If there actually isn’t a line containing exactly two points in the plane,then it follows that for every line defined by two arbitrary points there has to exist a third point lying on the same line.

I want to prove that if that’s the case,then all the points have to be collinear,which would contradict the initial hypothesis that not all points are collinear!This is where i use induction

1)Prove it for the smallest number possible (here that number is 3).It is quite easy to understand that if for any two out of three points there is a third lying on the same line,thenall three points are collinear!

2)Suppose the statement is true for n points,meaning that if for any two out of n points there exists a third lying on the same line,all points are collinear.

3)Finish the proof by proving it for n+1 points.Let’s suppose we have n+1 points on the plane with the property that for every two out of n+1 points there exists a third lying on the same line.If we prove that all n+1 points are collinear,we should be done!

Indeed,since all the possible choices of two points !!(out of n points)!! are all contained in the number of all the possible choices of two points out of n+1 points,it follows that n points are certainly collinear according to our assumption.Now suppose that the remaining point isn’t lying on the same line.If we draw a line from any of the n collinear points that passes through the last point ,then that line will only contain two points,which is impossible according to what we are trying to prove!

To sum up, we have proven that if there isn’t a line containing exactly two points then all the points have to be collinear,which is contradictory to the initial assumption of the problem!Therefore,there does exist one line containing exactly two points of the plane!

First of all,i would like to apologise for trying to be very descriptive in my solution,but i have been told from my teacher that this proof is wrong.However,looking at it over and over again i can’t seem to find the mistake!Please someone tell me whether my solution is indeed mistaken (or correct hopefully) and why.Also,please excuse any possible misspellings.I would like to thank all of you in advance.I will be waiting for a response.

]]>The standard proof of the Sylvester-Gillai for the Euclidean plane that you hinted at (which fairly easily generalizes to the real projective plane), uses the fact that the Euclidean plane is metric space (has a real-valued point distance function satisfying the triangle inequality). The metric space properties are non-topological (they are the “geometry” in “geometry counts”). This standard proof can only conceivably be generalized to other metric spaces.

However, I was wondering if there might not be a topological proof of Sylvester-Gillai. Specifically, I noticed that the line separation theorem (a line divides the rest of the plane into two disjoint halfplanes, such that a segment between two points off the line crosses the line iff the points are in different halfplanes). Plane separation is a topological property, and holds for the Euclidean plane and the real projective plane, but not for the Fano plane, just like Sylvester-Gillai. I was wondering if there might not be a topological proof of Sylvester-Gillai based on plane separation.

]]>“Conclusion: geometry is important.”

I guess, especially as a parlor game, to those who enjoy playing it.

]]>https://en.wikipedia.org/wiki/Sylvester–Gallai_theorem

The beautiful proof was found by Tibor Gallai in the 1930s.

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