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From: David C. Ullrich on 15 Jun 2010 12:58 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: Bastian Erdnuess on 15 Jun 2010 13:24 David C Ullrich 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 (*)). So (*) should be xz + yw + x'y' + z'w' = x'z' + y'w' + xy + zw ? Add w'z and x'y on both sides. Factor for y and z on the left side and for x' and w' on the right side. Replace all four terms in the parentheses by the appropriate formula. Factor everything out and compare. Cheers, Bastian
From: Rob Johnson on 15 Jun 2010 13:25 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 ? Are you sure this is the equation you want to show? Perhaps I am missing something, but it appears that the way this equation is written, it can be shown simply using (i) and (ii). However the grouping is assumed in (*), we can use (ii) to transform the left-hand side of (*) to wz + (xy + (x'y' + w'z')) = (xy + (x'y' + w'z')) + wz using (i) = ((x'y' + w'z') + xy) + wz using (i) = ((w'z' + x'y') + xy) + wz using (i) Now, we can use (ii) again to get the right-hand side of (*); for example, to get the grouping the same as it was in the first line: ((w'z' + x'y') + xy) + wz = (w'z' + x'y') + (xy + wz) using (ii) = w'z' + (x'y' + (xy + wz)) using (ii) >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). 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 15 Jun 2010 14:18 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 > > (*) wy + xz + XY + WZ = WY + XZ + xy + wz [correcyed -wgd] > > 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). HINT wy+xz+xY+wZ = wY+xZ+xy+wz ie. (w-x)(y-z) = (w-x)(Y-Z) + wY+xZ+XY+WZ = WY+XZ+xY+wZ ie. (w-x)(Y-Z) = (W-X)(Y-Z) => wy+xz+XY+WZ + xY+wZ+wY+xZ = WY+XZ+xy+wz + xY+wZ+wY+xZ => wy+xz+XY+WZ = WY+XZ+xy+wz ie. (w-x)(y-z) = (W-X)(Y-Z) That the summands hold true is easy, e.g. wy+xz+xY+wZ = wY+xZ+xy+wz <=> w(y+Z) + x(z+Y) = w(Y+z) + x(Z+y) --Bill Dubuque
From: Rob Johnson on 16 Jun 2010 05:05
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' Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |