From: Randy Poe on
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. In that case the points
in your model have positions, but nevertheless the naturals
themselves don't have physical positions or exist as geometric
entities.

Do you have any idea what I'm saying? 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.

- Randy

From: Bob Kolker on
Lester Zick wrote:
> They are if they're associated with points and points define line
> segments.

And what if they aren't? The integers are the integers regardless of how
they are interpreted.

Bob Kolker

>
> ~v~~
From: Virgil on
In article <45feac8a(a)news2.lightlink.com>,
Tony Orlow <tony(a)lightlink.com> wrote:

> Bob Kolker wrote:
> > Tony Orlow wrote:
> >
> >>
> >> One may express them algebraically, but their truth is derived and
> >> justified geometrically.
> >
> > At an intuitive level, but not at a logical level. The essentials of
> > geometry can be developed without any geometric interpretations or
> > references.
>
> But how do you know they are essentials of anything without a reference
> to that to which they refer?

If a system isolated from those references allows one to produce exactly
the same set of theorems as one can get using those references, then the
the references themselves are irrelevant to the theory.
From: Virgil on
In article <45fead36(a)news2.lightlink.com>,
Tony Orlow <tony(a)lightlink.com> wrote:

> Virgil wrote:
> > In article <45fd9bfe$1(a)news2.lightlink.com>,
> > Tony Orlow <tony(a)lightlink.com> wrote:
> >
> >> Virgil wrote:
> >>> In article <45fd6045(a)news2.lightlink.com>,
> >>> Tony Orlow <tony(a)lightlink.com> wrote:
> >>>
> >>>> Virgil wrote:
> >>>>> In article <45fc6fd6(a)news2.lightlink.com>,
> >>>>> Tony Orlow <tony(a)lightlink.com> wrote:
> >>>>>
> >>>>>
> >>>>>> Except that linear order (trichotomy) and continuity are inherent in
> >>>>>> R.
> >>>>>> Those may be considered geometric properties.
> >>>>> If one defines them algebraically, as one often does, are they still
> >>>>> purely geometric?
> >>>>>> Tony Orlow
> >>>> One may express them algebraically, but their truth is derived and
> >>>> justified geometrically.
> >>> How does one prove geometrically what is only defined algebraically?
> >
> >> example?
> >
> > 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.

So that TO is saying that it cannot be proved geometrically, which was
the whole point I was trying to make. Once one has isolated a theory
from its geometric interpretations, no geometry is needed, or relevant,
in its proofs.
>
> >
> > That the cardinality of the symmetric group on n elements is n!.
>
> Isn't that demonstrable geometrically?

Can you do it, TO?
>
> >
> > That a polynomial over the reals of degree n has no more than n real
> > zeroes.
>
> Also geometrically demonstrable.

Show me!
>
> >
> > That a square matrix always 'satisfies' its characteristic polynomial.
> >
> > Etc.
>
> Not sure what that means, but excuse me if I take "square" to be a
> geometrical concept....

Given a real matrix which can be multiplied by itself, show that it
satisfies its characteristic polynomial.

Find the conditions under which an exact sequence of Abelian groups
splits.

And there are thousands more.
From: Virgil on
In article <45feb69a(a)news2.lightlink.com>,
Tony Orlow <tony(a)lightlink.com> wrote:

> >
> >>> That the cardinality of the symmetric group on n elements is n!.
> >> Isn't that demonstrable geometrically?
> >
> > No. There is nothing geometrical in that statement.
> >
>
> Arrange them in a convex polygon as vertices, and draw lines between
> them....

Arrange what in a convex polygon? If you arrange the elements that way,
how do you represent a permutation?
>
> >>> That a polynomial over the reals of degree n has no more than n real
> >>> zeroes.
> >> Also geometrically demonstrable.
> >
> > Please describe such a demonstration.
> >

> >>> That a square matrix always 'satisfies' its characteristic polynomial.
> >>> Etc.
> >> Not sure what that means, but excuse me if I take "square" to be a
> >> geometrical concept....
> >
> > It's not, in this context.
> >
> > - Randy
> >
> >
>
> Oh. Define the geometrical definition of square, that's not "analytical"....
>
> - Tony

TO has it backward, as usual. One can "square" things analytically,
without any geometric interpretation, whenever one has a multiplication
operation that is analytic but not geometric.

In this case the "square" is a Cartesian product of {1,2,3,...,n} with
itself and the the matrix itself is a mapping from that "square" to some
field of numbers.
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