Z is not naturally ordered?
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From: William Elliot on 8 Jun 2010 04:59 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.
From: Tegiri Nenashi on 8 Jun 2010 11:51 On Jun 7, 8:02 pm, Ostap Bender <ostap_bender_1...(a)hotmail.com> wrote: > On Jun 7, 5:24 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > > Te giri ne nashi? Govorite po-russki? Yes indeed:-)
From: James Dolan on 8 Jun 2010 12:37 in article <aa29bc5a-8dd2-4122-86ef-42b235a218e5(a)g39g2000pri.googlegroups.com>, tegiri nenashi <tegirinenashi(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? whether or not you can define the standard order on z in terms of an algebraic structure on z depends on which algebraic structure you're considering. exercise: the standard order on z can't be defined in terms of the standard abelian group structure on z. the standard order on z can be defined in terms of the standard commutative ring structure on z. -- jdolan(a)math.ucr.edu
From: Tegiri Nenashi on 8 Jun 2010 13:57 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 implies details are not as succinct as equivalences 1 and 2. Here is more intuition. Let + be binary idempotent associative operation. 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)!
From: Tegiri Nenashi on 8 Jun 2010 14:10
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? |