Am Dienstag, 3. Oktober 2017 00:33:48 UTC+2 schrieb burs...@gmail.com: > What do you want to fix? If pi is already there, > than a 0-step process is sufficient, the process says: > > hi I am at pi > > Or if you want you can use a 1-step process, one > that starts with Euler number e: > > hi I am at e > now I add pi-e to myself > hi I am at pi > > I guess you mean Q-series or something. Yes pi is > irrational, no element from Q. And a Q-series will > never hit pi on its way. Here is a proof: > > Proof: Assume a Q-series would hit pi on its way. > Then there would be an index n, such that sn=pi, > the partial sum up to n summands would equal pi. > > But each partial sum of a Q-series is from Q, and > pi is not from Q, so we would get a contradiction > saying pi is from Q, since it would be sn=pi. > > So by proof by contradiction the > Q-series cannot hit pi. > > > Am Montag, 2. Oktober 2017 23:32:25 UTC+2 schrieb netzweltler: > > Am Montag, 2. Oktober 2017 22:09:44 UTC+2 schrieb burs...@gmail.com: > > > Well this is probably the greatest nonsense somebody > > > ever posted on sci.math. You know, you didn't say > > > rational number line. > > > > > > So when it is the real number line, pi is of course > > > there. There is of course a point on the real number > > > line that is pi. > > > > It doesn't make sense to discuss the "number line" as long as the problem under discussion hasn't been fixed. > > > > > > > > Am Montag, 2. Oktober 2017 19:54:44 UTC+2 schrieb netzweltler: > > > > Yes. pi is already there and we can exactly locate its position on the number line, but you cannot locate a point on the number line representing pi if this point would be the result of a stepwise process - neither a finite process nor an infinite.