3 ms·
... yes, I understand. There's nothing novel here. I feel like I'm taking fucking crazy pills.
by stackghost 2mo ago
... yes, I understand.
There's nothing novel here. I feel like I'm taking fucking crazy pills.
- hyperhello 2mo agoMath 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 agoYou'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 agothink this way: EITHER all points are on the same line OR 2 of the points are on the line only for them