The Definition of Points
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From: hagman on 20 Mar 2007 05:10 On 20 Mrz., 00:43, Lester Zick <dontbot...(a)nowhere.net> wrote: > However I would like to suggest a couple of folk remedies for the > unwary. First off if like Hilbert you don't intend to define points > and lines don't use the terms. Well I define geometries and - in the context of such a geometry - the terms "point" and "line" *are* defined. Therefore you won't keep me from using these terms in such a context. > And second if you intend to use terms > don't pretend they're undefined because the predicates you use them in > conjunction with define them quite effectively. Isn't that exactly what I and others stated? As a matter of speach, you start with previously undefined words, cast a few axioms at them and - voila - they are defined. And the definition as given is not circular. > And third if you > decide to use the terms don't like Bob try to justify their usage with > nonsensical circular definitional regressions. I had not seen Bob's post. However, assuming you have a complete ordered field named R at hand, one might - define a point to be a pair (a,b) in RxR - define a line to be a triple (a,b,c) in RxRxR such that a^2+b^2=1 and a>0 or (a=0 and b>0) - define the incidence relation as (x,y) is_on (a,b,c) iff a*x+b*y +c=0 One then proceeds to show that all of e.g. Hilbert's axioms hold and thus we have a model of a Euklidean plain. In this model, the definition of lines does not use points and vice versa. However, in the coursee of showing the validity of geometric axioms for this model, you have shown that "two lines determine a point", i.e. for two lines (a,b,c) and (a',b',c') there is at most one point (x,y) such that (x,y) is_on (a,b,c) and (x,y) is_on (a',b',c'). Here's a different model: - define a point to be a pair (a,b) in RxR - define a line to be a certain set of points(!), i.e. a subset of RxR of the form {(x,y) in RxR | a*x+b*x+c=0 } for some a,b,c in R with (a,b) != (0,0) - define incidence as element containment This is another model and is indeed isomorphic to the one above (how surprising). Here, lines are composed of points. And two (non-parallel) lines determine (not define!) a point, namely the unique element of their intersection. Thus: Lines are composed of points. And lines determine points. Where is the problem?
From: hagman on 20 Mar 2007 06:11 On 20 Mrz., 00:30, Lester Zick <dontbot...(a)nowhere.net> wrote: > > > I'm saying that a > >model is just a model. The properties of the model do not cause > >the thing it's modeling to have those properties. > > Oh great. So now the model of a thing has properties which don't model > the properties of the thing it's modeling. So why model it? > > ~v~~ You are trying to walk the path in the wrong direction. E.g. 0:={}, 1:={{}}, 2:= {{},{{}}}, ... is a model of the naturals: The Peano axioms hold. However, in this model we have "0 is a set", which does not follow from Peano axioms. Thus the model has some additional properties. What's wrong with that? However, the model shows that the Peno axioms are consistent (provided the set theory we used to construct the model is).
From: Eckard Blumschein on 20 Mar 2007 11:32 On 3/19/2007 9:48 PM, Virgil wrote: > Since 1 = 1.0 = 1.00 = 1.000 accepted = 1.000..., why not? because infinity is a gulf (as uttered by G. Cantor) BTW, Bob each time reminds me of someone who intends to make something a better place and is thinking about it every single day. The latter may not even be true for Bob.
From: Randy Poe on 20 Mar 2007 11:58 On Mar 19, 7:30 pm, Lester Zick <dontbot...(a)nowhere.net> wrote: > On 19 Mar 2007 11:51:47 -0700, "Randy Poe" <poespam-t...(a)yahoo.com> > wrote: > > > > >On Mar 19, 2:44 pm, Lester Zick <dontbot...(a)nowhere.net> wrote: > >> On 19 Mar 2007 08:59:24 -0700, "Randy Poe" <poespam-t...(a)yahoo.com> > >> wrote: > > >> >> > That the set of naturals is infinite. > > >> >> Geometrically incorrect. Unless there is a natural infinitely greater > >> >> than the origin, there is no infinite extent involved. > > >> >The naturals don't have physical positions, since they are not > >> >defined geometrically. > > >> They are if they're associated with points and points define line > >> segments. > > >By "associated with points" I assume you mean something > >like using points to model the naturals. > > Why do you assume that, Randy? Because I can't think of another meaning for this phrase. As usual, you throw out language which is all but meaningless, and it is up to people to guess by successive approximations what it is you mean. > I mean if I raise an issue and you > willy-nilly recast it in terms amenable to you You didn't "throw out an issue", you threw out an undefined phrase. This caused discussion to grind to a halt until a common language can be re-established. Is my guess wrong? Fine. I assume you know what you meant (oops, there I go making assumptions again). If so, then explain it. - Randy
From: Tony Orlow on 20 Mar 2007 13:35
Bob Kolker wrote: > Tony Orlow wrote:> >> >> How do you know the conclusions are correct, > > You mean how do you know conclusions correctly follow from the axioms. > You look. Proofs are precisely the evidence that the conclusion follows > from the premises. You know that's not what I mean. How do you measure the accuracy of the premises you use for your arguments? You check the results. That's the way it works in science, and that's the way t works in geometry. If some set of rules you define leads you to conclude that the volume of a sphere is equal to the volume of two spheres of the same radius as the first, well, you probably want to go back and reexamine your premises and make sure you didn't err somewhere in the derivation of your conclusions. > > Look any any standard treatise on first order logic for a definiton of > proof. Checking to see that a proof indeed shows the desired conclusion > follows from the axioms is a purely mechanical algorithmeic proceedure. > It does not involve intelligence. -Finding- a proof does. Checking a > proof can be done by a trained gorilla or Lester Zick on one of his > better days. > > Bob Kolker > Bob - wake up. How do we know relativity is correct? Because it follows from e=mc^2? Oy! Tony |