tag:blogger.com,1999:blog-9106890883561797850.post4693074878465238383..comments2024-01-23T09:03:14.169-08:00Comments on Llamas and my stegosaurus: Negative DimensionalUnknownnoreply@blogger.comBlogger2125tag:blogger.com,1999:blog-9106890883561797850.post-69702728976176428862007-07-06T03:40:00.000-07:002007-07-06T03:40:00.000-07:00It's really three half-planes in the plane whose i...It's really three half-planes in the plane whose intersection forms a triangle. I think there is something useful in that fact, that could somehow be extended to the negative-dimensional case. <BR/>But I don't think your suggested form of the equation can be consistent. The intersection of three lines in the plane doesn't typically form a line, which is what that equation says.<BR/>Mandelbrot's approach uses a rule involving codimensions, which might be a better approach for some people.Dhttps://www.blogger.com/profile/15141040566131538098noreply@blogger.comtag:blogger.com,1999:blog-9106890883561797850.post-12932016781344472342007-07-05T08:09:00.000-07:002007-07-05T08:09:00.000-07:00So in two dimensions, two lines intersect in a poi...So in two dimensions, two lines intersect in a point:<BR/> 1+1-2=0<BR/>and a point and a line intersect in a -1 dimensional object:<BR/> 0+1-2=-1<BR/><BR/>But in 2 dimensions, three arbitrary lines typically form the boundary of a triangle. Perhaps<BR/> 1+1+1-2=1 ?Mike Stayhttps://www.blogger.com/profile/03408641732412584050noreply@blogger.com