JT
Posts:
1,267
Registered:
4/7/12


Re: Matheology sqrt(2): WM admits to unlistability of 0/1 sequences
Posted:
Dec 22, 2013 3:30 PM


Den söndagen den 22:e december 2013 kl. 01:27:55 UTC+1 skrev Zeit Geist: > Jonas.t write: > > > > > So is it possible the answer is never, and could the conclusion be that 0.999... not really equal to 1. > > > > So then what does it equal? > > > > What is 1  0.999... equal to? > > > > It must be positive if 1 ~= 0.999... . > > But, if x is positive, then 0.999... + x is greater than 1. > > > > ZG
The correct answer to that question is a string of infinitly many zeros followed by a one.
Now i do not consider infinity to be numerical or having a magnitude, a limit is another thing.
You see it is all very simple to be able to claim that 0.999... actually add up to 1. You must be able to prove that there is such x that 10^x actually equals zero and i do not see how you can.
The use of ALL seem a little fuzzy since creating all is not reachable. Take this computation when will it reach zero, the answer is never. There is no infinite sequense of 0.999... that will render a zero in the remainder, that is simple truth.
Bijective bases do not have that problem in any instance you just can add an A to tranform to raise the previous 9 to an A.
And since in bijective base 1 equals .A the problem could never occure as i see it 0.999... is the unfortunate result of t that 1/3 can not be expressed in using base 10 decimals. And it followed that 3*0.333... must have an answer.
The simple truth is that you simple can not partiton 1/9, 1/3 in base ten decimals exactly.
The workround fix was to just stipulate and claim that they will if there is an infinite number of following 3s or 9s.
But it simply not the truth. x=1; y=0.9 while (x>0) {x=xy; y=y/10;}
Will Always give a none zero remainder.

