
Re: Euclid's Elements Book 1 Proposition 1  something Euclid missed?
Posted:
Feb 19, 2014 4:47 PM


In message <0322c429a2304f819f983377b6e0a2f5@googlegroups.com>, Ken Quirici <kquirici@yahoo.com> writes >On Monday, February 17, 2014 6:20:09 PM UTC5, David Hartley wrote:
... >>I assume, from the discussions I've seen on >> >> the web, there are models which satisfy this but still have circles >> >> which don't intersect even though they "should". Further Googling would >> >> probably turn up an example but for now it's more fun trying to find one >> >> on my own.
Consider the relationship between circles and the equality of straight lines.
A circle is defined by a collection of equal straight line segments with one common endpoint.
Two straight line segments are equal if they are radii of a common circle, or if they are the first and last of a finite sequence of straight line segments where each adjacent pair are radii of a common circle.
That is all that Euclid's axioms and postulates tell us about the two concepts. As it stands the two definitions form a closed loop and are absolutely useless. Add in a postulate which forces certain circles to intersect and suddenly you have a powerful technique for constructing equal straight lines and the whole wonderful system becomes possible.
The following system models this. Take the standard model, keep the straight lines and points but define two line segments to be equal iff they are identical. Each circle has exactly one point on its circumferences, no two distinct straight line segments have the same length, but all the postulates are satisfied.
Pasch's axiom is also obeyed, as will be many of the other implicit axioms that have been spotted over the centuries  as long as they don't involve circles or length, (e.g. order properties). Yet proposition 1 fails completely, there are no equilateral triangles.
So that answers the original question. You can't prove, from Euclid's original axioms and postulates, that two circles which share a radius must intersect, even if you add Pasch's axiom.
 David Hartley

