>>13145389>>13148539>>13148505I think i proved by contradiction that 1 cannot be 0.999...
let's assume 0.999... = 1 and call "0.999..." as D because i don't know how to write the periodic symbol like this guy
>>13145409As such, 10*D = 9.999... = 10, so far so good
if D = 9 + D/10, then wouldn't it be fair to say that 0.9 = 0.8 + D/10?
if so, 0.7 + D/10 would be equal to 0.8; 0.6 + D/10 would be equal to 0.7; 0.5 + D/10 would be equal to 0.6, and so on until we get that 0.1 + D/10 = 0.2
As
0.9 + D/10 = 1
It follows
1 = 0.9 + D/10 = 0.8 + D/10 + D/10 = 0.8 + 2 (D/10) = (0.7 + D/10) + 2 (D/10) = 0.7 + 3 (D/10) = ... = 0.1 + 9 (D/10)
As 1 is assumed to be 0.9 + D/10, that means 0.1 equals 0.09 + D/100.
Hence
0.1 + 9 (D/10) = 0.09 + D/100 + 9 (D/10)
But
0.09 = 0.08 + 0.00999... = 0.08 + D/100
0.08 = 0.07 + 0.00999... = 0.07 + D/100
...
0.02 = 0.01 + 0.00999... = 0.01 + D/100
Hence
1 = 0.09 + D/100 + 9 (D/10) = 0.08 + D/100 + D/100 + 9 (D/10) = 0.08 + 2 (D/100) + 9 (D/10) = ... = 0.01 + 9 (D/100) + 9 (D/10) = (1/10^2) + 9 (D/10^2 + D/10^1)
You can see the pattern right?
1 = (1/10^m) + 9 * the_sum_of [ D/(10^n) ], with n beginning at 1 and ending at some m (i.e. n must be >= to 1 and m must be >= n to represent the pattern described so far)
Since
the_sum_of [ D/(10^n) ], with n beginning at 1 and ending at some m is = D/10 + D/100 + D/1000 + ... + D/10^m = D (1/10 + 1/100 + 1/1000 + ... + 1/10^m )
It can be rewritten as
(1/10^m) + 9 * D * the_sum_of [ 1/(10^n) ]
But
the_sum_of [ 1/(10^n) ] Is just 0.1 + 0.01 + 0.001 + ... + 1/10^m
which is the same as
0.9/9 + 0.09/9 + 0.009/9 + ... + 9 * 10^m/9
which is also the same as D/9
Now I can rewrite the equation as (1/10^m) + 9 * D * D/9 which is (1/10^m) * D^2, for some m equal or greater than 1 (because n must be >= 1 and m must be >= n)
But remember, this equation must be equal to 1 and m at least 1
It follows that
D^2/10^m = 1
Therefore
D, which is 0.999... , cannot be 1
Q.E.D.
