11–20 of 52 posts
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.
> Every finite set of points in the Euclidean plane that is not collinear has a line that passes through exactly two of the points. I can't make out the point here (no pun). Of course a line can pass through any two points. It could pass through three if those points were collinear but the statement says they're not. So what is the new fact?
I think what's going on here is that you've misunderstood the theorem's hypothesis. The hypothesis isn't that no three of the points are collinear; rather, it's the weaker statement that there isn't any one single line that all the points lie on. It's true that with your version of the hypothesis the theorem would be trivial; but with the actual hypothesis it is is nontrivial.
... of course there's no single line that all the points lie on. They've been defined to be non-collinear.
Edit: can't reply because of HN's stupid rate-limit mechanism, but to this:
>So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always find a line with exactly two points.
Of course you can. It's absolutely implied by the problem definition. My 9 year old could do this, given a ruler and a pencil, with 100% success rate. I absolutely do not believe this is a novel "theorem"
>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 the…
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.
>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 the…
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.
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.
>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 the…
Earlier quoted context omitted.
I think what's going on here is that you've misunderstood the theorem's hypothesis. The hypothesis isn't that no three of the points are collinear; rather, it's the weaker statement that there isn't any one single line that all the points lie on. It's true that with your version of the hypothesis the theorem would be trivial; but with the actual hypothesis it is is nontrivial.
>rather, it's the weaker statement that there isn't any one single line that all the points lie on ... of course there's no single line that all the points lie on. They've been defined to be non-collinear. Edit: can't reply because of HN's stupid rate-limit mechanism, but to this: >So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always…
So the theorem proves that no matter which way you arrange any finite set of points, except for all on the same line, then you can always find a line with exactly two points.
>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 the…
"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".
There's nothing novel here. I feel like I'm taking fucking crazy pills.
Earlier quoted context omitted.
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.
Earlier quoted context omitted.
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.
Consider a 3x3 grid. It satisfies this argument.
Earlier quoted context omitted.
"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.
. . .
. . .
. . .