Thread: non-Euclidean -help on proof

Projective stuff... Great

From what I can recall, a line passes through every two points.

For the total number of lines, define a_ij to indicate the line passing through points i and j. All lines have N+1 points; So the cardinality of (a_ij) is (N+1)^2. But also, the diagonal (a_ii) does not define a line; So exclude N lines from this list, to get a total of (N+1)^2-N=N^2+N+1 lines.

For the number of points, I think there should be more than just N^2+N+1... Check again plz.

I have seen this question posted before, but I can't find the post

Choose a point P and line L with P not lying on L (if you cannot do this it is not a plane but a line).

The lines through P are in 1-1 correspondence with the points on L. Any line through P meets in in just one point, and any point on L defines a line through P. So there are N+1 lines, each of which has N points other than P. So there are N(N+1)+1 points in the plane.

Now there are (N+1)(N^2+N+1) pairs of the form (P,T) with P a point and T a line through P. Since each line T has N+1 points P, each line occurs in this list (N+1) times. So the total number of lines is (N^2+N+1).

There's another way of seeing the latter. Use duality to observe that every property of the plane remains true if you interchange the words point and plane. Since there are N+1 points on each line, and we showed N+1 lines through each point, there are the same number of points as lines by duality.

thanks, guys

every property of the plane remains true if you interchange the words point and plane.

Oops! How could I forget something so important???

Maybe besides reputation points we also need dumbness points, I would score high

Originally Posted by Rebesques

Oops! How could I forget something so important???

Maybe besides reputation points we also need dumbness points, I would score high

There is, is called negative rep.
Just click on rep and say "do not approve".