
Re: Collinear
Posted:
May 7, 2013 6:24 AM


On 05/07/2013 04:39 AM, William Elliot wrote: > 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?
You're right. I'm nowhere near solving this right now.
>> 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. >
 Jesus is an Anarchist.  J.R.

