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>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

>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.

agnishom 2mo ago
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.
- stackghost 2mo ago
  >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.
  - isomorphic_duck 2mo ago
    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.
  - Cerium 2mo ago
    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.
  - agnishom 2mo ago
    Not really. Perhaps this will help: Can you have a set of points Y on a plane such that Y satisfies the following? Given any line passing through 2 points on Y, there is also a third point in Y that it passes through.
LegionMammal978 2mo ago
"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".
- stackghost 2mo ago
  ... yes, I understand. There's nothing novel here. I feel like I'm taking fucking crazy pills.
  - hyperhello 2mo ago
    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. . . . . . . . . .
    - stackghost 2mo ago
      >There's always a way to find only two points. As soon as the set of points are defined to be non-collinear in Euclidean space, this property must be true, purely from the definition of the problem. To suggest otherwise would be to violate either the problem definition or the axioms of Euclidean geometry.
      - hyperhello 2mo ago
        You're still not quite getting it, the statement isn't that trivial. It deals with the area between totally collinear (obviously impossible to find a lonely pair) and totally non-collinear (obviously impossible not to find a lonely pair).
  - NooneAtAll3 2mo ago
    think this way: EITHER all points are on the same line OR 2 of the points are on the line only for them
fxwin 2mo ago
> 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 wrong and you are misunderstanding what collinearity means. You could have a set of points where all but one are on the same line, and the set of a whole will be not collinear, while we can obviously draw a line that passes through more than two of them. A different way of stating the theorem is that any finite set of points has either a line passing through all points (i.e. the set is collinear) or there exists a line that passes through exactly two points. This dichotomy (why two and not three? Why can't we construct a set where any line passes through at least three points?) is not immediately obvious.
Mithriil 2mo ago
Say you've got a set of 5 points that are not all collinear (e.g. everytime you draw a line, you never get all 5 of them). Now every time you try to draw a line between two points, you always end up with a third one. Can this happen? This is what is stated here: there always exists at least a pair where this does not happen.

agnishom 2mo ago

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.

stackghost 2mo ago
>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.
- isomorphic_duck 2mo ago
  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.
- Cerium 2mo ago
  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.
- agnishom 2mo ago
  Not really. Perhaps this will help: Can you have a set of points Y on a plane such that Y satisfies the following? Given any line passing through 2 points on Y, there is also a third point in Y that it passes through.

LegionMammal978 2mo ago

"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".

stackghost 2mo ago
... yes, I understand. There's nothing novel here. I feel like I'm taking fucking crazy pills.
- hyperhello 2mo ago
  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. . . . . . . . . .
  - stackghost 2mo ago
    >There's always a way to find only two points. As soon as the set of points are defined to be non-collinear in Euclidean space, this property must be true, purely from the definition of the problem. To suggest otherwise would be to violate either the problem definition or the axioms of Euclidean geometry.
    - hyperhello 2mo ago
      You're still not quite getting it, the statement isn't that trivial. It deals with the area between totally collinear (obviously impossible to find a lonely pair) and totally non-collinear (obviously impossible not to find a lonely pair).
- NooneAtAll3 2mo ago
  think this way: EITHER all points are on the same line OR 2 of the points are on the line only for them

fxwin 2mo ago

> 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 wrong and you are misunderstanding what collinearity means. You could have a set of points where all but one are on the same line, and the set of a whole will be not collinear, while we can obviously draw a line that passes through more than two of them. A different way of stating the theorem is that any finite set of points has either a line passing through all points (i.e. the set is collinear) or there exists a line that passes through exactly two points. This dichotomy (why two and not three? Why can't we construct a set where any line passes through at least three points?) is not immediately obvious.

Mithriil 2mo ago

Say you've got a set of 5 points that are not all collinear (e.g. everytime you draw a line, you never get all 5 of them). Now every time you try to draw a line between two points, you always end up with a third one. Can this happen? This is what is stated here: there always exists at least a pair where this does not happen.