
Re: Euclid's Elements Book 1 Proposition 1  something Euclid missed?
Posted:
Feb 16, 2014 8:24 PM


On Saturday, February 15, 2014 11:33:45 PM UTC5, John Gabriel wrote: > On Saturday, February 15, 2014 10:48:52 PM UTC+2, Ken Quirici wrote: > > > On Saturday, February 15, 2014 1:23:27 AM UTC5, John Gabriel wrote: > > > > > > > On Saturday, February 15, 2014 4:00:41 AM UTC+2, Ken Quirici wrote: > > > > > According to Wikipedia, Book 1 Proposition 1 (constructing equilateral > > > triangle on given line segment) is missing a premise  that if two different > > > circles share a radius (that is, a line segment connecting > > > > > their centers is a radius for both circles). > > > > Nope. There is NO premise missing. The construction is perfectly clear and there are no doubts about the existence of the point of intersection. > > > > > > > Could somebody provide a proof of this premise? > > > > How to construct: > > > > http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI1.html > > > > Why does the triangle contain an equilateral triangle? >
I don't think I asked this question. Once you accept that the circles intersect then indeed as you point out, that the triangle is equilateral is clearly demonstrated by Euclid.
> > > Answer is very simple: By construction, both circle centres are joined through an equal radius. Since the circles must intersect, at either point of intersection, the distance to the centres will also be of length radius. > > > > This is axiomatic. > > > > > The premise that I think is missing is that the two circles must intersect  >Euclid uses this premise to find the vertex of the equilateral triangle whose >base is the common radius of the two circles  the vertex being either one of >the two intersecting points of the circles. > > > > As I said, there is NO missing premise. The link (aleph0) I provided also states something along those lines, but the author David Joyce is mistaken. > > > > > Imagine you start with the two centers, and the length of the radius. Now > > > start drawing the two circumferences. It's clear that they eventually > > > intersect, but how do you demonstrate this rigorously  that is, just using > > > Euclid's Definitions, Postulates, and Common Notions? It's obvious, but so is > > > that you can draw a circle given a center and radius, but Euclid felt it > > > necessary to include that in his 5 Postulates. So even if it's obvious that > > > the two circles intersect, it still needs to be somehow demonstrated OR > > > assumed. > > > > You can't start with two centers. The centres are already in place once you draw the diameter of the first circle!!!! There is NO missing premise. You are confusing yourself. It's actually far simpler than you think. > > > > > Try something else. Start with one circle and its center, already 'constructed' as the Postulate assert> it is possible to do. Now take your compass and put it on the circumference of this circle, and its > > > 'pencil' at the center of the first circle. Now start swinging the compass. It eventually hits the first circle's circumference. Obviously. But WHY. > > > > It MUST MEET the circumference because the distance from the centre to the circumference is ALWAYS of length RADIUS!!! There is no problem here! :) > > So simple that most would never question it. > > > > Look, it's easy to prove if you must have a proof: > > > > If your compass is set to ZERO length, it will never meet the other circle. Any LENGTH greater than zero will cross the path of the circle. > > > > > I'm beginning to think that Euclid takes a huge number of 'obvious' facts for granted. I'm beginning to think Euclid's elements can only REALLY be made rigorous by making the apparently simple geometry a subset of R^2 real analysis  the whole kitandcaboodle of calculus, continuity, &c. > > > > NONSENSE! Euclid's thinking is far clearer than any of the baboons who teach real analysis today. It is VERY rigorous and does not require extra proofs. Furthermore, there are no missing premises.
I think there are. Euclid depends on the continuity of the plane and of lines for all his constructions.
That the two circles intersect depends on showing that a line between two points, one outside the circle and one inside the circle, meets the circle. Not a straight line, any line. In particular, a portion of the circumference of the OTHER circle.
Doesn't it? It seems obvious to the intuition, but that intuition is based on prior concepts of the continuity of the plane and of lines on the plane.
Doesn't it?
Maybe I'm just as you imply confusing myself.
Regards,
Ken

