that > tape cell i = 1 iff i(th) TM halts on empty input and the other tape is > reserved for the input. That doesnt help. Its not a problem of liking or disliking, but of the fact that these are two completely different kinds of situation (i.e. TMs that are or are not restricted to finite inputs). Neither one is magical. > This is just an argument against the existence of machines with > any arbitrary starting configuration on their tape. But they do exist, and without contradiction. I think the first instance of such a TM that I can recall seeing is in Minskys old textbook. He mentions them in the context of what he calls real-computable functions, which turn out to be necessarily continuous. (The input tape contains representations of a natural number n and of the infinite binary expansion of the argument x of the function, say f. The output tape contains representations of the nth digit of f(x) and the argument x.) --r.e.s. === Subject: Re: No Set Contains Every Computable Natural <87ptcau3xj.fsf@p Discussion, linux) >> Another excellent point against starting with whatever tape you like. >>My use of anything is too broad, and maybe someone would say any >>sequence of values that is actually computable can be an initial tape... >>but then that goes back to requiring a finitely-lengthed input. >> I dont see how Alans point is applicable here. The behaviour >> of infinite-input TMs is irrelevant to the finite-input halting >> problem -- by its definition that halting problem would be >> solved only by a finite-input TM. > Its not just the act of allowing an infinite input, but allowing *any* > infinite input would lead to solvability of any problem, like the halting > problem. Let a TM have initial tape with cell T_i = 1 if and only if the > i(th) Turing Machine halts on an empty input. Then this TM can decide the > question: Given k, does the k(th) Turing Machine halt on empty input? > (This problem is undeciable.) Its still undecidable in the relevant sense. The halting problem is decidable iff there is a machine T such that, when we feed T a tape with two natural numbers m and n on it, T halts and outputs 1 iff blah blah blah. The characteristic tape for the halting problem is irrelevant. The problem is about feeding a machine a tape with two numbers on it, not two numbers *and* a characteristic function. None of this gives any reason to reject infinite tapes (tapes with an infinite number of ones) as input. -- Jesse F. Hughes The sole cause of all human misery is the inability of people to sit quietly in their rooms. -- Blaise Pascal === Subject: Some Help is Needed advance for helping me. Let G be a group and let g E G. If g^2 /= e and g^6 = e, then prove that g^4 /= e and g^5 /= e. What can you say about the order of g? I couldnt really figure out how to write not equal to so the symbol I used is /=. I understand that since g^2 /= e then g^4 /= e. Also the order of g is 6 So if anyone wants to guide me in the right direction it would greatly be appreciated. === Subject: Re: Some Help is Needed >advance for helping me. > Let G be a group and let g E G. If g^2 /= e and g^6 = e, then prove >that g^4 /= e and g^5 /= e. What can you say about the order of g? >I couldnt really figure out how to write not equal to so the symbol I >used is /=. >I understand that since g^2 /= e then g^4 /= e. > Ahh, no. Suppose G =Z/4Z. Then 1+1+1+1 = 0, but 1+1 /= 0. He meant that if g^6 = e and g^2 /= e, then g^4 /= e. This is in fact true. For if g^4 = e, then e = g^6 = (g^4)(g^2) = g^2. But g^2 /= e. Contradictcion, so g^4 /= e. >Also the order of g is 6 >So if anyone wants to guide me in the right direction it would greatly >be appreciated. > Looking at g^4 again ... e = g^6 = (g^4)(g^2). Now suppose > g^4 = e, and what can you conclude about g^2? === Subject: Re: Some Help is Needed > advance for helping me. > Let G be a group and let g E G. If g^2 /= e and g^6 = e, then prove > that g^4 /= e and g^5 /= e. What can you say about the order of g? > I couldnt really figure out how to write not equal to so the symbol I > used is /=. > I understand that since g^2 /= e then g^4 /= e. Ahh, no. Suppose G =Z/4Z. Then 1+1+1+1 = 0, but 1+1 /= 0. > Also the order of g is 6 > So if anyone wants to guide me in the right direction it would greatly > be appreciated. Looking at g^4 again ... e = g^6 = (g^4)(g^2). Now suppose g^4 = e, and what can you conclude about g^2? === Subject: Re: Some Help is Needed > advance for helping me. > Let G be a group and let g E G. If g^2 /= e and g^6 = e, then prove > that g^4 /= e and g^5 /= e. What can you say about the order of g? > I couldnt really figure out how to write not equal to so the symbol I > used is /=. > I understand that since g^2 /= e then g^4 /= e. > Also the order of g is 6 Does e represent the additive identity? If so, assume g^5=e. Then whats (g^5)^2? Doug === Subject: Re: help with solutions for three questions from the past contest ... > 1. Al and Bob are at opposite ends of a diameter of a silo in the > shape of a tall right circular cylinder with radius 150 ft. al is due > west of Bob. Al begins walking along the edge of the silo at 6 ft. per > second at the same moment that Bob begins to walk due east at the same > speed. The value closest to the time in seconds when Al first can see > Bob is what? answer: 48 Al travels a distance 6t around the silo (CW or CCW, makes no difference) to a point where his or her sightline to Bob is tangent to the circle; if Al and Bob then are d apart, you have a right triangle r, d, r+6t, and an angle a = arctan(r/d); and you also know that 6t = (a+pi/2)r. Then write an equation d = sqrt(r*r*((arctan(r/d)+pi/2+1)^2-1)), plug in an initial value of d, solve for a new value of d a few times, getting about 411.562, then solve for t about 48.007. Since we know 6t > 2pi*r/4, use that t and d=sqrt(r*r+(r+6t)^2) ~ 414 for the initial guess of d. > 2. if a, b, c, and d are nonzero numbers such that c and d are > solutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, find > a+b+c+d. ans: -2 This is simple algebra if you write x^2+ax+b=(x-c)(x-d), x^2+cx+d=(x-a)(x-b), expand, equate, substitute -(c+d) and cd into a+b+c+d, reduce, substitute -(a+b) and ab into a+b+c+d, reduce, equate, deduce b=d and a=c, substitute into x^2+ax+b= (x-c)(x-d), deduce a=1, hence a+b+c+d=-2. > 3. A boat with an ill passenger is 7.5 mi north of a straight > coastline which runs east and west. A hospital on the coast is 60 > miles from the point on shore south of the boat. If the boat starts > toward shore at 15 mph at the same time an ambulance leaves the > hospital at 60 mph and meets the ambulance, what is the total distance > (to the nearest 0.5 mile) traveled by the boatand the ambulance? ans: > 62.5 Dividing all the numbers by 7.5, the boat starts 1 unit N and 8 W (or E, makes no difference) from the ambulance; the ambulance travels 8t units E and the boat 2t units about SSE; draw the appropriate right triangles (eg, if 8=8t+u, you have a 1, 8, sqrt(65) triangle and a 1, u, 2t triangle); substitute u=sqrt(4t^2-1) into 65=1+(8t+u)^2, write an equation t_(i+1)=f(t_i) and approach a solution as in problem 1 to get t=0.833...; and t*75=62.5. There probably is a better way to solve this problem. -jiw === Subject: Re: help with solutions for three questions from the past contest questions. 1. Al and Bob are at opposite ends of a diameter of a silo in the shape of a tall right circular cylinder with radius 150 ft. al is due west of Bob. Al begins walking along the edge of the silo at 6 ft. per second at the same moment that Bob begins to walk due east at the same speed. The value closest to the time in seconds when Al first can see Bob is what? answer: 48 2. if a, b, c, and d are nonzero numbers such that c and d are solutions of x^2+ax+b=0 and a and b are solutions of x^2+cx+d, find a+b+c+d. ans: -2 3. A boat with an ill passenger is 7.5 mi north of a straight coastline which runs east and west. A hospital on the coast is 60 miles from the point on shore south of the boat. If the boat starts toward shore at 15 mph at the same time an ambulance leaves the hospital at 60 mph and meets the ambulance, what is the total distance (to the nearest 0.5 mile) travelled by the boat the ambulance? ans: 62.5 ********************************************************* Could you post each question separately next time . Replying to 3 makes this a large posting. Let x = 6t feet be the distance travelled by Al (A) and Bob (B) in t seconds. Sketch a tangent to the circle, (representing the cylinder), from the centre O, draw in the radius which is perpendicular to the tangent contact at C. pi - x/150 is the acute angle BOC The cosine of this angle is OC/OB =150/(150+x) = C, say Use a graphing calculator to graph y = cos (pi - x/150) and y =150/(150+x) With x from - 1 to 300, y=-1.1 to+1.1 intersection gives x=28 8.0448 hence t is close to 48 [No graphical calculator allowed? As a very rough method, a careful sketch 150 (1 - .342)/.342 =approx 288 ] ************************************** Quadratics can be written x^2 - (sum of roots)+(product of roots) = 0 c+d = -a cd = b a+b = -c ab = d b-c-d=-c b=d c=b/d=1 a=d/b=1 a+b+c+d=-(a+c)=-2 **************************** Take meets to mean arrives at the same time, and assume that boat intelligently steered towards optimum angle x west of due south. Sketch this for yourself. Let cosine x=c, sine x=s, tangent x=t, Distance travelled by boat = 7.5/c Miles Time taken by boat = 30/c Miles Miles = minutes for ambulance = 60 - 7.5 t Equating x and dividing by 7.5 4/c = 8 - t 4 = 8 c - s s = 4(2c - 1) s^2 = 16(4c^2 - 4c + 1) = 1 - c^2 65c^2 - 64c + 15 = 0 = (5c - 3)(13c - 5) Shortest time taken by boat corresponds to largest c, so select c = 3/5 which also gives t = 4/3 Distance travelled by boat = (7.5 x 5)/3 = 12.5 miles Minutes taken by boat= 4 x 12.5 miles = 50 but this is same as miles travelled by ambulance, so Total distance travelled by boat and ambulance = 62.5 [ By the way, the other solution to the quadratic also allows boat and ambulance to arrive at same time, but c = 5/13 with t = 12/5 lead to a boat distance of 19.5 miles and an ambulance distance of 78 miles. This is interpreted as the less useful solution where the boat is steered the east of south ] afterwards Ian Hutcheson === Subject: Brian Greenes new book I got news for Brian in my book Super Cosmos http://qedcorp.com/destiny/ under construction Battle of the Books I am Ralph Nader to Brians beating about the Bush. The Higgs ocean is my Vacuum Coherence Field calming the turbulent ZPF. It explains; 1. Einsteins GR 2. Dark energy and dark matter as virtual exotic vacuum stuff. Dark matter detectors will not click with right stuff. 3. Stabililty of spatially extended leptons and quarks and why they look increasingly point like in high energy scattering (partons). 4. Universal slope of Regge trajectories of hadronic resonances. alpha ~ (1 Gev)^-2 5. Hierarchy problem - Quantum Gravity Planck area is actually ~ (1 fermi)^2 The weave is, perhaps, not as fine as previously thought, at least in the current cosmic epoch. 6. Not to mention the unmentionable you know what. For I never use the Big Big U ... Chorus: What never? Well hardly ever .. http://math.boisestate.edu/gas/pinafore/captain.mp3 http://math.boisestate.edu/gas/pinafore/html/index.html Bought Brian Greenes book this morning - nice chapter on the Higgs fields called vaporizing the vacuum... Brian Greene: If the Higgs ocean is not found, it will require a major rethinking of a theoretical framework that has been in place for more than thirty years === Subject: Re: Im guessing this is an easy problem - but I cant solve it! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1MKlMj01590; >Working through calculus... Stuck on a problem of factorisation: Please >help! >question is: >find lim u--> 1 f(u) > where f(u) = (u^4 - 1) / > (u^3 -1) >Obviously - I cannot just replace u with 1 as this will give me a division >by zero. I tried replacing (u^2 - 1) (u^2 +1) on top as this seemed most >logical - but I cant figure out what to do with the bottom now. >Im probably going to kick myself for not figuring this one out - but Ive >spent hours on this and Im still lost. >Help! If this is a calculus problem, look up LHospitals Rule phil === Subject: Re: Category Theorys De-Emphasis of Sets is a Serious Mistake by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1MKlPS01696; Here is a list of some of the implicit or explicit errors of the Category Theory approach regardless of its direct negative effects on the Set-related Theories that I mentioned in the last posting. A. The assumption that an Object is more general than a General- ized Set. The claim is often made that there is a paradox in defining a Set of Sets, which is allegedly absent from an Object. But nothing is common to Objects since they are not defined in terms of a common characteristic, and if they were the Set Paradox would affect it since a collection of things with common characteristics is a Set! B. The assumption that Functions are more useful than Sets. I had an argument with a Belgian Computer-Oriented Logician about this some years ago (Professor Smets who to my recollection founded the famous IRIDIA in Belgium and France), and we didnt conclude that one was more useful than the other although each of us remained convinced personally that our favorite was probably more useful. However, the assumption that Functions are more useful than Sets in Real Analysis would involve the assumption for example that the Lebesgue Integral is more useful than the Lebesgue Measure (Measures being on sigma-algebras, which are types of Sets), which is silly. C. Like B above with Morphisms replacing Functions. D. Like B above with Functors replacing functions. E. The Mis-direction of Composition. The Category Theory claims priority for Composites of Morphisms, or in Functions Composites of Functions. The composition f o g (x) = f(g(x)) is not the same as fg(x) = f(x)g(x) or (f + g)(x) = f(x) + g(x) and so on, although one could always claim that the latter two are somehow special cases of the former with a slight generalization. In Rare Event Theory, it is shown that fg and f + g are far more important than f o g. This is perhaps the most likely to be criticized by the Mainstream researchers since it goes over to the whole question of matrix representations, but its true for Rare Event Theory. One may not want something to be true, but that doesnt change its truth. F. The failure of Category Theory to emphasize discovery of new ideas and new concepts in its preoccupation with organizing old ones. It does nothing with Anomalies, Paradoxes, Infinities, Causation, Correlation, Motion, and so on in this respect. This includes Category Theorys failure to match Rare Event Theorys discovery of 1 + y - x as a form across geometry, Fuzzy Multivalued Logics, and Probability-Statistics. You can organize forever, but looking for common interdisciplinary forms is unbeatable in my opinion. Osher Doctorow === Subject: Holograms & Strings As Above, So Below? Susskinds world hologram suggests something like Lp* = Lp^2/3(c/H)^1/3 H = R^-1dR/dt in FRW metric with Einstein cosmological constant / ~ (H/c)^2 = Einsteins Cosmological Constant alpha ~ Lp*^2 in Wittens string theory i.e. (h = c =1 convention) alpha ~ Lp^4/3//^2/3 In terms of ADS the world Hubble horizon is ~ /^-1 in area where the sign is for dark energy repulsion Each Bekenstein Bit has quantum of area Lp*^2 The Entropy of the expanding accelerating universe is then ~ (Lp*^2/)^-1 So is this plausible? It solves the Arrow of Time problem trivially by hooking it to the area of the Hubble horizon. The Second Law of Thermodynamics is explained as well as the origin of gravity and inertia and dark energy/matter in terms of Sakharovs Vision and P.W. Andersons Battle Tested tried and true More is different. We dont need no damn G-string vibrators in hyperspace, loops and weaves Of course those ideas may play a useful role anyway. More with less. No excess metaphysical baggage and no excess mathematical baggage. The Question is: What is The Question? ;-) I got news for Brian in my book Super Cosmos http://qedcorp.com/destiny/ under construction Battle of the Books I am Ralph Nader to Brians beating about the Bush. The Higgs ocean is my Vacuum Coherence Field calming the turbulent ZPF. It explains; 1. Einsteins GR 2. Dark energy and dark matter as virtual exotic vacuum stuff. Dark matter detectors will not click with right stuff. 3. Stabililty of spatially extended leptons and quarks and why they look increasingly point like in high energy scattering (partons). 4. Universal slope of Regge trajectories of hadronic resonances. alpha ~ (1 Gev)^-2 5. Hierarchy problem - Quantum Gravity Planck area is actually ~ (1 fermi)^2 The weave is, perhaps, not as fine as previously thought, at least in the current cosmic epoch. 6. Not to mention the unmentionable you know what. For I never use the Big Big U ... Chorus: What never? Well hardly ever .. http://math.boisestate.edu/gas/pinafore/captain.mp3 http://math.boisestate.edu/gas/pinafore/html/index.html Bought Brian Greenes book this morning - nice chapter on the Higgs fields called vaporizing the vacuum... Brian Greene: If the Higgs ocean is not found, it will require a major rethinking of a theoretical framework that has been in place for more than thirty years === Subject: Typo Greene Strings & Hologram alpha ~ Lp^4/3//^2/3 should be alpha ~ Lp^4/3//^1/3 === Subject: Re: advice on graduate studies > I am an undergrad who will be graduating this semester and > will be going to grad school in fall((if they let me)) > and i have a few questions. > I have heard that you should not only think about what you like > but also what is still a growing field. > I really enjoy analysis.((classical,functional,Fourier..)) > I really dislike (at least so far) algebra,number theory ,combinatorics. > Now my question is : are the above named analysiss still a > good field to get into or should i look more at say applied > analysis.Someone once told me that unless you are a genious > that harmonic analysis is not a good field to get into. > Is this because the field is so old and developed? > Would the same apply to classical and functional analysis? I have done some PDE theory and enjoyed working in the > Sobolev spaces with the functional analysis techniques. > If i was to go into PDEs would I expect more of the same > (which i enjoyed) or does it turn into > a grab bag of techniques ? my limited experience tells me that some topics in harmonic analysis are essential in the study of partial differential equations. if you choose to work in harmonic analysis problems but find yourself stalled, you can find your way into PDE, which btw, is a rather popular field. > I have never done any measure theoretic probability > but considering my interests would this be another one > to think about? imo, it would be better you explore subjects for which you already have adequate exposure, like PDE. of course, you should take one or two courses on measure theory in grad school. > Any replies would be greatly appreciated don === Subject: Re: advice on graduate studies > I have heard that you should not only think about what you like but also > what is still a growing field. I really enjoy analysis. > ((classical,functional,Fourier..)) That, of course, depends on your intentions... if you want to study mathematics because you love mathematics and yadda yadda, then study what you like most. But if you actually care about, say, landing a career out of your studies, then considering the Ôhot topics would be useful. Some poeple would, of course, say that you should always have a career in mind, but I dont think a university is supposed to be a job-training institution (but that argument is for a different thread.) > I really dislike (at least so far) algebra,number theory ,combinatorics. Ah, pity... but to each their own. > Now my question is : are the above named analysiss still a > good field to get into or should i look more at say applied > analysis.Someone once told me that unless you are a genious > that harmonic analysis is not a good field to get into. > Is this because the field is so old and developed? > Would the same apply to classical and functional analysis? The pluaral of analysis should be analyses. I would kind of have to agree with what that person told you... standard and classic algebra and analysis are the core fields of math and are populated with some of the best. They are also (I think) the oldest fields and most developed, so it takes a lot of (a combination of) work and ingenuity to get somewhere new in those fields. And, of course, since they are so far developed into abstractness, the applications (and hence funding) are harder to come by. Some fields like applied analysis can study problems and write papers with titles that sound applied-like, like physics related topics. The fields you mentioned that you dont like (like number theory and combinatorics) have gained a lot of interest from fields like computer science because of the host of algorithmic problems to study in those fields... I was interested in discrete math and optimization, and when I applied for grad school (to math departments) to ended up in a computer science department. I dont give a darn about databases, operating systems, graphics, etc, but I love the math that I can study here and I get a lot more funding here than I any most student I know (and yet I know I am quite inferior to a majority of them.) > I have done some PDE theory and enjoyed working in the > Sobolev spaces with the functional analysis techniques. > If i was to go into PDEs would I expect more of the same > (which i enjoyed) or does it turn into > a grab bag of techniques ? Every field will have its standard techniques for solving problems, but those will only go so far. > I have never done any measure theoretic probability > but considering my interests would this be another one > to think about? Someone else is probably more qualified to reply to this :P Good luck, and remember that much of the above is just one persons opinions. J === Subject: Re: A newbies question -- about real number > Two real numbers r1 = 0.89, r2 = 0.889999999..... (with infinite 9s), * Yes. earle * === Subject: Re: Mathcad Upgrade Question I actually, finally purchased Mathcad 2001 -- NOT 2001i -- and installed it. As far as I can tell, checking my processes and my registry, c-dilla is not on here. Steve O. >Dont use mathcad. I installed version 2001i this weekend. After >installition I noticed that c-dilla had also been introduced to my registry. >Google c-dilla to find out what this program does. Anyhow I have spent 2 >days rebuilding my hard drive because of the damage that Mathcad did. >> I am thinking of upgrading from an old version of Mathcad (Version 7) >> to the most recent. But past experience with upgrades has taught me >> to be a bit wary, so 2 questions: >> 1. Did Mathsoft do anything to screw up the product, in terms of >> usability or features, since I last purchased it (back in 1997)? >> 2. Does the latest product come with any kind of registration >> feature, like Mathematica has, that locks your software into one >> computer, and possibly locks you out of the software if you upgrade >> your hardware? (I respect the rights of companies to make money off >> their software, but Im a firm believer that the only fair deal is >> one-user/one-app. If I have three computers -- and I do -- I should >> not have any worries or complications with loading the same software >> package on all three of them.) >> Steve O. >> Standard Antißame Disclaimer: Please dont ßame me. I may actually >*be* an idiot, but even idiots have feelings. === Subject: Re: Advice on future with Math > Im an upperclass math major that was/is planning on attempting a masters or > phd in math. Here lately, Ive wondered if this is a great idea. > My first question is this: How much should you study (reading the material, > working problems) for, say, an abstract algebra or advanced calculus > junior/senior level class? I talked to a graduate student that said > he studied for a couple of his first year grad classes at least 4 hours > a day on weekdays and about 8 hours on Saturday and Sunday. Needless to > say, I do nothhing like that. Should I be doing that or is that just > an insane amount of time to be spending on it? It seems that if youre > good at something, it shouldnt require so much work. it depends - some people are dumber than others. for instance i recall working even days on single analysis exercises. although, i had quite a number of scummy nitwits in charge of supplying me with the basics. > At times I have difficulty understanding things that, after finding out > where I went wrong, seem pretty simple. I assume others dont have > this problem that is the learning process. i would say it is somewhat normal. > because normally they can respond right away with what some math > concept meant or, less often, help me where I am having a problem > understanding part of a proof. Is it normal to not follow a proof > or just a sign that Im just weak mathematically (were again talking > junior/senior level advanced calc/abstract algebra here)? depends. if you seem to have trouble understanding points and details most of the time, that is likely a sign of a mediocre preparation. on the other hand, if you experience such difficulty ocassionally, then it is a matter of spending more time and being persistent. > I have a bad tendency to be hot/cold when it comes to math. It seems > like Ill have the highest/near highest score on an exam and then > really screw up on the next exam, getting something in the bottom > half of the grades. I wish that I could attribute it to something like > when I first started school: I would do no work until after I screwed up > on the first exam, and then work harder after that and do quite well. > Now it seems that Ill do well on the first exam and then screw up on > the next one. I really dont think that I will have studied > much less for the second one but I guess it could be a possibility. When I > do bad on the other exam, I begin to wonder if I just got lucky on the > problems that were given on the good exam and would have done lousy if a > few different problems were picked. do you study by yourself or do you have group studies? group studies seem to be ineffective in higher math, except when one has successfully solved problems independently prior to the group study session. > Other times, I wonder if I am just not good at proofs, thus making me > about the crappiest grad school candidite out there. quality of students and schools are not high these days. worse case scenario you are in the norm. > I sit and stare at the homework problems and often cant get anywhere. bad sign - if you cant get anywhere in routine exercises, you may be missing something significant. do you spend sufficient time studying math? or do you like partying a lot? is it possible that there is an assortment of incompetent teachers in charge of provideing you with the basics? > After I see an answer, it doesnt look too bad but it just seems that > theyre impossible at times. I had a bad proofs > class that was designed for just about anyone wanting to take it, so it was > really easy. Im not sure if I can use that excuse at this point though. My > pre-calculus education was really bad since I was from a really poor school (I > often have no clue what certain things that were assumed as Ôknowledge from > high school) but again this doesnt seem to matter too much in the proof-based > classes. the future seems grim for you - i doubt any good graduate school will take in a student with a weak backround. quit math while you are not so far behind, unless you really love it for its own sake. > Well, that was kind of long-winded. I hope that I got the point across. Any > advice would be helpful. === Subject: Re: Advice on future with Math with how I study or that I am just not quite good enough for more advanced math. === Subject: vertex covering, integer programming, and linear optimization covering problem, namely, given a graph G = {E, V}, what is the smallest number of vertices we must choose in V, so that every edge contains at least one of the chosen vertices. What I would like is a way to obtain a 2-approximation (using linear optimization, and then explaining how the rational solutions lead to integral solutions). Any assistance would be appriciated. --David Jerrisfield === Subject: Re: p_(n+1) < 2 p_n > Let p_n be the nth prime. Is there a simple proof that p_(n+1) < 2 p_n > for all n? It follows from the fact that for all n >= 2, there is a prime p such that n < p < 2n. Heres Erdos proof: http://en.wikipedia.org/wiki/Proof_of_Bertrand%27s_postulate -- --Tim Smith === Subject: p_(n+1) < 2 p_n Let p_n be the nth prime. Is there a simple proof that p_(n+1) < 2 p_n for all n? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: JSH: Understand now? Frustration? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1N1UDb24831; Using P(x) = (x+8a)(x+b), where ab=1, and considering when (8a + b) is >an integer can give you a perspective on what Ive been saying, I >hope. Since b = 1/a, by your requirement, then (8a + b) = (8a + 1/a). Are you > requiring that this expression be an integer? >>To be fair, he said consider those cases where (8a+b) _is_ an integer. >>In a previous post he chose (8a+b) = 9, ab=1 which has the two solutions >> [b=8, a=1/8] and [b=1, a=1] >>so both (8a + b) and (8a + 1/a) are indeed integer even though a=1/8 is not. >>KeithK >> I dont believe I was being unfair. I was simply asking for clarification. James >> has a history of posting contradictory requirements and leaving the burden of >> choosing what to work with to the reader. >Ive been arguing with posters like Arturo Magidin for YEARS, and >throughout that period, Ive found my attempts to clarify were >hampered by attempts by other posters like here C. Bond to confuse, >and I thank KeithK for catching one as he did. Didnt Arturo win every single argument? >The problem isnt with C. Bond using 8a + 1/a as that is correct, as >long as you consider the possibility of Ôa being a factor of 1 in >some more inclusive ring than the ring of algebraic integers. >Circular arguments have gone on and on and on for some years here, as >people like Arturo Magidin, have successfully dodged being pinned down >on this issue as they repeatedly relied on certain numbers not being >algebraic integers to support their claims. So far as I know Arturo was right every time. Are you saying he wasnt? >Notice, Im NOT saying that Ôa is an algebraic integer when its >irrational! >If you say that 1/a is not an algebraic integer, when irrational, yes, >you are correct. >But my point is that the label algebraic integer does not contain >all numbers such that 1/a is contained and -1 and 1 are the only >integer units. Theres a more inclusive ring, where -1 and 1 are the >only integer units, where 1/a is a unit, and Ôa is not an algebraic >integer. >Here it takes effort if some poster comes in to confuse you, relying >on the *assumption* that if 1/a is not an algebraic integer then its >more like a fraction like 1/2 than 1/1. >But think of x=(1+sqrt(-3))/2, and someone giving you 1/x, if you were >mostly experienced with integers. >Think of how easy it might be for someone to confuse you with >2/(1+sqrt(-3)) if you werent careful and it was new to you. Well, x = (1+sqrt(-3))/2 is an algebraic integer _and_ a unit. >Thats has to do with why at times Ive just gotten totally pissed and >called Arturo Magidin evil, as its just so frustrating to deal with >people so adept at obscuring the truth. I think you must mean, adept at math. What Arturo said was almost invariably true. What you the truth while you denied it as long as possible. >My hope is that at least some of you will play with >x^2 + (8a + b)x + 8, with ab = 1, and (8a + b) an integer >so that you can see why *logically* a possibility must still be there >when Ôa and Ôb are irrational, even if Ôa is not an algebraic >integer. If ab = 1 and 8a + b is an integer, then 8a + 1/a = m for some integer m. This implies 8a^2 - m a + 1 = 0, so a is an algebraic _number_, and in fact it equals (m +/- sqrt(m^2 - 32))/16. This can be an algebraic integer if m = 9 or m = -9 and not otherwise I think. If m = 7, for example, a = (7 +/- sqrt(17))/16, and in that case 8a + b = 7, but ... so what? Similarly, if m = 11, a = (11 +/- sqrt(89))/16, 8a + b = 11. So what? m = 13: a = (13 +/- sqrt(137))/16, 8a + b = 13. So what? etc. Do you see something strange or mysterious about this? Clearly there are infinitely many cases where a and b are irrational, ab=1, and 8a + b is an integer. All such a and b must be algebraic _numbers_. Usually they are both irrational. So? Are you thinking that for some reason there is a need to invent a new ring for numbers like a? Why? Why do you think the existing ring of algebraic numbers isnt sufficient? Lets take m = 1. Then a = (1 +/- sqrt(-31))/16. Do you want both of these roots to be in your new ring? Because if you do, then their sum is also. That means that 1/8 is in your new ring, which means 8 is a unit. Do you want that? If you only want one of the roots to be in the new ring, how do you pick which one? >If you can get a handle on that possibility, then maybe discussions >can be more fruitful about those numbers lost in the shufße that >irrational Ôa here represents. >You know, they are *most* numbers given their cardinality. As I noted above a and b are both algebraic numbers. Therefore their cardinality is the same as that of the integers, ie, countable. There are no more of them than there are integers or rationals. Most makes no sense. The real mystery with this particular thread is, what the heck is your point? - Whipple >James Harris === Subject: Re: 1st order Diff Eqns by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1N1UEu24872; >> http: //www.sosmath.com/diffeq/first/bernouilli/bernouilli.html I tried to follow these messages about perplex patterns but I am a little lost. > Could someone summarize what has already been done about this subject, > that is how many patterns have been found for n from 1 to 20 (or > something like that) and also how many one can expect to discover. ... sci.math thread with details of the above. Brießy, throughout last year B.S. Rangaswamy reported several perplex patterns (found without using a computer, if Im not mistaken) for n=5, 6, 7, and 8; in December posted some programs, while Ralph Furmaniak posted a more general but slightly slower program. (At n=10, his takes 8 minutes on his machine and mine takes 1 minute on a 750 MHz Pentium.) Below are the number of solutions I find for n=1 to 12. I think no one knows how many or if any exist for n>12. -jiw n sols(n) Examples (unsquared) 1 3 {1} {2} {3} 2 4 {9 4} {8 7} {4 8} {6 8} 3 7 {29 22 12} {21 22 12} {11 17 14} ... 4 1 {46 35 36 81} 5 16 {304 167 221 136 263} ... 6 17 {865 668 932 476 472 738} ... 7 13 {1913 2636 2333 3134 2343 2643 3114} ... 8 12 {7918 5141 8034 9615 7042 8839 5341 6454} ... 9 11 {30746 20596 23361 17818 13924 25616 22873 10596 25719} 10 9 {72927 57786 36719 94789 56828 65462 84523 61608 54432 98038} 11 8 {290369 218896 198022 105629 212771 121426 234925 261263 135884 248686 128108} 12 2 {957511 358784 827807 934038 475632 868708 628471 391023 767854 403042 504407 411636} {683281 798836 829893 903962 855838 531773 996627 509885 664623 992182 833332 411643} === Subject: [TOC] Mathematical Models and Methods in Applied Sciences - Vol 14 No 2 Mathematical Models and Methods in Applied Sciences View table-of-contents and abstracts at http://www.worldscinet.com/m3as.html Contents: An Explicit Subparametric Spectral Element Method Of Lines Applied To A Tumour Angiogenesis System Of Partial Differential Equations J. Valenciano and M. A. J. Chaplain Limit Behaviour Of A Dense Collection Of Vortex Filaments H. Bessaih and F. Flandoli Study Of A Mathematical Model For Stimulated Raman Scattering T. Boucheres, T. Colin, B. Nkonga, B. Texier and A. Bourgeade Linear Parabolic Equations In Locally Uniform Spaces Jose M. Arrieta Numerical Solution Of Two-Factor Models For Valuation Of Financial Derivatives Ana Berm.9cdez and Mar.92a R. Nogueiras For more information, go to http://www.worldscinet.com/m3as.html === Subject: Re: I got low score on math test, please advise me and take a look > Tim, > I had run into a few professors throughout my college years who were > as arrogant and utterly unthoughtful (in the analytical way, not in > the lovey dovey way) as you are; luckily, I do not find quite so many > in graduate school. Allow me to explain: Your students are > purchasing a service from you, if the course title is Calculus and the > description in the handbook is ÔIntroduction to Calculus. Topics > include convergent sequences, limits, differentiation [... insert > additional Calculus topics here ...] then your students are > purchasing a course in Calculus. Your comments about marking students > wrong when it takes additional time to read their answer and your > personal vendetta against students squeezing work in the margin when > there is a back page are fine for your own time - in fact, you can > even explain to your students that you, personally, like this and > dont like that. But the second that you take points off of a > students grade for work that you dont like, you are commiting fraud > and breaking the law in the worst way. Which is a worse crime, taking off points for work that does not meet the requirements that I make clear the first day of class, or passing a student who has made no effort to learn the subject and has, in fact, not learned the material? Personally, I think the fraud lies in the seond option. I am sorry, but I do not allow students to purchase their grades. They must honestly earn whatever grade they receive. expectations): A rough guide to what work merits which letter grade is as follows. Understanding the solution to every assigned homework problem and most of the concepts very well by test day will be enough to earn a ÔC. The ability to do a large majority of problems and understanding the solution to every problem, assigned or not, and understanding the concepts will earn a ÔB. Being able to do nearly every problem, assigned or not, and understanding the relationships between concepts will be enough to earn an ÔA. Also, I dont think I said I take off points for sqeezing your work into the margins. I do not. However, if you are unclear or I cannot follow the work (which might be because it was squeezed into the margins), I could take off a point or two. > You see, not only have you, mid course, decided that Calculus will no > longer be the topic of discussion, but (addressing the Ôin the worst > way comment) your misleading of the students and change of topic may > very well destroy a students whole future. Imagine the scenario > where a student is genetically predispositioned to write in margines, > but is Godel smart in mathematical logic and wants to study at > Princeton University. A grade of ÔF in the ÔTest taking, my way > class that the student inadvertently signed up for (entitled > ÔCalculus) could very well destroy these chances. If one wishes to study math at Princeton, one should be prepared to be published. Incorrectly written proofs, poorly constructed arguments, and messiness will not get someone published. In my classes, I have always been told to write every homework problem, every test answer, even my notes as if they were going to be published by a scholarly journal. I do not expect that of my students. To be honest, the classes I am teaching are not the students who will be going on further in math. Although, if they do, I applaud them and hope I was helpful in letting them know what to expect. > I dont expect you to understand the above argument, but perhaps > youll understand this and copy the behavior. Whenever I ran into a > professor who thought so highly of themselves that they dared to > dictate the format that a student could write their test answers, > above and beyond the standard Ôshow your work, write so that I can > read it and circle your final answer (but never Ôlook to see if there > is a back page, if so: do not write in margin if not: blah blah > blah etc.), they were always very untallented theoretically. That > is, they would never be able to understand the Ôtheoretical scenario > that I proposed above, they would never be able to understand the fact > that they are Ôtheoretically changing the course topic and commiting > fraud (false advertising). This lack of tallent invariably carried > over into mathematics (or computer science, depending upon the course) > as well. I never ran into a very smart professor who did this sort > of thing. Perhaps youll want to be thought of as smart, too, and get > down from your high horse. I never change the course topic. We spend time learning whatever is listed in the catalog. My requirements are plain and simple, I require clear, well thought out responses. If I cannot follow your work, it has been from my experience, that the student is trying to drown me in knowledge with the hope I will forget what answer I am supposed to find. > Please note, I think that this ÔSpockie Hendrick guy is a pathetic > whiney excuse for a student who refuses to read his handbook which, > almost certainly, states that a student cant drop a corse past the > drop date for any reason whatsoever. This post has nothing to do with > ÔSpockie Hendrick ... This is only addressed to Tim and his arrogant > thought processes that tell him that his way of test taking is right > and must be spread around the world. I never said my method of test taking was right. In fact, I dont have a method of test taking. All I ask is that students take the time, and actually care enough, to make their responses clear. Perhaps this example will help you understand what I mean. I gave a test recently and one of the questions was, State one of the forms of the Law of Cosines. It was worth three points. For this class sides are labelled with lower case letters and angles with upper case, such as side a is opposite angle A, and that was clearly stated at the beginning of the test. How many points should I assign for a response like this: A^2 = a^2 + b^2 - 2*A*C*cos(a) B^2 = a^2 + c^2 - 2*A*C*cos(b) c^2 = a^2 + b^2 - 2*a*b*cos(C) Yes, the correct answer is one of those. But did the student actually know the answer? Or, was he just guessing hoping that I would see the correct answer and consider it 100% correct? Finally, let me tell you a little about what my students think of me. Where I teach, students are not afraid to give porr evaluations at the end of the semester. They are also quick to remind you that it is their tuition paying my salary. However, I have never received a poor evaluation. I do often receive comments like, high expectations, but they are made clear from day one, A tough, but fair instructor, I learned a lot, whether I wanted to or not. Also, I am a vibrant instructor. I often try to get the students to learn to like math. When I am teaching stats, we go to a fair and collect data on things like what number comes up on a Big-6 wheel, or which horse wins the watergun race most often. Also, we had a Vegas night, where we looked at expected values and how even though the roulette wheel might come up red 4 times in a row, that does not change the probability on the next spin. And what the sucker bets are in craps. And, even though it has little to do with stats, they learn how to count cards in blackjack. Perhaps one of my greatest compliments is that I have had a few students ask me what class I was teaching the next semester so they could take it. In fact, I have taught one student in three different class, Calculus I, Intro. to Statistics, and Discrete Math. Appearently he doesnt think I am draconian in my requirements. If you think I am just making these things up, I can give you email addresses of some of my past students to get their comments. Many times I have been told I make math fun for them, or I have been the best instructor they have ever had (that last one kind of makes me sad because I know the professors they have had and I know how good they are). Maybe, just maybe, I actually expect my students to learn, to be able to present clear thoughts. Maybe I wont just give points for a correct answer when why the answer is correct and how you got there matters even more. Hell, I can use Maple or MatLab or Mathematica or Derive or any other CAS to get answers, if that is all I wanted. If that makes me arrogant, then color me so. - Tim -- Timothy M. Brauch Graduate Student Department of Mathematics Wake Forest University email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: I got low score on math test, please advise me and take a look >> assuming that your report about progress feedback is accurate, a > rejection of the request above is a good glimpse into an incompetent > institution. what kind of justification were you offered? some > bureaucratic non-sense? >>My guess is that justification offered was THE RULE THAT YOU CANT > DROP A >>COURSE AFTER A CERTAIN DATE. >> in other words, bureaucratic non-sense. dont you think that such rule >> should have a provision stateing that students need some performance >> feedback prior to the drop deadline? >How do you know that he didnt get any performance feedback before the >deadline? > i believe he said that somewhere. now, can you answer my question - > perhaps in a direct manner this time? Fine. RULES ARE RULES. Hows that? Doug === Subject: Re: I got low score on math test, please advise me and take a look > Tim, > I had run into a few professors throughout my college years who were as > arrogant and utterly unthoughtful (in the analytical way, not in the lovey > dovey way) as you are; luckily, I do not find quite so many in graduate > school. Allow me to explain: Your students are purchasing a service from > you, if the course title is Calculus and the description in the handbook is > ÔIntroduction to Calculus. Topics include convergent sequences, limits, > differentiation [... insert additional Calculus topics here ...] then your > students are purchasing a course in Calculus. Your comments about marking > students wrong when it takes additional time to read their answer and your > personal vendetta against students squeezing work in the margin when there > is a back page are fine for your own time - in fact, you can even explain to > your students that you, personally, like this and dont like that. But the > second that you take points off of a students grade for work that you dont > like, you are commiting fraud and breaking the law in the worst way. > You see, not only have you, mid course, decided that Calculus will no longer > be the topic of discussion, but (addressing the Ôin the worst way comment) > your misleading of the students and change of topic may very well destroy a > students whole future. Imagine the scenario where a student is genetically > predispositioned to write in margines, but is Godel smart in mathematical > logic and wants to study at Princeton University. A grade of ÔF in the > ÔTest taking, my way class that the student inadvertently signed up for > (entitled ÔCalculus) could very well destroy these chances. > I dont expect you to understand the above argument, but perhaps youll > understand this and copy the behavior. Whenever I ran into a professor who > thought so highly of themselves that they dared to dictate the format that a > student could write their test answers, above and beyond the standard Ôshow > your work, write so that I can read it and circle your final answer (but > never Ôlook to see if there is a back page, if so: do not write in margin > if not: blah blah blah etc.), they were always very untallented > theoretically. That is, they would never be able to understand the > Ôtheoretical scenario that I proposed above, they would never be able to > understand the fact that they are Ôtheoretically changing the course topic > and commiting fraud (false advertising). This lack of tallent invariably > carried over into mathematics (or computer science, depending upon the > course) as well. I never ran into a very smart professor who did this sort > of thing. Perhaps youll want to be thought of as smart, too, and get down > from your high horse. > Please note, I think that this ÔSpockie Hendrick guy is a pathetic whiney > excuse for a student who refuses to read his handbook which, almost > certainly, states that a student cant drop a corse past the drop date for > any reason whatsoever. This post has nothing to do with ÔSpockie Hendrick > ... This is only addressed to Tim and his arrogant thought processes that > tell him that his way of test taking is right and must be spread around the > world. >> my website states my case and has jpg files of the four pages of the >> test please take a look and advise me or give me opinions >> http://www.johncho.us >Aside from the all the politics of whether the grade was fair and if the >test should be returned in time, I have a few comments. >First, for all the instructors out there, how many of you take of 1/4 of >a point? To me, if you make the problem worth 8 points, anything less >than perfection is 7. And, when I am assigning point values to >problems, I usually look over the problem and pick out the two or three >concepts being tested in that question and give each of those concepts a >value of 1 or 2 or even 3 points. I just cant igaine trying to grade >with fractional points. But, that is just my grading style and Im not >really saying anything is wrong with giving fractional points,I just >find it uncommon. >Secondly, if I am grading and I cannot follow the work (perhaps because >it is not neatly written), I have a hard time giving full credit. That >might sound harsh, but I have often found that students write two or >three possible answers for the problem in their mess and never clearly >mark which one they want me to grade. Also, judging from your scans of >the page, it looks like you had the back of each page left blank (which >is pure speculation on my part). I despise it when students try to >squeeze in something in the margins, leaving the whole back side of a >page blank. You arent going to use that page for anything else, why >squeeze things together? Now, if you really want to get your instructor >mad, you could ask him (or her) to please write the comments on your >test neater. But that would serve no purpose otehr than to point you >out as a snot-nosed brat. >Okay, about the grading. Personally, I dont know what topics were >stressed in class so I dont feel like I can make a judgement call. For >example, I gave a test in Trigonometry II on Tuesday (which I am in the >process of grading and wont be given back until Monday). For this test >the material I stressed was changing degrees to radians, law of sines, >and law of cosines. On one question I asked about the area of a >triangle. If a student accidentally used the formula area = base * >height, I will not take off many points probably one out of 10 points, >because they were not being tested on geometry (and had they asked me >the correct formula, I would have told them, since that wasnt the topic >they were being tested on). Or, maybe another example, if someone >computed 2+3=6 in part of a larger problem, multiplying instead of >adding, and they correctly carried that mistake through the problem >(that is if they had used 5 they would have gotten the correct answer) I >would probably only take off one point out of 10. >Those are my thoughts. Your grade, suck it up. If you ask the >instructor to look over the test again, he or she might find some places >when you should hae lost more points that you did. At least, if someone >quibles over two or three points I say I will look it over, and I do and >make sure I *took off* all the points I should have. > - Tim >-- >Timothy M. Brauch >Graduate Student >Department of Mathematics >Wake Forest University >email is: >news (dot) post (at) tbrauch (dot) com Anonymous, It is my opinion that your comments about Tim being arrogant and unthoughtful and every other rude thing you said about him were way out of line. For example, I see no evidence in his post that he assigns grades based on anything other than the subject matter at hand. For example, Tim said: Secondly, if I am grading and I cannot follow the work (perhaps because it is not neatly written), I have a hard time giving full credit. That might sound harsh, but I have often found that students write two or three possible answers for the problem in their mess and never clearly mark which one they want me to grade. Note that he finds a hard time giving full credit when he cannot follow the work or when students write more than one answer. How could he give such answers full credit? A large percentage of students bluff on exams by writing any irrelevant thought that comes into their heads to the point where their test papers are completely incoherent. Instructors usually do their best to find something good in the answers and go out of their way to give partial credit. I see no evidence that Tim isnt one of these caring teachers. Also, when he said things like he despised when students squeezed work into the margins, notice that he did not say such a thing affected the students scores. Tim gave further evidence that he is a fair and caring teacher when he said he doesnt take off very many points for errors that arent related to the material presently being taught. Tims last comment may seem harsh--that he might even deduct points if a student quibbles over his or her score and he finds a legitimate mathematical reason to deduct those points. But I believe he may be justified in this behavior. A significant percentage of students (especially at large universities) are immoral and do not care about learning the material. They only want to buy their grades. Such students often waste the time of the instructors by asking for regrades (even when they havent read their exam over closely)! I have seen students insist to instructors that they did a problem correctly and deserved full credit, when in fact they had no idea what the question meant, much less how to solve it. When students purchase the services of the instructor, how much services should they actually get? Unfortunately no contract is made and nothing is written down. I believe it is reasonable for a professor to set certain rules concerning re-grades and coherent test solutions. Without these rules, some students may use up too much of the professors time and he wont be able to give deserved time to other more respectful students. Anyway, your comments about Tim commiting fraud and breaking the law in the worst way are completely ridiculous. You dont have enough information to judge in this particular matter for one thing. For another, Tims post in no way supports such a judgement. Anyway, do you know exactly what the law states in such matters? For example, is every student in a class of 50 entitled to write so illegibly and incoherently on tests that it takes the professor three hours to grade it? If they were entitled to such things, the price of their education would go way up. -Leoanrd Blackburn === Subject: Re: I got low score on math test, please advise me and take a look >>assuming that your report about progress feedback is accurate, a >>rejection of the request above is a good glimpse into an incompetent >>institution. what kind of justification were you offered? some >>bureaucratic non-sense? >> My guess is that justification offered was THE RULE THAT YOU CANT > DROP A >> COURSE AFTER A CERTAIN DATE. >in other words, bureaucratic non-sense. dont you think that such rule >should have a provision stateing that students need some performance >feedback prior to the drop deadline? > How do you know that he didnt get any performance feedback before the > deadline? i believe he said that somewhere. now, can you answer my question - perhaps in a direct manner this time? > He only said that he didnt get THIS TEST back before the deadline. > How do you know if this is the first test? How do you know if hes had > other assignments, or how many? > This troll has only told us what he wants us to know. And whatever > respect I had for him, I lost when he started sending me unsolicited > e-mail. > Doug === Subject: Re: I got low score on math test, please advise me and take a look > How do you know that he didnt get any performance feedback before the > deadline? > He only said that he didnt get THIS TEST back before the deadline. > How do you know if this is the first test? How do you know if hes had > other assignments, or how many? In all fairness, he _did_ say, IIRC, that he hadnt had any other feedback in one of his other posts. Not that Im justifying him. My opinion is that he is being entirely unreasonable. If there are rules, there are rules. Sure, its nice if theyll make an exception for a good reason every now and then. I got an extension on an application once because I didnt receive the forms until the day it was supposed to be postmarked. But there is no obligation to do things like that. Its not that theyre being *unreasonable* if they dont. On the contrary, its often wise to avoid making exceptions so as not to set a precedent, which can cause all sorts of trouble. Jonathan Christensen --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). === Subject: Re: Simple idea, mathematics and common-sense > [SNIP] >Decimal notation is a convenience, not a necessity, for representation >of numbers. The fact that some numbers are not so representable is >merely an inconvenience, not a disaster. >> But thats where my story begins. > The idea behind exponential notation is rather more than a convenience, it > facilitates the definition of real numbers. I you dont believe this then > try doing arithmetic using unitary notation. > Tony Thomas The *definitions* of real numbers and of arithmetical operations in the field of real numbers are quite independent of the algorithms we use to implement or approximate them. While decimal and related notations are extremely useful in the execution of those arithmetical operations, they are in no way essential to their definition, which is the point I was trying to make. === Subject: Re: Simple idea, mathematics and common-sense [SNIP] > Decimal notation is a convenience, not a necessity, for representation > of numbers. The fact that some numbers are not so representable is > merely an inconvenience, not a disaster. >But thats where my story begins. The idea behind exponential notation is rather more than a convenience, it facilitates the definition of real numbers. I you dont believe this then try doing arithmetic using unitary notation. Tony Thomas === Subject: Re: Simple idea, mathematics and common-sense > [...] > Mathematicians may think that seeing is > believing, but I think in mathematics, logic > is king. > James Harris I didnt read your whole discussion, but agree with you on the one thing, above. All of whats been referred to as Mathematics is representational, and its representations derive in nervous system function, which reduces to individually-unique experiencing of the one- way ßow of energy from order to disorder that is whats =described= by 2nd Thermo [WDB2T]. So this or that Maths is =not= universally- applicable, but is Ôonly that which derives in this or that nervous systems Ôclimbing of the WDB2T that it, locally, experiences. Its not Ôhopeless, because whats here can, itself, be comprehended and Formalized, but it can only be Formalized in a way thats applicable to the functioning of =all= nervous systems [and all of physical reality]. k. p. collins === Subject: Re: Simple idea, mathematics and common-sense >> Yes, I can talk it all out rigorously and in a heavily mathematical >>format, >> James, >> I have been following your threads for some time now and I have seen >> on more occasions than I can count a request for you to do just this. > Really? Yes, really ... and Im not a mathematician so you cant even accuse me of lying (since, if I understand your appeals to the gallery correctly, it is only mathematicians that lie and the rest of us dont) >> I seem to recall that when you have deigned to acknowledge these >> requests it wasnt ...[talked] out rigourously and in a heavily >> mathematical format it was more of a diatribe against mathematicians >> and the evil society that they control. > Hey, theres this silly error, and Ive explained it, but > mathematicians seem either willing to ignore it, or engage in rather > vicious personal attacks or arguments which I see as designed to hide > the issue. As another poster has pointed out, you have acknowledged an error, apologised for it and then repeated it! >> I have seen you claim that clearly stated definitions are wrong, that >> (as you did in this post) there is some inherent ambiguity and that >> professional mathematicians (as well as the very capable amatuers) >> are too stupid to understand but I have *NEVER* seen you post >> anything that even I would consider to be rigourous. > Well Im tired, so Im pushing more on to people like yourself, so > consider No James, you are *NOT* pushing anything on to me - I have no desire and I see no need to do your work for you. If you have a problem then *YOU* solve it. That may be asking a bit much since you seem unable to follow and/or acknowledge the repeated assistance that you have been given by others. > P(x) = (x + 8a)(x + b), where ab=1, and (8a + b) is an odd integer. > You should have the expertise to play with that and figure out the > error in core yourself! Since I dont understand what error it is that you claim I cant possibly figure it out. *YOU* claim there is an error, therefore *YOU* must prove it. When you have ...[talked it] out rigourously and in a heavily mathematical format Im sure your claims will be examined. >> Having now said that you can do this, when can I expect to see it? > Play with P(x), and see if youre smart enough to see the mistake in > reasoning with current teaching about the ring of algebraic integers. Again, this is *YOUR* problem, not mine. It also isnt an answer to the question so I will ask again - when can I expect to see your claims ...[talked] out rigourously and in a heavily mathematical format ? I am content to sit and laugh at you and with everyone else. I especially like it when you use the F word lots and lots of times and even more so when you have yet another go at Professor Ulrich. On the subject of having a go at people, why do you keep referring to Arturo Magidin in your posts as being one of those who oppose you in this newsgroup? Arent you the one who told, using gutter language, Arturo not to participate any more? And isnt it Arturo who honourably kept his word and has since not responded directly to you? This is actually a great loss to many of us since Arturo has what I think is a gift for explaining things in a manner that even I can understand. >> Im sure that if I have any difficulty then either you or somebody >> else will be able to clarify. >> Ivan. > I certainly hope that a quadratic isnt too much for you! > If so, then hey, it may bug you, but I may start talking about evil > math society, and people who refuse to acknowledge the truth again!!! No, it doesnt bug me at all - since I have faced the truth, that I am a JSH addict, I am able to properly enjoy your bizarre rants. > Theres no more room for excuses, even for people like you Ivan. People like me? What are people like me like? > Either actually be a mathematician, and acknowledge mathematical > truth, or just wait for the consequences of being rogue and against > the best interests of society at large. Ummmmmm James ? Arent *YOU* the one who has repeatedly boasted that you are not a mathematician ? Dont you think you should really take your own advice and ...be a mathematician, and acknowledge mathematical truth... ? > Yup, if mathematicians refuse to teach correct mathematics, then > they are fighting against humanity itself. > And I think, will be treated accordingly, once that becomes common > knowledge. > Why fight against humanity, against progress and truth? Ummmmmm ... I dunno, why ? > James Harris Ivan. === Subject: Re: Simple idea, mathematics and common-sense I had an error in the example at the end - it doesnt make any conceptual difference, but below is the corrected version, plus another example: >Mathematicians may think that seeing is believing, but I think in >mathematics, logic is king. I could not invent a more perfect example of getting things > backward. YOU think that if you cannot see a factor, it cannot > be there: hence you think it MUST be true that 7 factors out of > two of the linear terms of your cubic. You cannot conceive that > there might be any other way to do it, a way that you cannot > easily see explicitly. But there can. We have told you how > repeatedly. You need some abstract ideas to understand it. ÔAbstract > means: something BEYOND seeing is believing. You are essentially > stuck with ancient fossil math, like bones in the LaBrea tarpits, > obsolete. You yourself provide a perfectly good examply, with > the roots of x^2 + 7*x + 2, and THEN YOU MISINTERPRET IT, and go off > on ridiculous tangents. You want nothing but seeing-is-believing > math, and then you accuse us of your own error! > You saw Deckers original example. Evidently you *still* dont > believe it. You saw Ramsays example. You know that it showed > something that was far from obvious. You saw it, but now you > seem to be saying you still dont believe it. You say logic is > king, but when your own logic defies hard evidence, you think > the error must be in the evidence! > It reminds me of the Rodney King tape. You apparently would have > agreed with the first jury: seeing is not believing! Those policemen > were not beating an unarmed helpless man on the ground with nightsticks! > Like them, apparently, you prefer to believe your own preconceptions. > Heres another example, perhaps more relevant. Consider > x^2 - 6*x + 6. The corrected version follows: > There are two roots: r1 = 3 + sqrt(3) and r2 = 3 - sqrt(3). > Note that r1 * r2 = 6 = 2*3. > It is easy to see that r1 has an algebraic integer factor > in common with 3. You can see that *BY INSPECTION*, the only > method that you actually trust. It is sqrt(3). Right? > Its not so easy to see what factor r1 has in common with 2, > is it? Cant get that one by inspection, right? So if we agree > with your thinking, perhaps we conclude that it must be either > 2 or 1. Clearly it is not. Nor do we need to invent an object ring, > and declare that (3 +sqrt(3))/2 is an object. 3 + sqrt(3) > DOES have an algebraic integer factor in common with 2. > See if you can figure out what it is. Note that real mathematicians > do not have trouble with this. They have ways of finding such things that > go beyond your own seeing-is-believing method. You need to stretch > your very limited imagination. Learn a little 19th century algebraic > number theory and you might be able to see it also. Let us know if > you cant get it. > Nora B. Here is another example of this kind: Let Q(a) = a^2 - 4*a + 6. The roots are r1 = 2 + sqrt(-2) and r2 = 2 - sqrt(-2). Again their product is 6 = 2*3. It is clear (by inspection!) that each of them has a factor in common with 2. Not so easily seen are the factors they have in common with 3. No, Mr. Harris, not 3 itself in either case. Galois would not be surprised. Dedekind would not be surprised. Harris ? Hard to say. He might need to invoke Objects! Nora B. James Harris === Subject: Re: Simple idea, mathematics and common-sense > Think Im crazy here? Yes! No doubt about it! === Subject: Re: Simple idea, mathematics and common-sense In sci.logic, James Harris lead to a few complexities, but its been fruitful to explain, or try > to explain, as I work to figure out why these simple ideas either > excite derision, anger or confusion, and I think I have it figured > out. > Im sure many of you are put off by mathematics, so I assure you up > front that what Ill be talking about will mostly be *very* simple, > and there will only be a few slightly complicated things at the end. > First of all, Im going to talk about a case where mathematicians gave > up because they couldnt see something, and assumed that because they > couldnt see something it didnt exist! > You know how with simple quadratics like x^2 + 3x + 2, its easy > enough to see factors of 2 in the roots? > I mean, its just (x^2 + 3x + 2) = (x+2)(x+1), and there they are. > However, if its something like x^2 + 7x + 2, you can use the > quadratic formula and get the roots to find > x = (-7 +/- sqrt(41))/2 > and who can see factors of 2 in that thing? I think part of the problem here -- and Ill admit to not being entirely certain where the prime failure occurs -- is that, if one has an integer such as 6, one can uniquely factor it into primes: 6 = 2 * 3 with the note that, since 1 and -1 are units of the integer ring, the factorization is unique up to multiples thereof, so one can also have 6 = (-2) * (-3) Simple as pi(e). Now, if we go into the realm of algebraic integers, though, Im not even sure what a prime is therein, although there are an awful lot of units. In fact, any roots of the equations x^n + a_{n-1} * x^{n-1} + ... + a_1 * x + 1 = 0 x^n + a_{n-1} * x^{n-1} + ... + a_1 * x - 1 = 0 where a_i are algebraic integers, will be a unit. (It doesnt matter whether the equation is reducible or not.) I believe Arturo has demonstrated that all units must solve equations of this form, as well. A pair of such units is 2 + sqrt(3) and 2 - sqrt(3). Multiply them together and one gets 1. The defining equation for this pair is x^2 - 4x + 1 = 0. Of course not every algebraic number is a unit, fortunately; sqrt(2) in particular is provably not a unit, as its irreducible equation is x^2 - 2 = 0. But sqrt(2) is not prime either, as sqrt(2) = 2^(1/4) * 2^(1/4). One might apply a categorization by declaring that one has algebraic integers of order n, where n is the degree of the irreducible equation with integer coefficients, of which the integer is a root. One then might be able to prove that, for any algebraic integer of degree n, there exists a unique (within units) factorization into other algebraic integers of degree either < n, or <= n, I dont know which. In such a system, sqrt(2) is an algebraic prime of order 2. Eric W. Weissteins site is of little help here, although he does have an entry: http://mathworld.wolfram.com/PrimeAlgebraicNumber.html but all it does is define the term; it gives no examples. There is the possibility that all algebraic primes are solutions of an irreducible equation x^n + a_{n-1} * x^{n-1} + ... + a_1 * x + p = 0 x^n + a_{n-1} * x^{n-1} + ... + a_1 * x - p = 0 where p is an integer prime. However, such primes are subject to the order consideration above, since sqrt(2) is such a number. I shall have to ask this question in a separate thread. [rest snipped] -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Simple idea, mathematics and common-sense [.snip.] > Simple as pi(e). Now, if we go into the realm of algebraic integers, > though, Im not even sure what a prime is therein, although there > are an awful lot of units. In fact, any roots of the equations > x^n + a_{n-1} * x^{n-1} + ... + a_1 * x + 1 = 0 > x^n + a_{n-1} * x^{n-1} + ... + a_1 * x - 1 = 0 > where a_i are algebraic integers, will be a unit. > (It doesnt matter whether the equation is reducible or > not.) I believe Arturo has demonstrated that all units > must solve equations of this form, as well. Its a classical result, going back a long way, and an easy exercise. That any such is a unit follows by noting that it divides 1; that any unit is such follows by taking a unit a, looking at its irreducible monic polynomial over Q, and considering 1/a; it satisfies the reverse polynomial. i.e., if a satisfies x^n + a_{n-1}*x^{n-1} + ... + a_1*x + a_0 then 1/a satisfies a_0*x^n + a_1*x^{n-1} + ... + a_{n-1}*x + 1 This is true of any algebraic number; now, easily, the first is irreducible over Q if and only if the second is irreducible over Q, so if both a and 1/a are algebraic integers, we must have a_0=1 or a_0=-1. Arturo Magidin, sans .sig === Subject: Re: Simple idea, mathematics and common-sense > In sci.logic, James Harris > lead to a few complexities, but its been fruitful to explain, or try >to explain, as I work to figure out why these simple ideas either >excite derision, anger or confusion, and I think I have it figured >out. >Im sure many of you are put off by mathematics, so I assure you up >front that what Ill be talking about will mostly be *very* simple, >and there will only be a few slightly complicated things at the end. >First of all, Im going to talk about a case where mathematicians gave >up because they couldnt see something, and assumed that because they >couldnt see something it didnt exist! >You know how with simple quadratics like x^2 + 3x + 2, its easy >enough to see factors of 2 in the roots? >I mean, its just (x^2 + 3x + 2) = (x+2)(x+1), and there they are. >However, if its something like x^2 + 7x + 2, you can use the >quadratic formula and get the roots to find >x = (-7 +/- sqrt(41))/2 >and who can see factors of 2 in that thing? > I think part of the problem here -- and Ill admit to not being > entirely certain where the prime failure occurs -- is that, > if one has an integer such as 6, one can uniquely factor it into > primes: > 6 = 2 * 3 Thats not the issue here. Its simple enough, but I have seen *posters* like this person come forward and confuse the issue enough times that I think Id better step in and make sure that the REAL issue isnt easily obscured. P(x) = (x+8a)(x+b), where ab=1, and consider 8a+b an integer If a=b=1, notice you have (x+8)(x+1), which is one of two basic possibilities with integers. Another possibility is a=1/2, b=4, which is the second basic possibility. That is, Ôa here can be a fraction, like 1/2, or it can be more like an integer than a fraction, like actually being 1 with *integers* in the other case. Remember that making an issue between integers versus irrationals is a major part of the logical mistake that mathematicians made. I say that just because human beings LOVE being able to count something out on their fingers, its not a mathematical constraint! However, current mathematical dogma is that *if* Ôa and Ôb are irrational then the first type possibility is eliminated so it must be the second possibility. Part of the problem is that if you imagine the first type possibility with ab = 1, where again Ôa and Ôb are *irrational* the mathematicians have a label for Ôb which is algebraic integer, but Ôa cannot be an algebraic integer. So they dont have a label for it! You know how important naming is with human beings, and it makes it that much harder for me to explain these ideas without a label. Ive named numbers like Ôa objects, but while mathematicians successfully trash my work or dismiss it, you can see that using the name Ive given might not resolve things. Thats it. Thats the issue as what mathematicians teach in this area doesnt follow from mathematics. It doesnt follow from logic or any axioms. Its just some human notion that has settled into dogma. Some mathematicians come to the logically specious conclusion because they cant stick the label algebraic integer on it for that reason Ôa is in no way an integer like 1 but is instead more like a fraction like 1/2. There is nothing in mathematics to support that conclusion. Its just a human preference attached to the label algebraic integer. James Harris === Subject: Re: Simple idea, mathematics and common-sense > ab = 1, >> where again Ôa and Ôb are *irrational* the mathematicians have a >> label for Ôb which is algebraic integer, but Ôa cannot be an >> algebraic integer. >> So they dont have a label for it! >> You know how important naming is with human beings, and it makes it >> that much harder for me to explain these ideas without a label. >> Ive named numbers like Ôa objects Take b = sqrt(2). Ôb is *irrational*. Thus a=1/sqrt(2) is an object. - William Hughes > I meant with with 8a + b odd. > Why not try to consider the idea versus just being a smart-aleck? > If you were concerned with the truth, you might have noted that with > 8a+b even, its clear that Ôa is a fraction like 1/2, like with your > example where it is 1/sqrt(2), versus being a smart-aleck by tossing > out that special case as if it were definitive. > Now then, do you understand why your example doesnt apply for 8a+b > odd? > Do you understand how your post is a smart-alecky one, which invites > the kind of response that I just gave it? > James Harris The current conditions are i) ab=1 ii) 8a + b is an odd integer iii) a and b are irrational then a is an object. Take 8a + b = 5 thus a1= (5 + sqrt(-7))/16 and a2= (5 - sqrt(-7))/16 are objects. Thus a1*a2=1/8 is an object. - William Hughes === Subject: Re: Simple idea, mathematics and common-sense In sci.logic, James Harris ab = 1, >>where again Ôa and Ôb are *irrational* the mathematicians have a >>label for Ôb which is algebraic integer, but Ôa cannot be an >>algebraic integer. >>So they dont have a label for it! >>You know how important naming is with human beings, and it makes it >>that much harder for me to explain these ideas without a label. >>Ive named numbers like Ôa objects >> Take b = sqrt(2). Ôb is *irrational*. Thus a=1/sqrt(2) is an object. >> - William Hughes > I meant with with 8a + b odd. [A1] You now might have the requirement that 8a + b = 2n + 1, n an integer. This, together with ab = 1, results in the quadratic 8a + 1/a = 2n + 1 or 8a^2 - (2n + 1)a + 1 = 0 which has as roots a = ( (2n + 1) +/- sqrt(4n^2 + 4n + 1 - 32) ) / 16 If n = 4 or n = -5, a is rational. No other rational a are possible. (The proof is left as an exercise.) [A2] It is also possible you meant 8a + b is an algebraic integer not divisible by 2, i.e., the quantity V = 4a + b/2 is not an algebraic integer. The problem is, if b is an algebraic integer, then 8a is an algebraic integer, and the problem divides into two subparts. [i] 4a is an algebraic integer. Therefore b/2 is not an algebraic integer. [ii] 4a is not an algebraic integer. Im not sure what conclusions can be drawn here although I strongly suspect b/2 = V - 4a is an algebraic integer. However, I cant prove it without a bit of schlogging. > Why not try to consider the idea versus just being a smart-aleck? > If you were concerned with the truth, you might have noted that with > 8a+b even, its clear that Ôa is a fraction like 1/2, [B] a = ( (2n) +/- sqrt(4n^2 - 32) ) / 16 = (n +/- sqrt(n^2 - 8)) / 8 For n = 3 or -3, a is rational. > like with your > example where it is 1/sqrt(2), versus being a smart-aleck by tossing > out that special case as if it were definitive. > Now then, do you understand why your example doesnt apply for 8a+b > odd? > Do you understand how your post is a smart-alecky one, which invites > the kind of response that I just gave it? > James Harris -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Simple idea, mathematics and common-sense > ab = 1, >> where again Ôa and Ôb are *irrational* the mathematicians have a >> label for Ôb which is algebraic integer, but Ôa cannot be an >> algebraic integer. >> So they dont have a label for it! >> You know how important naming is with human beings, and it makes it >> that much harder for me to explain these ideas without a label. >> Ive named numbers like Ôa objects Take b = sqrt(2). Ôb is *irrational*. Thus a=1/sqrt(2) is an object. - William Hughes > I meant with with 8a + b odd. And here we all thought that it was only JSH who was odd. > Why not try to consider the idea versus just being a smart-aleck? You cannot expect people to understand conditions which you do not state. He considered the idea that you presented subject to the conditions you stated. That is hardly being smart-aleck. > If you were concerned with the truth, you might have noted that with > 8a+b even, its clear that Ôa is a fraction like 1/2, like with your > example where it is 1/sqrt(2), versus being a smart-aleck by tossing > out that special case as if it were definitive. > Now then, do you understand why your example doesnt apply for 8a+b > odd? It doesnt apply for a^2 + b^2 = 1, either. So what? > Do you understand how your post is a smart-alecky one, which invites > the kind of response that I just gave it? The only one trying to be smart-alecky here is JSH, but he doesnt quite bring it off. > James Harris === Subject: Re: Simple idea, mathematics and common-sense ab = 1, where again Ôa and Ôb are *irrational* the mathematicians have a >label for Ôb which is algebraic integer, but Ôa cannot be an >algebraic integer. So they dont have a label for it! You know how important naming is with human beings, and it makes it >that much harder for me to explain these ideas without a label. Ive named numbers like Ôa objects > Take b = sqrt(2). Ôb is *irrational*. Thus a=1/sqrt(2) is an object. > - William Hughes I meant with with 8a + b odd. Why not try to consider the idea versus just being a smart-aleck? If you were concerned with the truth, you might have noted that with 8a+b even, its clear that Ôa is a fraction like 1/2, like with your example where it is 1/sqrt(2), versus being a smart-aleck by tossing out that special case as if it were definitive. Now then, do you understand why your example doesnt apply for 8a+b odd? Do you understand how your post is a smart-alecky one, which invites the kind of response that I just gave it? James Harris