email@example.com wrote: > Den tisdagen den 30:e juli 2013 kl. 11:58:10 UTC+2 skrev jonas.t...@gmail.com: >> If you create a hexagon with the radius of one, and recursively double the number of vertices of the hexagon(have same radius as side length) >> >> >> >> I guess the sum of side lengths/radius close in on Pi infinitly, so i imagine that that they will indeed come closer as you double up the vertices recursively. But is there a digit there they won't be the same regardless the number of digits. >> >> >> >> Maybe this is a valid way to calculate Pi?
With the correction, yes. Archimedes used it.
> > Side lengths/ diameter it should be... sorry >
-- Nam Nguyen in sci.logic in the thread 'Q on incompleteness proof' on 16/07/2013 at 02:16: "there can be such a group where informally it's impossible to know the truth value of the abelian expression Axy[x + y = y + x]".