Z is not naturally ordered?
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From: Tim Little on 8 Jun 2010 23:19 On 2010-06-08, Tegiri Nenashi <tegirinenashi(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? No, in mathematics. This is a mathematics newsgroup, after all. > May I ask where the unary relation "non-negative" is coming from? One example: The commutative ring structure requires a unique multiplicative identity, which we shall call 1. There is a unique minimal subset of Z containing 1 that is closed under addition. We can call that set the "positive" integers. There is also a unique additive identity, which we can label 0. An integer is then "non-negative" iff it is either 0 or in the set of positive integers. - Tim
From: Tim Little on 8 Jun 2010 23:38 On 2010-06-08, Tegiri Nenashi <tegirinenashi(a)gmail.com> wrote: > I'm not familiar with this. It is simply that your construction > appears to be too elaborate for such simple concept as standard > order. Well, the integers are often defined in terms of structures on or extensions of the natural numbers, so it makes perfect sense to me for the standard order on the integers to be defined in terms of the naturals. BTW, the standard order on the reals is in a sense only "accidentally" definable by "x <= y iff there exists z such that x + z*z = y". It is not true in the logically prior rationals, for example, nor does it extend to the complex numbers. > Once again, I was asking to exhibit a first order formula that > defines the standard order on Z. In the monologue that upset you so > much, it struck me that the formula with existential quantifier > ranging over four variables seems to cover all the cases. Yes, that would suffice. - Tim
From: Arturo Magidin on 9 Jun 2010 00:06 On Jun 8, 5:11 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > On Jun 8, 12:03 pm, Arturo Magidin > > > I'm confused. The "details" that were available on request was that of > > defining the order of Z, algebraically, by defining the nonzero > > natural numbers to be the positive class; that is, details on how one > > defines the order. But you seem to note, elsewhere, that you are > > familiar with this, so surely you don't need the details. > > I'm not familiar with this. Then how exactly were you able to point out that it is defined by "x <= y if and only if y-x is nonnegative" ? > It is simply that your construction > appears to be too elaborate for such simple concept as standard > order. *That* construction, yes; but then that is not the standard way to define the order among the integers. But you didn't talk about "standard order", you were talking about "natural orders" and continue to fail to explain just what that might refer to, or why you believe that Z does not have such an order (the latter, of course, would necessarily be contingent on an explication of the former). > > So... the fact that one can define the order of Z via the notion of > > "positive class" (using the ring structure, not just the additive > > structure), "implies details are not as succinct as equivalences 1 and > > 2"? > > > Well, I don't see how they are related; nor do I see how (1) and (2) > > are in any sense "equivalent." > > But what was your proof at the bottom? Your (1) and (2) were not equivalent because (2) used the notion of addition for the natural numbers, whereas (1) was a definition applicable to any lower semilattice. The natural numbers have standard, natural notion of addition that precedes, in the usual development, the notion of order; the natural numbers, however, do *not* have a standard, natural notion of meet that precedes the notion of order. So they cannot be equivalent. Further, I did not prove that *your* (1) and (2) were "equivalent". What I showed was that a suitable modification of (2) (with N replaced with an arbitrary lower semilattice, and "+" replaced with meet) yield the same relation on the lowere semilattice. But I needed to *change* (2) in order to even be able to *ask* the question of whether they defined the same relation on a semilattice. Meanwhile, you continue to be frustratingly silent on just what you think the meet operation on N is supposed to be that you have before you define the order. > > In the case of N, one usually defines the order using the additive > > semigroup structure as you note in (2): we *define* x<y to mean "there > > exists z in N such that x+z = y". (If your N includes 0, this defines > > the usual less-than-or-equal ordering; if your N does not include 0, > > this defines the associated strict ordering; but the distinction does > > not matter here). > > I'm not sure why we repeatedly going over this again: Because you insist on not making yourself clear, and instead go off in random musing about semilattice structures. How these are supposed to be related to N and Z which do not have natural semilattice structures that precede the order is apparently clear only to you. > -- In the case of N, one can define the standard order via "there > exists z in N such that x+z = y" > -- In the case of (lower semi-)lattice, one can define the partial > order via "there exists z in L such that x v z = y" > > Also what is your emphasis on "define"? If you have a semilattice, you can use the semilattice structure to define an order. What is the problem with that statement? >I'm routinely use Prover9 > (hence "v" and "^" symbols for lattice operations), which doesn't > understand the concept of "define". Well, computer programs don't "understand" any concepts anyway. But why does this matter, or why should I care what you use, routinely or not? > A formula like > > x < y <-> x v y = y. > > works just fine. But is useless unless you either know what "<" means or you know what "v" means. If you don't know either, then the formula is vacuous nonsense. > > *Once* you have this order on N, it turns out that you have a total > > ordering so you can define a (boring) lattice on N using this order. > > But while N naturally comes equipped with a "successor" function, from > > which we define addition, from which we define the order, it does not > > seem to me to come with a natural "lattice" structure that you could > > base the order on in the first place. > > I share the feeling that lattices define an order relation the most > "natural" way. I don't expect this view to carry over to non- > idempotent algebras (N, Z, R, etc). I'm not expecting to uncover non- > vacuous lattice structure there, as you seems to imply. Yet you begin this thread by stating the standard, non-lattice-derived definition of order in N, compare it with some mysterious lattice- structure-derived definiiton (it is the lattice structure which is mysterious), make random comments about whether or not Z is "naturally ordered", and now apparently "share" a feeling that nobody has expresssed but yourself (who do you share it with? Yourself?) If I seem to "imply" anything, please rest assured that I am only implying my utter bewilderment as to just what the hell your point may have been in the past, or is in the present, because I have no clue. > > But this still does not tell me what this has to do with Z or what it > > is you are calling "naturally ordered". > > Once again, I was asking to exhibit a first order formula that defines > the standard order on Z. YOU NEVER SAID THAT. Were we supposed to read your mind and find out this is what you were allegedly "asking"? The standard construction of Z is as equivalence classes of ordered pairs of natural numbers. If you already know what N is and what the order of N is, then you can define the order of Z as follows: [(a,b)] < [(c,d)] <=> a+d < b+c The order on the left is the order you are defining among integers; [(a,b)] is the equivalence class of the ordered pair (a,b). The order on the right and the sum on the right is the sum and order of natural numbers. >In the monologue that upset you so much, it >struck me that the formula with existential quantifier ranging over >four variables seems to cover all the cases. The "upset" comes from feeling that you are just wasting my time and ignoraing anything I say, and would rather just muse on your own. If that's the case, then why post at all? -- Arturo Magidin
From: Aatu Koskensilta on 9 Jun 2010 00:39 Arturo Magidin <magidin(a)member.ams.org> writes: > On Jun 8, 5:11�pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > >> Once again, I was asking to exhibit a first order formula that >> defines the standard order on Z. > > YOU NEVER SAID THAT. > > Were we supposed to read your mind and find out this is what you were > allegedly "asking"? > > The standard construction of Z is as equivalence classes of ordered > pairs of natural numbers. If you already know what N is and what the > order of N is, then you can define the order of Z as follows: > > [(a,b)] < [(c,d)] <=> a+d < b+c > > The order on the left is the order you are defining among integers; > [(a,b)] is the equivalence class of the ordered pair (a,b). The order > on the right and the sum on the right is the sum and order of natural > numbers. This doesn't really answer Tegiri's question. He appears to be asking for a first-order formula R(x,y) such that for all integers a,b M |= R(a,b) iff a < b where M is a structure consisting of the integers with some unspecified designated elements, relations, functions, and so on. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Arturo Magidin on 9 Jun 2010 01:44
On Jun 8, 11:39 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Arturo Magidin <magi...(a)member.ams.org> writes: > > On Jun 8, 5:11 pm, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > > >> Once again, I was asking to exhibit a first order formula that > >> defines the standard order on Z. > > > YOU NEVER SAID THAT. > > > Were we supposed to read your mind and find out this is what you were > > allegedly "asking"? > > > The standard construction of Z is as equivalence classes of ordered > > pairs of natural numbers. If you already know what N is and what the > > order of N is, then you can define the order of Z as follows: > > > [(a,b)] < [(c,d)] <=> a+d < b+c > > > The order on the left is the order you are defining among integers; > > [(a,b)] is the equivalence class of the ordered pair (a,b). The order > > on the right and the sum on the right is the sum and order of natural > > numbers. > > This doesn't really answer Tegiri's question. He appears to be asking > for a first-order formula R(x,y) such that for all integers a,b > > M |= R(a,b) iff a < b > > where M is a structure consisting of the integers with some unspecified > designated elements, relations, functions, and so on. Right. So R(a,b) could be Er,s,t,u( r,s,t,u in N & (r,s) in a & (t,u) in b & r +u = s+t) Is this not first order? -- Arturo Magidin |