Z is not naturally ordered?
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From: Arturo Magidin on 8 Jun 2010 14:33 On Jun 8, 12:57 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > On Jun 7, 7:53 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > > On Jun 7, 7:24 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > > > > I'm aware of two (partial)) order definitions (and looking forward to > > > expand my perspective): > > > 1. (Semi)lattice: x < y <-> x ^ y = y. > > > 2. N: x < y <-> exists z: x + z = y. > > > (There is no pun intended with the name "natural order" applied in > > > second case) > > > > Now it comes with a bit of surprise that there is no natural order > > > definition for Z. > > > What do you mean by "natural order"? A definition like 2? Yes, 2 does > > not define an order relation on Z. > > > However, when people talk about the "natural order" of Z, they don't > > refer to (2). > > > Generally, given a set S, a partial order on S is a binary relation # > > such that: > > > (i) For all s in S, s#s; [reflexivity] > > (ii) For all s and t in S, if s#t and t#s, then t=s [anti-simmetry]; > > (iii) For all s, t, u in S, if s#t and t#u, then s#u. [transitivity].. > > > If in addition we have > > (iv) For all s and t in S, either s#t or t#s [totality] > > then we say the order is a "total order". > > > A "strict partial order" on a set S is a binary relation < such that > > (1) For all s in S, s<s does not hold [areflexivity]; > > (2) For all s, t, u in S, if s<t and t<u, then s<u [transitivity] > > > If in addition we have > > (3) For all s and t in S, exactly one of > > s<t; s=t; t<s > > holds [trichotomy] > > then we have a "total strict order". > > > If # is a partial order, then we can define a strict order by letting > > s<t if and only if s#t and s=/=t. > > > If < is a strict partial order, then we can define a partial order by > > letting s#t if and only if s=t or s<t. > > > Lattices are special kinds of partially ordered sets, in which every > > pair of elements has a least upper bound and a greatest lower bound. > > > The integers have "natural" (or canonical) total order, which can be > > either derived from the order of the natural numbers, or that can be > > defined "algebraically" by defining the non-zero natural numbers to be > > the "positive class". Details available on request. > > This What is "this"? I wrote several things. Which one are you refering to? >implies details are not as succinct as equivalences 1 and 2. By (1) you mean ordering using a "(semi)lattice order" on N, (1) x<y iff x/\y = x. [note you defined it incorrectly, since you said x<y iff x/\y =y; but that means that you want the greatest common lower bound of x and y to be y, which means that you want y to be less than or equal to x]. By (2) you mean the definition (2) x<y if and only if there exists z such that x+z=y. First: what is the primitive (semi)lattice structure that you think you have on N on which you base (1)? For (1) to make sense, a (semi)lattice structure must pre-exist your definition. What is that structure on N? Where did it come from? Second: what "equivalence"? The two definitions are not "a priori" equivalent on N, because you do not have an a priori semi-lattice structure on N. > Here is more intuition. > > Let + be binary idempotent associative operation. What do you mean by "idempotent"? >Then, > > x + y = y <-> exists z x + z = y. > > So we can view lattice definition (#1) to be the same as the one > defined for natural numbers (#2)! What is the "lattice" structure you think you can define on N without relying on the usual order? Or how do you define it in terms of the "+" operation? -- Arturo Magidin
From: Arturo Magidin on 8 Jun 2010 14:34 On Jun 8, 1:10 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > On Jun 8, 9:37 am, jdo...(a)math.UUCP (James Dolan) wrote: > > > exercise: > ... > > the standard order on z can be defined in terms of the standard > > commutative ring structure on z. > > Exercise in what, google skills? "commutative ring" "standard partial > order" seems to lead the same way as the other replies, order defined > as nonnegative difference: > > x < y <-> y - x is non-negative > > May I ask where the unary relation "non-negative" is coming from? "Is in N". -- Arturo Magidin
From: Ostap Bender on 8 Jun 2010 14:58 On Jun 8, 1:59 am, William Elliot <ma...(a)rdrop.remove.com> wrote: > On Mon, 7 Jun 2010, Ostap Bender wrote: > > On Jun 7, 5:24 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > >> I'm aware of two (partial)) order definitions (and looking forward to > >> expand my perspective): > >> 1. (Semi)lattice: x < y <-> x ^ y = y. > >> 2. N: x < y <-> exists z: x + z = y. > >> (There is no pun intended with the name "natural order" applied in > >> second case) > > >> Now it comes with a bit of surprise that there is no natural order > >> definition for Z. Or there is? Perhaps, some algebras (other than Z > >> introduce order in some other way? > > > What do you mean by Z? The set of integers? > > Yes. Then even my house cleaner is aware of a very simple and total ordering not only on Z but even on R, more or less. :-)
From: Tegiri Nenashi on 8 Jun 2010 15:35 On Jun 8, 10:33 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Jun 8, 12:57 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > > > > > On Jun 7, 7:53 pm, Arturo Magidin <magi...(a)member.ams.org> wrote: > > > > On Jun 7, 7:24 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > > > > > I'm aware of two (partial)) order definitions (and looking forward to > > > > expand my perspective): > > > > 1. (Semi)lattice: x < y <-> x ^ y = y. > > > > 2. N: x < y <-> exists z: x + z = y. > > > > (There is no pun intended with the name "natural order" applied in > > > > second case) > > > > > Now it comes with a bit of surprise that there is no natural order > > > > definition for Z. > > > > What do you mean by "natural order"? A definition like 2? Yes, 2 does > > > not define an order relation on Z. > > > > However, when people talk about the "natural order" of Z, they don't > > > refer to (2). > > > > Generally, given a set S, a partial order on S is a binary relation # > > > such that: > > > > (i) For all s in S, s#s; [reflexivity] > > > (ii) For all s and t in S, if s#t and t#s, then t=s [anti-simmetry]; > > > (iii) For all s, t, u in S, if s#t and t#u, then s#u. [transitivity]. > > > > If in addition we have > > > (iv) For all s and t in S, either s#t or t#s [totality] > > > then we say the order is a "total order". > > > > A "strict partial order" on a set S is a binary relation < such that > > > (1) For all s in S, s<s does not hold [areflexivity]; > > > (2) For all s, t, u in S, if s<t and t<u, then s<u [transitivity] > > > > If in addition we have > > > (3) For all s and t in S, exactly one of > > > s<t; s=t; t<s > > > holds [trichotomy] > > > then we have a "total strict order". > > > > If # is a partial order, then we can define a strict order by letting > > > s<t if and only if s#t and s=/=t. > > > > If < is a strict partial order, then we can define a partial order by > > > letting s#t if and only if s=t or s<t. > > > > Lattices are special kinds of partially ordered sets, in which every > > > pair of elements has a least upper bound and a greatest lower bound. > > > > The integers have "natural" (or canonical) total order, which can be > > > either derived from the order of the natural numbers, or that can be > > > defined "algebraically" by defining the non-zero natural numbers to be > > > the "positive class". Details available on request. > > > This > > What is "this"? I wrote several things. Which one are you refering to? > > >implies details are not as succinct as equivalences 1 and 2. Among other things, I forgot question mark at the end of my sentence. I referred to the very last "Details available on request". > By (1) you mean ordering using a "(semi)lattice order" on N, > > (1) x<y iff x/\y = x. > > [note you defined it incorrectly, since you said x<y iff x/\y =y; but > that means that you want the greatest common lower bound of x and y to > be y, which means that you want y to be less than or equal to x]. My apologies. "An economy of thinking" taken to the extreme leads to sloppy writing where a reader is supposed to figure out all mistakes and typos. For example I routinely use "<" where it means to be "<=", assuming that almost nobody uses strict order anyway. You seems to have caught the typo where I got the order the wrong way. > By (2) you mean the definition > > (2) x<y if and only if there exists z such that x+z=y. > > First: what is the primitive (semi)lattice structure that you think > you have on N on which you base (1)? For (1) to make sense, a > (semi)lattice structure must pre-exist your definition. What is that > structure on N? Where did it come from? I'm not saying one can define order on N via x < y <-> x + y = y. because addition operation + is not idempotent (i.e. x + x = x). Sorry for the confusion. > Second: what "equivalence"? The two definitions are not "a priori" > equivalent on N, because you do not have an a priori semi-lattice > structure on N. These definitions are equivalent on semi lattice. However, since existential definition also works for N, I suppose it is more general. Hope that clarifies the matter.
From: Tegiri Nenashi on 8 Jun 2010 15:47
On Jun 8, 10:34 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Jun 8, 1:10 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > > > On Jun 8, 9:37 am, jdo...(a)math.UUCP (James Dolan) wrote: > > > > exercise: > > ... > > > the standard order on z can be defined in terms of the standard > > > commutative ring structure on z. > > > Exercise in what, google skills? "commutative ring" "standard partial > > order" seems to lead the same way as the other replies, order defined > > as nonnegative difference: > > > x < y <-> y - x is non-negative > > > May I ask where the unary relation "non-negative" is coming from? > > "Is in N". Well, the puzzling flavor of James' message makes me expect him coming out with succinct first order formula involving variables over commutative ring, and operations "*" and "+". Please note that in R definition of standard order is also quite simple: x < y <-> exists z x + z*z = y. |