Is AC really needed? Put any total ordering on the complex plane, for example by ordering everything first by magnitude and then by going around circles counterclockwise starting from the positive real axis. This isn't a well-ordering on C of course but it induces one on every finite subset of C. So for some p(x) over C we can just take its smallest root with respect to this ordering. I don't believe we need AC to guarantee a choice function in this circumstance.
Compare the usual proof that well-ordering implies AC: let X_alpha be a collection of sets indexed by I and let X be the union over I of the X_alpha. Put a well-ordering on X and then define a function f : I --> X by taking f(alpha) to be the minimal element of X_alpha as a subset of X. This constructs a choice function. But we don't actually use exactly that X is well-ordered, just that there is an ordering under which every subset we're interested in has a minimal element.
While elite music of lasting aesthetic quality has been produced, the main reason people listened to elite music in the past (and, indeed, the main reason people listen to 'classical' music now) is a wish to identify themselves as elite - 'I listen to posh music so I am posher than you'.
This, if I may say, needs to be elucidated. The "why" a particular classical piece was composed many be somewhat less-than-noble, but let's not count this against it. Why is the word 'classical' in quotation marks? Perhaps you mean to include music under different period titles (baroque, etc) in your remarks this way.
Rather more mystifying to me is why you believe that the principal reason for people today listening to these 'classical' pieces is an attempt to identify themselves with a higher social segment. If there is some definitive survey or study on this subject, this would certainly be very interesting to those of us so naive that we thought the "lasting aesthetic quality" of those pieces which have lasted was the principal reason.
Is it really true that most people who listen to classical pieces do so not because they actually enjoy the music but because they want to be seen listening to it?
This seems somehow incongruous to me but I can't explain why except by appealing to my prejudices. For example, Vivaldi's "Spring" concerto is perennially popular with TV advertisors; but, is this because they think it elevates the standing of their product or because this piece is justifiably popular and they want listeners to think of their product when they hear it? (I would remark that most of Vivaldi's compositions were for popular consumption and were almost lost until rediscovered, at which time they became very popular, indeed.) The same remarks apply to Orff's "Carmina Burana" and those who score motion pictures.
A better try, if you will: around Christmas there is always a production of Handel's "Messiah" here - I imagine this (and more) is true of most large cities. They even have a special night where the audience is permitted/encouraged to sing along with the choruses. I've not worked up enough courage to go to one of those performances, but do those who do go believe that the words 'sing along' are associated with haute culture?
A different approach: Do people in high social strata listen to such music to keep up pretenses? I can't say this has been my experience. On the contrary, I've seen people champion music of "classical" style but questionable (in others' opinions) "aesthetic value" simply because they do in fact like it. Others have undergone periods of extreme infatuation with certain compositions to the potential annoyance of those around them. So I don't think the converse claim holds.
But that "I want to be posh" motivation as the primary one -- damn my lack of cynicism but I just can't see it.
The best option, I think, is to submit it to the sci.math newsgroup. While the AMS would send a polite rejection notice and your local math dept. would at best inform you of where the first error occurs, the denizens of the newsgroup spend their time voluntarily reading through this sort of stuff and are quite able (and often willing) to explain, sometimes at length, what has gone wrong.
Alternatively but unlikely to be as helpful is the AMS (www.ams.org) (or your nation's mathematical society); they publish at least one journal and likely get all sorts of unsolicited mail for review.
Of course, even if the result is true and the reasoning essentially correct the proof may be so illegible that the referee cannot accept it, but only recommend that it be accepted after revision. So the math department of your local university may be better: they also get mail of this type all the time and may even have someone designated to respond to some of it.
The last of the options which should not be recommended is web-publishing. It is unlikely anyone would even see unless one of the above courses is taken with direction to the site.
"And, what if the standard of refutation? Is it enough to claim "oh, this proof is all just handwaving", or "this proof is worthless, it uses a physicists approach", or does any detractor need to precisely pinpoint where the error is "on page 13, where they get from equation 63 to 64, they effectively multiply both sides with zero"?"
There is a saying sometimes employed by cruel mathematicians to describe illucid 'proofs': "This isn't right. This isn't even wrong."
Having said that, it surely would be nice to see 'exactly where it goes wrong'. For simple arguments a standard "the first error is on page t, line s" will suffice. More convenient is to find a claim to which an explicit counterexample can be constructed, that is, to show that the proof (if valid) leads to a contradiction - of course this works best when the proof is not reductio ad absurdum:-P
Finally, we should always remember that the burden of "proof" is on the "prover" - it's not anyone else's responsibility if the paper isn't even written in what we'd call mathematics. (Naturally it has no chance of being published in a print journal if it's in that state, which is a necessary condition for the awarding of the prize.)
(The rest is a slightly tangential discussion of two common problems arising from extremely imprecise methods, aka "handwaving".)
Here's a gratuitous example:
Prop. There are more real numbers than integers. "Proof." Consider the interval [n,n+1) for arbitrary integer n. In this interval there is one integer, but a slew of reals, eg, n+1/2, n+pi/4, and so forth. So, there are more reals than integers.
Of course this proposition is true but the proof is nonsense, since we can "derive" by mimickery
Prop. There are more rational numbers than integers.
which is just false.
This is a pretty simplistic example and doubtless the author of the purported proof of RH is using much more sophisticated handwaving, for which I'll produce another analogy:
Prop. 2 has no proper factorisation. "Proof." The only integers dividing 2 are 1, -1, 2, -2, the former two being units and the latter two the number and an associate.
This proposition falls into the "not even wrong" category. Why? -- because to write down the word "factorisation" begs the question: in what ring (ie, algebraic context)? Absolutely, 2 is irreducible in the integers. But not in the Gaussian integers:
Prop. 2 is not irreducible in the Gaussian integers. Proof. We see easily that 2 = (1+i)(1-i). Each factor of the RHS has norm 2, so neither is a unit, and in fact 1+i is a Gaussian prime, and 1-i = (-i)(1+i) is also a Gaussian prime.
The error in the first proposition about 2 was that it made no reference to the context of the discussion. This is the sort of handwaving that occurs in the "proof" of RH. We can see it really makes a difference - in one context the prop. was true, and in another false; sometimes even to utter the word "factorisation" is to already condemn oneself to the "not even wrong" bin.
"I think I can say with pretty good authority that he'd say no to [a match with human and machine against human and machine]."
Not only did he not say "no" to it, he more or less invented it. Advanced chess matches have become semi-annual since Kasparov and Topalov first played in Leon in 1998 (a recap by the always-excellent ChessBase team).
"For any product of N primes there is at least one gallois field,"
Just a few corrections for posterity's sake:
First, it is certainly _not_ true that for _any_ k primes p1,... pk there exists a Galois field F with p1*...*pk elements. The next paragraph self-indulgent argument to this effect which uses considerable machinery against a simple typo. Apologies in advance.
(Let F be a Galois field. For any field, the prime subfield is isomorphic either to Q or to Zp for some prime p; here obviously Q is not contained in F, so Zp is the prime subfield. In particular Zp is contained in every subfield of F. So, suppose k > 1 and pick any p's not all the same; then write |F| = q^n * a with a > 1, a,q coprime and q not equal to p. Then there exists a Sylow q-subgroup - call it S - of F; then the group S* is a finite multiplicative subgroup of a field, so is cyclic; every cyclic group of a certain order is isomorphic, so S* is isomorphic to GF(q^n)*, and in particular S is isomorphic to GF(q^n). But this latter has characteristic q not equal to p. Zp is not a subfield of this; its prime subfield is Zq. Contradiction!)
Obviously what was meant is that for any integer prime p and integer n >= 1, there exists a Galois field GF(p^n) with p^n elements.
Furthermore, while it is true that there exists "at least one", more is true: there exists _only_ one up to isomorphism (this one is easy; as above, the multiplicative subgroup is cyclic and...)
(Finally, I am neither a cryptographer nor a cryptanalyst, so I am unqualified to speak on the balance of the parent message.)
Slashdot Mirror
Is AC really needed? Put any total ordering on the complex plane, for example by ordering everything first by magnitude and then by going around circles counterclockwise starting from the positive real axis. This isn't a well-ordering on C of course but it induces one on every finite subset of C. So for some p(x) over C we can just take its smallest root with respect to this ordering. I don't believe we need AC to guarantee a choice function in this circumstance.
Compare the usual proof that well-ordering implies AC: let X_alpha be a collection of sets indexed by I and let X be the union over I of the X_alpha. Put a well-ordering on X and then define a function f : I --> X by taking f(alpha) to be the minimal element of X_alpha as a subset of X. This constructs a choice function. But we don't actually use exactly that X is well-ordered, just that there is an ordering under which every subset we're interested in has a minimal element.
While elite music of lasting aesthetic quality has been produced, the main reason people listened to elite music in the past (and, indeed, the main reason people listen to 'classical' music now) is a wish to identify themselves as elite - 'I listen to posh music so I am posher than you'.
This, if I may say, needs to be elucidated. The "why" a particular classical piece was composed many be somewhat less-than-noble, but let's not count this against it. Why is the word 'classical' in quotation marks? Perhaps you mean to include music under different period titles (baroque, etc) in your remarks this way.
Rather more mystifying to me is why you believe that the principal reason for people today listening to these 'classical' pieces is an attempt to identify themselves with a higher social segment. If there is some definitive survey or study on this subject, this would certainly be very interesting to those of us so naive that we thought the "lasting aesthetic quality" of those pieces which have lasted was the principal reason.
Is it really true that most people who listen to classical pieces do so not because they actually enjoy the music but because they want to be seen listening to it?
This seems somehow incongruous to me but I can't explain why except by appealing to my prejudices. For example, Vivaldi's "Spring" concerto is perennially popular with TV advertisors; but, is this because they think it elevates the standing of their product or because this piece is justifiably popular and they want listeners to think of their product when they hear it? (I would remark that most of Vivaldi's compositions were for popular consumption and were almost lost until rediscovered, at which time they became very popular, indeed.) The same remarks apply to Orff's "Carmina Burana" and those who score motion pictures.
A better try, if you will: around Christmas there is always a production of Handel's "Messiah" here - I imagine this (and more) is true of most large cities. They even have a special night where the audience is permitted/encouraged to sing along with the choruses. I've not worked up enough courage to go to one of those performances, but do those who do go believe that the words 'sing along' are associated with haute culture?
A different approach: Do people in high social strata listen to such music to keep up pretenses? I can't say this has been my experience. On the contrary, I've seen people champion music of "classical" style but questionable (in others' opinions) "aesthetic value" simply because they do in fact like it. Others have undergone periods of extreme infatuation with certain compositions to the potential annoyance of those around them. So I don't think the converse claim holds.
But that "I want to be posh" motivation as the primary one -- damn my lack of cynicism but I just can't see it.
The best option, I think, is to submit it to the sci.math newsgroup. While the AMS would send a polite rejection notice and your local math dept. would at best inform you of where the first error occurs, the denizens of the newsgroup spend their time voluntarily reading through this sort of stuff and are quite able (and often willing) to explain, sometimes at length, what has gone wrong.
Alternatively but unlikely to be as helpful is the AMS (www.ams.org) (or your nation's mathematical society); they publish at least one journal and likely get all sorts of unsolicited mail for review.
Of course, even if the result is true and the reasoning essentially correct the proof may be so illegible that the referee cannot accept it, but only recommend that it be accepted after revision. So the math department of your local university may be better: they also get mail of this type all the time and may even have someone designated to respond to some of it.
The last of the options which should not be recommended is web-publishing. It is unlikely anyone would even see unless one of the above courses is taken with direction to the site.
"And, what if the standard of refutation? Is it enough to claim "oh, this proof is all just handwaving", or "this proof is worthless, it uses a physicists approach", or does any detractor need to precisely pinpoint where the error is "on page 13, where they get from equation 63 to 64, they effectively multiply both sides with zero"?"
:-P
There is a saying sometimes employed by cruel mathematicians to describe illucid 'proofs': "This isn't right. This isn't even wrong."
Having said that, it surely would be nice to see 'exactly where it goes wrong'. For simple arguments a standard "the first error is on page t, line s" will suffice. More convenient is to find a claim to which an explicit counterexample can be constructed, that is, to show that the proof (if valid) leads to a contradiction - of course this works best when the proof is not reductio ad absurdum
Finally, we should always remember that the burden of "proof" is on the "prover" - it's not anyone else's responsibility if the paper isn't even written in what we'd call mathematics. (Naturally it has no chance of being published in a print journal if it's in that state, which is a necessary condition for the awarding of the prize.)
(The rest is a slightly tangential discussion of two common problems arising from extremely imprecise methods, aka "handwaving".)
Here's a gratuitous example:
Prop. There are more real numbers than integers.
"Proof." Consider the interval [n,n+1) for arbitrary integer n. In this interval there is one integer, but a slew of reals, eg, n+1/2, n+pi/4, and so forth. So, there are more reals than integers.
Of course this proposition is true but the proof is nonsense, since we can "derive" by mimickery
Prop. There are more rational numbers than integers.
which is just false.
This is a pretty simplistic example and doubtless the author of the purported proof of RH is using much more sophisticated handwaving, for which I'll produce another analogy:
Prop. 2 has no proper factorisation.
"Proof." The only integers dividing 2 are 1, -1, 2, -2, the former two being units and the latter two the number and an associate.
This proposition falls into the "not even wrong" category. Why? -- because to write down the word "factorisation" begs the question: in what ring (ie, algebraic context)? Absolutely, 2 is irreducible in the integers. But not in the Gaussian integers:
Prop. 2 is not irreducible in the Gaussian integers.
Proof. We see easily that 2 = (1+i)(1-i). Each factor of the RHS has norm 2, so neither is a unit, and in fact 1+i is a Gaussian prime, and 1-i = (-i)(1+i) is also a Gaussian prime.
The error in the first proposition about 2 was that it made no reference to the context of the discussion. This is the sort of handwaving that occurs in the "proof" of RH. We can see it really makes a difference - in one context the prop. was true, and in another false; sometimes even to utter the word "factorisation" is to already condemn oneself to the "not even wrong" bin.
"I think I can say with pretty good authority that he'd say no to [a match with human and machine against human and machine]."
Not only did he not say "no" to it, he more or less invented it. Advanced chess matches have become semi-annual since Kasparov and Topalov first played in Leon in 1998 (a recap by the always-excellent ChessBase team).
"For any product of N primes there is at least one gallois field,"
... pk there exists a Galois field F with p1*...*pk elements. The next paragraph self-indulgent argument to this effect which uses considerable machinery against a simple typo. Apologies in advance.
Just a few corrections for posterity's sake:
First, it is certainly _not_ true that for _any_ k primes p1,
(Let F be a Galois field. For any field, the prime subfield is isomorphic either to Q or to Zp for some prime p; here obviously Q is not contained in F, so Zp is the prime subfield. In particular Zp is contained in every subfield of F. So, suppose k > 1 and pick any p's not all the same; then write |F| = q^n * a with a > 1, a,q coprime and q not equal to p. Then there exists a Sylow q-subgroup - call it S - of F; then the group S* is a finite multiplicative subgroup of a field, so is cyclic; every cyclic group of a certain order is isomorphic, so S* is isomorphic to GF(q^n)*, and in particular S is isomorphic to GF(q^n). But this latter has characteristic q not equal to p. Zp is not a subfield of this; its prime subfield is Zq. Contradiction!)
Obviously what was meant is that for any integer prime p and integer n >= 1, there exists a Galois field GF(p^n) with p^n elements.
Furthermore, while it is true that there exists "at least one", more is true: there exists _only_ one up to isomorphism (this one is easy; as above, the multiplicative subgroup is cyclic and...)
(Finally, I am neither a cryptographer nor a cryptanalyst, so I am unqualified to speak on the balance of the parent message.)