
Re: It is a very bad idea and nothing less than stupid to define 1/3 = 0.333...
Posted:
Oct 2, 2017 7:01 PM


That pi would be never hit by Qsteps, I guess I dont need to explain this induction step:
sn in Q and an in Q ==> sn+1 in Q
So if you add rational number summands to you get rational number partial sums. But still who
suggested pi irrational first is an interesting question. I guess it was not Leonardo de Pisa (Fibonnaci).
Here is another hint:
"There is a claim on the wikipedia article on irrational numbers that Aryabhata wrote that pi was incommensurable (5th century) but the question had to be asked as soon someone realized there was such numbers... (that was 5th century Before Christ)" When was ? first suggested to be irrational? https://math.stackexchange.com/questions/177546/whenwaspifirstsuggestedtobeirrational
(No use to ask bird brain John Gabriel, he knows nothing, he still waits that somebody bangs his anal warts for his claim S=Lim S)
Am Dienstag, 3. Oktober 2017 00:41:50 UTC+2 schrieb burs...@gmail.com: > Because pi is an incommensurable ratio, a Greek notion, > it can never by a rational number. But I don't know > whether they really considered pi already as such. > > There is this story with pythagoras and sqrt(2) > irrational. But I am also loss what concerns the > "lost book by Fibonacci", I must have lost it as well: > > Were ratios of incommensurable magnitudes > interpreted as irrational numbers prior to Fibonacci? > https://hsm.stackexchange.com/questions/5582/wereratiosofincommensurablemagnitudesinterpretedasirrationalnumbersprio/5585 > > > 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 0step process is sufficient, the process says: > > > > hi I am at pi > > > > Or if you want you can use a 1step process, one > > that starts with Euler number e: > > > > hi I am at e > > now I add pie to myself > > hi I am at pi > > > > I guess you mean Qseries or something. Yes pi is > > irrational, no element from Q. And a Qseries will > > never hit pi on its way. Here is a proof: > > > > Proof: Assume a Qseries 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 Qseries 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 > > Qseries 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.

