
Re: Collinear
Posted:
May 7, 2013 4:39 AM


On Tue, 7 May 2013, David Bernier wrote: > On 05/07/2013 03:48 AM, William Elliot wrote:
> > Let C be a collection of n points with the property that > > any line L with two points of C on it, has a third point > > of C on it. > > > > How is it that C is collinear, ie all points of C are > > on a single line? > > I think this can be proved by induction on 'n' the > number of points. > > For n >= 3, let P(n) denote the statement: > "In any collection C of n points in the plane, > if C is such that any line L with two points > of C on it has a third point on it (i.e., L), > then the set C consists of collinear points." > > P(3) is obvious. > > Then, for any n >= 3, we want to prove: > P(n) implies P(n+1). > > If we can do that, we're done. > > Sketch: assume P(n) and let C be any collection > of n+1 points in the plane. Let Q be some point > in C and let C' = C \ {Q}. > > Then C' has n points. By the induction hypothesis, > P(n) is assumed. > xxxx I dunno ...
Why whould C' have the desired property? > I would try to show that > P(3) implies P(4), so try to show P(4) using > the property (assumed) P(3) ...
P(4) is easy without induction. Pick a,b from S and thus there's some third c in S with collinear { a,b,c }. Now for d, the fourth point, the line ad must contain either b or c. Thus the lines ad and abc coincide and S is collinear.
With more todo, P(5) can proved directly without induction.

