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From: David C. Ullrich on 16 Jun 2010 10:23 Thanks to evervone who helped. Sorry about the typo (I was interrupted by Something Important just as I was about to check things over) - glad I included the explanation of where the problem came from so you could figure out the actual question! If you're curious how I could be so dense: The basic idea, add something to both sides so you can factor and then use the given identities, was clear. But I was studiply overlooking the symmetry: I added things to the left side, factored and replaced, expecting to get something that looked like the right side. Dumb - I should have realized that similar manipulations on the right side would also be needed. Like I said, it was trivial from other axioms that were going to be introduce anyway. The real project is to characterize the set of positive elements of an ordered field (no hints please, I want to work it out myself). So: I could get from semiring to ring using some ordered0fieldish assumptions, but it seemed like it _should_ follow just from the semiringish assumptions. It does. Much nicer, thanks. (Why? Because constructing the _positive_ reals using Dedekind cuts (sets of positive rationals) works out much more elegantly than constructing the reals using Dedekind cuts; negative numbbers introduce unfortunate complications in multiplication. No fair working this out and publishing it, btw... you heard it here first.) On Tue, 15 Jun 2010 11:58:25 -0500, David C. Ullrich <ullrich(a)math.okstate.edu> wrote: >Say P is a structure with an addition and >multiplication satisfying > > >(i) x+y = y+x >(ii) x+(y+z) = (x+y)+z >(iii) If x+z = y+z then x = y. >(iv) (P, multiplication) is an abelian group >(v) x(y+z) = xy+xz. > >I gather that although the definitions are not >quite standard, this might be called a semi-ring >plus cancellation. > >Now suppose that w + x' = w' + x and y + z' = y' + z. >Does it follow that > >(*) wz + xy + x'y' + w'z' = w'z' + x'y' + xy + wz ? > >If we had subtraction then this would be clear, >since (*) just says that (w-x)(y-z) = (w'-x')(y'-z') >and we're given that w-x = w'-x' and y-z = y'-z' >(btw if (*) is _not_ the same as (w-x)(y-z) = (w'-x')(y'-z') >then it's a typo in (*)). Seems like the cancellation >property (iii) should be a substitute for subtraction, >but I've gone around in circles with no luck. > >(If anyone's curious, this has to do with showing >that the obvious definition of multiplication is >well-defined if we apply the Grothendiek construction >to try to get a ring from the semi-ring. The question >doesn't really matter, since in the context I have >in mind there are more axioms that help). > >David C. Ullrich.
From: Rob Johnson on 16 Jun 2010 10:50 In article <20100615.151745(a)whim.org>, Rob Johnson <rob(a)trash.whim.org> wrote: >In article <r6cf16hkg4tv8a8dhs2ncro0nmg80puf7m(a)4ax.com>, >David C. Ullrich <ullrich(a)math.okstate.edu> wrote: >>Say P is a structure with an addition and >>multiplication satisfying >> >> >>(i) x+y = y+x >>(ii) x+(y+z) = (x+y)+z >>(iii) If x+z = y+z then x = y. >>(iv) (P, multiplication) is an abelian group >>(v) x(y+z) = xy+xz. >> >>I gather that although the definitions are not >>quite standard, this might be called a semi-ring >>plus cancellation. >> >>Now suppose that w + x' = w' + x and y + z' = y' + z. >>Does it follow that >> >>(*) wz + xy + x'y' + w'z' = w'z' + x'y' + xy + wz ? >> >>If we had subtraction then this would be clear, >>since (*) just says that (w-x)(y-z) = (w'-x')(y'-z') >>and we're given that w-x = w'-x' and y-z = y'-z' >>(btw if (*) is _not_ the same as (w-x)(y-z) = (w'-x')(y'-z') >>then it's a typo in (*)). Seems like the cancellation >>property (iii) should be a substitute for subtraction, >>but I've gone around in circles with no luck. >> >>(If anyone's curious, this has to do with showing >>that the obvious definition of multiplication is >>well-defined if we apply the Grothendiek construction >>to try to get a ring from the semi-ring. The question >>doesn't really matter, since in the context I have >>in mind there are more axioms that help). > >Now that I have seen the reply from Bastian Erdnuess, I understand >that the question is really to show that > > wy + xz + w'z' + x'y' = w'y' + x'z' + wz + xy (*) > >After having worked this through, I realize that it is the samw as >Bill Dubuque's answer, but I will post it anyway since a second way >of looking at something is sometimes helpful. > > (wy + xz + w'z' + x'y') + (x'y + w'z + w'y + x'z) > > = (wy + x'y) + (xz + w'z) + (w'z' + w'y) + (x'y' + x'z) > > = (w'y + xy) + (x'z + wz) + (w'z + w'y') + (x'y + x'z') > > = (xy + wz + w'y' + x'z') + (w'y + x'z + w'z + x'y) > >Now we just cancel the right summand from the far-left and far-right >sides to get (*): > > wy + xz + w'z' + x'y' = xy + wz + w'y' + x'z' I started thinking about this as being a 4 component catalyst, where x'y -> w'y -> w'z -> x'z -> x'y and realized that this could work with one cancelling term and four steps instead of four cancelling terms and one step: (wy + xz + w'z' + x'y') + x'y = (wy + x'y) + xz + w'z' + x'y' = (w'y + xy) + xz + w'z' + x'y' [w + x' = w' + x] = xy + xz + (w'z' + w'y) + x'y' = xy + xz + (w'z + w'y') + x'y' [z' + y = w + y'] = xy + (xz + w'z) + w'y' + x'y' = xy + (x'z + wz) + w'y' + x'y' [x + w' = x' + w] = xy + wz + w'y' + (x'y' + x'z) = xy + wz + w'y' + (x'y + x'z') [y' + z = y + z'] = (xy + wz + w'y' + x'z') + x'y Cancel the x'y to get (*). Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: Bill Dubuque on 16 Jun 2010 16:45 rob(a)trash.whim.org (Rob Johnson) wrote: >David C. Ullrich <ullrich(a)math.okstate.edu> wrote: >> >>Say P is a structure with an addition and >>multiplication satisfying >> >> (i) x+y = y+x >> (ii) x+(y+z) = (x+y)+z >> (iii) If x+z = y+z then x = y. >> (iv) (P, multiplication) is an abelian group >> (v) x(y+z) = xy+xz. >> >>I gather that although the definitions are not quite standard, >>this might be called a semi-ring plus cancellation. >> >>Now suppose that w + x' = w' + x and y + z' = y' + z. >>Does it follow that >> >>(*) wz + xy + x'y' + w'z' = w'z' + x'y' + xy + wz ? >> >>If we had subtraction then this would be clear, >>since (*) just says that (w-x)(y-z) = (w'-x')(y'-z') >>and we're given that w-x = w'-x' and y-z = y'-z' >>(btw if (*) is _not_ the same as (w-x)(y-z) = (w'-x')(y'-z') >>then it's a typo in (*)). Seems like the cancellation >>property (iii) should be a substitute for subtraction, >>but I've gone around in circles with no luck. >> >>(If anyone's curious, this has to do with showing >>that the obvious definition of multiplication is >>well-defined if we apply the Grothendiek construction >>to try to get a ring from the semi-ring. The question >>doesn't really matter, since in the context I have >>in mind there are more axioms that help). > > Now that I have seen the reply from Bastian Erdnuess, I understand > that the question is really to show that > > wy + xz + w'z' + x'y' = w'y' + x'z' + wz + xy (*) > > After having worked this through, I realize that it is the samw as > Bill Dubuque's answer, but I will post it anyway since a second way > of looking at something is sometimes helpful. > > (wy + xz + w'z' + x'y') + (x'y + w'z + w'y + x'z) > > = (wy + x'y) + (xz + w'z) + (w'z' + w'y) + (x'y' + x'z) > > = (w'y + xy) + (x'z + wz) + (w'z + w'y') + (x'y + x'z') > > = (xy + wz + w'y' + x'z') + (w'y + x'z + w'z + x'y) > > Now we just cancel the right summand from the far-left and far-right > sides to get (*): > > wy + xz + w'z' + x'y' = xy + wz + w'y' + x'z' By removing the motivational remarks from my proof, and by inlining the lemmas, you've completely obfuscated the essence of the matter. In case it was not clear, my proof is just a special case of the well-known proof that congruences can be multiplied, e.g. LEMMA a = A, b = B (mod n) => ab = AB (mod n) PROOF n|a-A,b-B => n | (a-A)b + A(b-B) = ab - AB As I mentioned in [1], this well-known identity also lies at the foundation of the product rule for limits, derivatives, differences, etc. Once one masters this basic idea it is a purely mechanical task to translate it into other contexts, just as I did in this thread. But stripping all this structure from the proof - as you did above - leaves you pulling the proof out of a hat. Why did you choose those particular expressions to add together? Why does it work? Good proofs provide insight not simply verification of a binary truth value. Recall that I've stressed this point to you many times here e.g. [2] when you inlined invocations of Bezout's Lemma into an obfuscated proof of the Rational Root Test. As I said there: Of course one can always "inline" lemmas, abstractions...to produce completely "elementary" versions of theorems. A well-known case of such is the completely self-contained proof of unique factorization of integers by Klappauf, Lindemann, Zermelo [3]. This inlines the Euclidean/Division algorithm descent into a direct inductive proof. By eliminating all the internal structure supporting the argument such proofs tremendously obfuscate the real essence of the matter. In fact it was the discovery of such hidden structure (esp. ideals and modules) in elementary number theory that led Dedekind to invent most of the major classical algebraic structures. Without the introduction of such abstraction there would be little hope of conquering the complexity inherent in modern number theory --Bill Dubuque [1] sci.math, Mar 10 2004, Proof that Multiplication Modulo n is Associative? http://groups.google.com/group/sci.math/msg/0d66427464e5b7a7 http://google.com/groups?selm=y8zbrn4xzfz.fsf%40nestle.ai.mit.edu [2] sci.math, May 18, 2009, Irrationality of sqrt (n^2 - 1) http://groups.google.com/group/sci.math/msg/5ce760473f5d4399 http://google.com/groups?selm=y8z3ab1hnm4.fsf%40nestle.csail.mit.edu [3] http://www.math.uiuc.edu/~dan/ShortProofs/uf.html
From: Rob Johnson on 16 Jun 2010 17:44 In article <l2caaqupuz6.fsf(a)shaggy.csail.mit.edu>, Bill Dubuque <wgd(a)nestle.csail.mit.edu> wrote: >rob(a)trash.whim.org (Rob Johnson) wrote: >>David C. Ullrich <ullrich(a)math.okstate.edu> wrote: >>> >>>Say P is a structure with an addition and >>>multiplication satisfying >>> >>> (i) x+y = y+x >>> (ii) x+(y+z) = (x+y)+z >>> (iii) If x+z = y+z then x = y. >>> (iv) (P, multiplication) is an abelian group >>> (v) x(y+z) = xy+xz. >>> >>>I gather that although the definitions are not quite standard, >>>this might be called a semi-ring plus cancellation. >>> >>>Now suppose that w + x' = w' + x and y + z' = y' + z. >>>Does it follow that >>> >>>(*) wz + xy + x'y' + w'z' = w'z' + x'y' + xy + wz ? >>> >>>If we had subtraction then this would be clear, >>>since (*) just says that (w-x)(y-z) = (w'-x')(y'-z') >>>and we're given that w-x = w'-x' and y-z = y'-z' >>>(btw if (*) is _not_ the same as (w-x)(y-z) = (w'-x')(y'-z') >>>then it's a typo in (*)). Seems like the cancellation >>>property (iii) should be a substitute for subtraction, >>>but I've gone around in circles with no luck. >>> >>>(If anyone's curious, this has to do with showing >>>that the obvious definition of multiplication is >>>well-defined if we apply the Grothendiek construction >>>to try to get a ring from the semi-ring. The question >>>doesn't really matter, since in the context I have >>>in mind there are more axioms that help). >> >> Now that I have seen the reply from Bastian Erdnuess, I understand >> that the question is really to show that >> >> wy + xz + w'z' + x'y' = w'y' + x'z' + wz + xy (*) >> >> After having worked this through, I realize that it is the samw as >> Bill Dubuque's answer, but I will post it anyway since a second way >> of looking at something is sometimes helpful. >> >> (wy + xz + w'z' + x'y') + (x'y + w'z + w'y + x'z) >> >> = (wy + x'y) + (xz + w'z) + (w'z' + w'y) + (x'y' + x'z) >> >> = (w'y + xy) + (x'z + wz) + (w'z + w'y') + (x'y + x'z') >> >> = (xy + wz + w'y' + x'z') + (w'y + x'z + w'z + x'y) >> >> Now we just cancel the right summand from the far-left and far-right >> sides to get (*): >> >> wy + xz + w'z' + x'y' = xy + wz + w'y' + x'z' > >By removing the motivational remarks from my proof, and by inlining >the lemmas, you've completely obfuscated the essence of the matter. >In case it was not clear, my proof is just a special case of the >well-known proof that congruences can be multiplied, e.g. I removed nothing because I wrote my article based on the typo noted by Bastian Erdnuess. I am sorry that you seem to find the need to deprecate my work, but I was just trying to answer David's question. I was not trying to detract from your post; in fact, I had not even read your post until just before I posted, at which point, I added a reference to your article. I am sorry that my post offended you so; however, as soon as I saw that you had posted, I expected to get a lashing for my post. Obviously we have different approaches, and you find yours far superior to mine. >LEMMA a = A, b = B (mod n) => ab = AB (mod n) > >PROOF n|a-A,b-B => n | (a-A)b + A(b-B) = ab - AB > >As I mentioned in [1], this well-known identity also lies at >the foundation of the product rule for limits, derivatives, >differences, etc. Once one masters this basic idea it is a >purely mechanical task to translate it into other contexts, >just as I did in this thread. But stripping all this structure >from the proof - as you did above - leaves you pulling the proof >out of a hat. Why did you choose those particular expressions >to add together? Why does it work? Good proofs provide insight >not simply verification of a binary truth value. > >Recall that I've stressed this point to you many times here >e.g. [2] when you inlined invocations of Bezout's Lemma into >an obfuscated proof of the Rational Root Test. As I said there: > >Of course one can always "inline" lemmas, abstractions...to produce >completely "elementary" versions of theorems. A well-known case of >such is the completely self-contained proof of unique factorization >of integers by Klappauf, Lindemann, Zermelo [3]. This inlines the >Euclidean/Division algorithm descent into a direct inductive proof. > >By eliminating all the internal structure supporting the argument >such proofs tremendously obfuscate the real essence of the matter. >In fact it was the discovery of such hidden structure (esp. ideals >and modules) in elementary number theory that led Dedekind to >invent most of the major classical algebraic structures. Without >the introduction of such abstraction there would be little hope >of conquering the complexity inherent in modern number theory Thanks for writing. I appreciate all opinions, even those which find my articles so offensive. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: Gerry Myerson on 16 Jun 2010 20:32
In article <g3nh16hp0mr5g3f0apdqnfq38b83750598(a)4ax.com>, David C. Ullrich <ullrich(a)math.okstate.edu> wrote: > (Why? Because constructing the _positive_ > reals using Dedekind cuts (sets of positive > rationals) works out much more elegantly > than constructing the reals using Dedekind > cuts; negative numbers introduce unfortunate > complications in multiplication. No fair working > this out and publishing it, btw... you heard it > here first.) No, I read it (or something very much like it) first on pages 25-26 of Conway's On Numbers And Games. He says the best path from the positive integers to the reals is via the positive rationals and then the positive reals, and the reason he gives is precisely that if you go from the rationals to the reals by Dedekind cuts you have to make lots of special cases depending on signs when you get to the multiplication. -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email) |