In article <firstname.lastname@example.org>, Joe Geluso <email@example.com> wrote: >On 27 Sep 2001 16:17:35 -0500, firstname.lastname@example.org (Dave Seaman) >wrote:
>>It doesn't have to be between, and betweenness is not related to order >>anyway.
>Thanks for the response. I don't understand some of the distinctions >made at other points in your post, but this statement befuddles me in >two ways.
>First, as I understand it, interpolation assumes two things: >a. The existence of a path between two points which may be >approximated in some acceptable way -- be it a straight line or a >nonlinear curve. >b. The desired point is between two known points on that path.
>What is wrong with this understanding?
I don't see anything wrong with your understanding. I notice that your understanding does not mention the word "order" at all.
>Second, the "betweenness theorem" on the following page
>seems to say that betweenness depends on the existence of order.
How so? The "betweenness theorem" simply says that if C is between A and B, then AC + CB = AC. It does not say anything about order, and it does not say either that A < B or A > B.
>How can we say that "3 is between 2 and 4" unless they are known to be >in some order -- either 2-3-4 or 4-3-2?
If "order" simply means that you can draw a (not necessarily linear) path through the three points, fine. I would like you to give me an example of three points that are not in some order according to your definition. I thought from your earlier posting that you were talking about whether the complex numbers form an ordered field, which is an entirely different question.
In order for a field F to be ordered, you must identify some subset P of F that is closed under addition and multiplication, and that satisfies the trichotomy law. There is nothing in the definition of an ordered field that has anything to do with interpolation.
You keep saying the complex numbers cannot be ordered. That is false. For example, we can define an order such that (a+bi) < (c+di) if (a<b) or (a=b and c<d). This is a perfectly valid order on the set C, but it does not make C into an ordered field. The set of all a+bi such that a>0 or (a=0 and b>0) is not closed under multiplication, since i is in the set but i^2 = -1 is not.
-- Dave Seaman email@example.com Amnesty International calls for new trial for Mumia Abu-Jamal