Comment by stackghost

2 hours ago

>Every finite set of points in the Euclidean plane that is not collinear has a line that passes through exactly two of the points.

Isn't this a tautology?

The problem definition states that the set of points is in Euclidean space, which from Euclid's Axioms means we can draw a line between any two points. The set of points is defined to be not collinear, thus we cannot draw a line passing through more than two of them. This is just simple logic.

That is not what was meant. Here is a better rephrasing:

Let X be a set of points not all of which are collinear. Then, there are two points a, b in X such that the line l passing through X only passes through a and b.

  • >Let X be a set of points not all of which are collinear. Then, there are two points a, b in X such that the line l passing through X only passes through a and b.

    I don't see how this rephrasing changes anything. Of course there are two points a and b because again, the definition of the problem leads naturally, obviously, and definitionally to this result.

    • Not all points being collinear does NOT mean that all 3-tuples of points are non-collinear! The hypothesis of the theorem is the former. And what it proves is that there is at least one such 3-tuple.

    • The other thread above helped me. You can have as many collinear points as you want as long as at least one point in the set is non-collinear.

      Consider a 3x3 grid. It satisfies this argument.

"The set is not collinear" here means "there is no straight line passing through all the points simultaneously", not "there is no straight line passing through some three points".

  • ... yes, I understand.

    There's nothing novel here. I feel like I'm taking fucking crazy pills.

    • Math is like that. But try to put any number of points in some configuration where you can't find some line with only two on it. In this diagram, you can't do an axis-aligned line with more or less than three -- but you can go diagonal and cross only two points. There's always a way to find only two points.

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