If a finite set of n>2 points in the plane are not
all on one line, then prove that there exists a line through exactly two of the points.
A simple elegant proof can be achieved by mathematical induction, as follows :-
The theorem obviously holds for n=3. We now have to prove that if it holds for (n-1), it will hold for n : -
If the situation holds for (n-1) points, then there exists at least one line which goes only through 2 points. If the next point ( the "n"th point) is added outside of one of the existing "2 point lines", the theorem will continue to hold, as the "2 point lines" will not have been disturbed by the additional point. If , on the other hand, the "n"th point is added on a "2 point line" ( including its continuation) , then there will have been created at least one new "2 point line", connecting this point with one of the other (n-3) points. This holds, unless a point can be found which lies on the common intersection of all previously existing infinite length straight lines, but such common intersection cannot exist, as no 3 point interconnecting lines can have one common intersection ( if they don't lie on the same straight line) . Q.E.D
Simpler Solution
