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


Re: Matheology sqrt(2): WM admits to unlistability of 0/1 sequences
Posted:
Dec 22, 2013 12:04 AM


In article <983305e278304b858847009d5fcaeb9c@googlegroups.com>, jonas.thornvall@gmail.com wrote:
> Den söndagen den 22:e december 2013 kl. 04:03:50 UTC+1 skrev Virgil: > > > > > > > > > > While such constant sequences themselves converge, their sums do not, > > > > so > > > > > > I, for one, do not claim either infinite sum actually has a value. > > > > > > > > > > > > > > Is a number converging towards a limit, the same as the number equals the > > > > > limit. > > > > > > > > "A number" does not converge to a limit a all. A sequence of numbers or > > > > a function can do that but "a number" cannot. > > > >  > > I know but 0.999... will approach 1 even if it never actually get there
0.999..., when regarded as a number, is either already equal to 1 or, according to some, never will be. I regard it, as a number, to equal 1.
It is only if it is regarded as an abbreviation for the fairly obvious infinite sequence of decimals, 0.9. 0.99, 0.999, ..., each of finite length that "it" can approach anything.
, that > was why i was a bit confused. I do not consider 0.999... a number but you do > if converge is not correct what should be used "approach" i have no idea.
Note that every RATIONAL number which does not have a terminating (finite) decimal expansion will have a repeating decimal expansion, and that every real number that has either type of decimal expansion, terminating or repeating, is a rational, and only irrational real numbers can ever have decimal expansions which will never become repeating AND will never terminate.
Since 0.999... is clearly a repeating decimal expansion, it must represent some rational number. The only possible question is which one. 

