#45 flawed logic saying 9≠10 is correct but then extending this to 0.9recurring is incorrect since there are an infinite number of 9s in one. Add .9recurring to any of your examples for it to be correct.
0.999...ad infinitum does NOT Equal 1!
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Aug 29, 2023
0
#63
#47 wrong ⅔ is widely accepted as 0.6recurring you are just rounding it so that you cannot add ⅓ to find that 3/3=0.9recurring=1
As n approaches Infinity
((10^n)-1))/(10^n) shall not reach 1.
As is true of 0.999 ad infinitum, which some here have argued against
Once again,
I WON!
Aug 29, 2023
0
#66
0.9999 repeating infinitely and 1 are the same thing because there is just no room for some number in the middle
((10^n)-1)/(10^n) never diverges to 1.
Case closed!
That theory is validly presented.
One_Zeroth wrote:
As n approaches Infinity
((10^n)-1))/(10^n) shall not reach 1.
As is true of 0.999 ad infinitum, which some here have argued against
As 'n' approaches infinity , value of the above expression becomes infinity/infinity = 1
-1?
Infinity - ( a finite constant )= infinity
Therefore infinity -1 = infinity
Aug 30, 2023
0
#75
How exactly does @One_Zeroth explain ⅓ being 0.333recurring if 0.999recurring is not 1?
The OP already lost this argument a couple months ago...
https://www.chess.com/forum/view/off-topic/999-repeating-equals-1-89355481
DeltaCrimson wrote:
One_Zeroth wrote:
As n approaches Infinity
((10^n)-1))/(10^n) shall not reach 1.
As is true of 0.999 ad infinitum, which some here have argued against
As 'n' approaches infinity , value of the above expression becomes infinity/infinity = 1
infinity/infinity does not only equal 1, it is also 2, and 3^(1/6) or any positive real.
Still wrong as ever.
Using the " ... " notation to indicate repeating digits, if ...
0.111111... = 1/9
0.222222... = 2/9
0.333333... = 3/9 = 1/3
0.444444... = 4/9
0.555555... = 5/9
0.666666,,, = 6/9 = 2/3
0.777777... = 7/9
0.888888... = 8/9
... then ...
0.999999... = 9/9 = 1
Make sense?