Z is not naturally ordered?
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From: Arturo Magidin on 9 Jun 2010 01:52 On Jun 9, 12:44 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > 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) should be "r+u < s+t", of couse. -- Arturo Magidin
From: Tegiri Nenashi on 9 Jun 2010 12:51 On Jun 8, 8:38 pm, Tim Little <t...(a)little-possums.net> wrote: > On 2010-06-08, Tegiri Nenashi <tegirinena...(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. It is a matter of taste of course, but I like definition purely in algebraic terms. > 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. Hmm, it is straightforward to repair it with division x < y <-> exists s,t : x + s/t*s/t = y however, generality of the "Lagrange trm amended" formula which extended to lattices and naturals would be lost. Or couldn't we just get rid of division to have x < y <-> exists s,t,u,w,z : x*z*z + s*s + t*t + u*u + v*v = y*z*z Unfortunately, it doesn't work for lattices; even smaller assertion x v y = y <- exists s exists z (x^(z^z)) v (s^s) = y^(z^z). has counterexample 2-element lattice... The case of complex numbers is hopeless because there is no standard order?
From: Arturo Magidin on 9 Jun 2010 13:12 On Jun 9, 11:51 am, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > On Jun 8, 8:38 pm, Tim Little <t...(a)little-possums.net> wrote: > > > On 2010-06-08, Tegiri Nenashi <tegirinena...(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. > > It is a matter of taste of course, but I like definition purely in > algebraic terms. > > > 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. > > Hmm, it is straightforward to repair it with division > > x < y <-> exists s,t : x + s/t*s/t = y Tim's point is that this does not hold true for rationals (nor for complex numbers, though of course it depends on how you order the complex numbers). In the rationals, (1/3) < (1/2), but there does not exist a rational r such that (1/3)+r^2 = y. -- Arturo Magidin
From: Tegiri Nenashi on 9 Jun 2010 14:09
On Jun 9, 10:12 am, Arturo Magidin <magi...(a)member.ams.org> wrote: > On Jun 9, 11:51 am, Tegiri Nenashi <tegirinena...(a)gmail.com> wrote: > > > > > On Jun 8, 8:38 pm, Tim Little <t...(a)little-possums.net> wrote: > > > > On 2010-06-08, Tegiri Nenashi <tegirinena...(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. > > > It is a matter of taste of course, but I like definition purely in > > algebraic terms. > > > > 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. > > > Hmm, it is straightforward to repair it with division > > > x < y <-> exists s,t : x + s/t*s/t = y > > Tim's point is that this does not hold true for rationals (nor for > complex numbers, though of course it depends on how you order the > complex numbers). In the rationals, (1/3) < (1/2), but there does not > exist a rational r such that (1/3)+r^2 = y. Easily repairable. Consider (3/7) + x/y = (11/13) where 3/7 and 11/13 are arbitrary nonnegative rational numbers. Rewriting it: (3/7)*y + x = (11/13)*y Next, represent x and y in terms of four squares. We will have quantification over eight variables, and the question is of course if 1. It holds in cases of N,Z,R (it doesn't work for lattices:-( 2. If shorter identity exists |