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From: Bill Dubuque on 17 Jun 2010 10:41 rob(a)trash.whim.org (Rob Johnson) wrote: >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. >> >>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 > > I removed nothing because I wrote my article based on the typo noted > by Bastian Erdnuess [...] in fact, I had not even read your post > until just before I posted Ok, then I'm quite interested in the motivation behind your proof. Why did you choose those particular expressions? Why add them, etc? As you said, a second way of looking at it may be helpful, so I'm interested in understanding the key idea(s) of your "second way".
From: David C. Ullrich on 17 Jun 2010 13:57 On Thu, 17 Jun 2010 10:32:56 +1000, Gerry Myerson <gerry(a)maths.mq.edi.ai.i2u4email> wrote: >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. Drat. Thanks for the information. How'd he manage to steal my idea years before I had it?
From: Bill Dubuque on 17 Jun 2010 15:04 Gerry Myerson <gerry(a)maths.mq.edi.ai.i2u4email> wrote: > 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. It's an interesting (though tedious) exercise to prove that the two different ways yield isomorphic structures. I recall seeing the details worked out in a textbook, but the title escapes me right now. --Bill Dubuque
From: master1729 on 17 Jun 2010 11:27 David Ullrich wrote : > On Thu, 17 Jun 2010 10:32:56 +1000, Gerry Myerson > <gerry(a)maths.mq.edi.ai.i2u4email> wrote: > > >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. > > Drat. Thanks for the information. > > How'd he manage to steal my idea years before I had > it? > > > > since your a genius he must have a time machine.
From: Rob Johnson on 17 Jun 2010 15:45
In article <l2ck4pxiuvr.fsf(a)shaggy.csail.mit.edu>, Bill Dubuque <wgd(a)nestle.csail.mit.edu> wrote: >rob(a)trash.whim.org (Rob Johnson) wrote: >>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. >>> >>>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 >> >> I removed nothing because I wrote my article based on the typo noted >> by Bastian Erdnuess [...] in fact, I had not even read your post >> until just before I posted > >Ok, then I'm quite interested in the motivation behind your proof. >Why did you choose those particular expressions? Why add them, etc? >As you said, a second way of looking at it may be helpful, so I'm >interested in understanding the key idea(s) of your "second way". As I mentioned in a follow-up message, I had viewed this problem as a catalytic process; that is, add something that effects a change but itself remains unchanged. Then we can remove the catalyst using the cancellation property (iii). In the equations above, I tried to exhibit this process using parentheses. The four partial processes (annotated in the follow-up) are given by (wy + x'y) = (w + x')y = (w' + x)y = (w'y + xy) [1] Equation [1] converts wy to xy while rotating catalyst x'y to w'y (xz + w'z) = (x + w')z = (x' + w)z = (x'z + wz) [2] Equation [2] converts xz to wz while rotating catalyst w'z to x'z (w'z' + w'y) = w'(z' + y) = w'(z + y') = (w'z + w'y') [3] Equation [3] converts w'z' to w'y' while rotating catalyst w'y to w'z (x'y' + x'z) = x'(y' + z) = x'(y + z') = (x'y + x'z') [4] Equation [4] converts x'y' to x'z' while rotating catalyst x'z to x'y Combining these 4 equations, we convert wy + xz + w'z' + x'y' to xy + wz + w'y' + x'z' while rotating the catalysts as follows x'y -> w'y -> w'z -> x'z -> x'y. Therefore, the full catalyst, x'y + w'z + w'y + x'z, remains unchanged. Thus, (wy + xz + w'z' + x'y') + (x'y + w'z + w'y + x'z) [5] match elements from each group to get = (wy + x'y) + (xz + w'z) + (w'z' + w'y) + (x'y' + x'z) [6] use the catalytic processes described above to get = (w'y + xy) + (x'z + wz) + (w'z + w'y') + (x'y + x'z') [7] separate left and right elements from each group to get = (xy + wz + w'y' + x'z') + (w'y + x'z + w'z + x'y) [8] To get [6], use the commutativity and associativity of addition. To get [7], use [1]-[4]. To get [8], use the commutativity and associativity of addition. Those were the key ideas in what I had done. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |