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The Sylvester–Gallai Theorem

25 points by surprisetalk ago | 18 comments

hyperhello |next [-]

> 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?

Sniffnoy |root |parent |next [-]

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.

stackghost |root |parent [-]

>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 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"

hyperhello |root |parent [-]

His statement helped me. It's not that every three points are non-collinear, it's that any three points are non-collinear. A set of points all lying on a line is the only exception; you can have every point lying on a line except for one, or two, or whatever you want. In a square grid of sixteen points, there are lots of sets of four collinear points for example, but not all sixteen, and that's what counts.

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.

glimshe |root |parent |next |previous [-]

Try to come up with a set non-colinear points where NO line passes through two and ONLY TWO points and you'll see the value of the statement.

You may think "I'm sure I can arrange these points in a way where EVERY line will cross three or more points" but you will fail if you try unless ALL points are colinear.

scythe |root |parent [-]

This is true for finite sets. For infinite sets, the Sierpinski triangle is a counterexample.

chenb4425 |root |parent |next |previous [-]

It’s that the line passes through exactly two points, which if you think about it is not exactly obvious.

math_loser |root |parent |next |previous [-]

> So what is the new fact?

For all arbitrarily sized (but finite) sets of not collinear points, there's always a line that passes through exactly two points in the set.

tzs |root |parent |previous [-]

It can help to think about theorems like this by restating them as a puzzle asking for a counterexample.

Given N points, N > 2, can you arrange them in a Euclidean plane so that (1) they are not all on the same line, and (2) every line that goes through two of the points must also go through at least one more of the points?

The theorem says that you cannot do this.

Nail2680 |next |previous [-]

I might be too stupid to understand why this is interesting and useful. If it helps I am a working physicist, and a lot of pure math is lost on me. I think I followed this, but I don't know why one would care or this would be interesting.

emil-lp |next |previous [-]

Futility closet is fantastic!

stackghost |previous [-]

>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 |root |parent |next [-]

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 |root |parent [-]

>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.

Cerium |root |parent |next [-]

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.

isomorphic_duck |root |parent |previous [-]

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.

LegionMammal978 |root |parent |previous [-]

"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 |root |parent [-]

... yes, I understand.

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