Virgil
Posts:
8,833
Registered:
1/6/11


Re: Matheology sqrt(2): WM admits to unlistability of 0/1 sequences
Posted:
Dec 22, 2013 12:41 AM


In article <2d1af107cc9c4643be5783b64d37b6d7@googlegroups.com>, jonas.thornvall@gmail.com wrote:
> > There are two ways of viewing 0.999..., either as a number carried out > > to infinitley many decimal places or as the sequence of finitely long > > decimals 0.9, 0.99,0.999, ... having the infinitiely long decimal, > > 0.999... as its limit. > > > > For either interpetation, the final value differs from 1 by the amount 0. > > No there is no instance of iterated .999... following a zero that will add up > to 1.
Does jonas.thornvall claim that there is any numerical difference between 0.999... And s = sum_(n in N) 9/10^n ? > > Not even after infinitly many added nines to form an infinite sequense > whatever that can be, this is easily seen as the difference after infinitly > many added nines still will be infinitly many zeros followed by a 1.
In my world, infinity many zeros following a decimal point make any susequenct trailing digits irrelevant.
Or does jonas.thornvall wish to propose some nouveau theory of infinitesmals to this discussion? > > You will approach 1 forever, but you will never get there.
Depends on whether one take the high road or the low road, you clearly have taken the wrong road.
> But if we use a bijective base ten a number like 9A, .999A , 9999A, 99999999A > all equals 1 however .999 .999... will not in a bijective base.
What in Niffflehiem is a "bijective" base? The ones I am familiar with are all merely positive integers greater than 1. And since only one of your "9A, .999A , 9999A, 99999999A" above contains a point, whether decimal or bijective, three of them do NOT equal 1. 

