.99999999... does not exist. It (along with infinity) is merely a convenient concept.
All numbers are concepts. Some numbers, like 1, have stronger links to reality than others, but we are looking at mathematics here, not the real world. If you're going to throw away numbers which can't concretely exist, then you're throwing away pi, e, i, zero, and, frankly, almost all of mathematics.
1/9 =/= .11111111...
As you add more digits .1111.... approaches 1/9, but no matter how far you go, it will never equal it.
Silly, you can't "add more digits" to a static number.
.111... is the decimal representation of 1/9, just as 1.000... is the decimal representation of 1/1.
However if you are dissatisfied with his proof, I will be glad to offer you another one:
x = 0.9999...
Multiply both sides by ten:
10x = 9.9999...
Subtract x from both sides:
10x - x = 9.9999... - 0.9999...
9x = 9.0000...
Divide by nine:
x = 1.0000...
And another one:
0.9999... = 0.9 + 0.09 + 0.009 + 0.0009 + ...
= 9·0.1 + 9·0.01 + 9·0.001 + 9·0.0001 + ...
= 9·10-1 + 9·10-2 + 9·10-3 + 9·10-4 + ...
n=∞
= Σ 9·10-n
n=1
n=N
:= lim Σ 9·10-n
N→∞ n=1
= lim ( 1 - 10-N )
N→∞
= lim 1 - lim 10-N
N→∞ N→∞
= 1 - 0
= 1
And another one:
0.9999... ≤ 1.
Assume
0.9999... ≠ 1 (*).
Then
0.9999... < 1,
so there must be some positive number P so that
0.9999... + P = 1.
But for ANY positive P,
0.9999... + P > 1,
which is a contradiction, and definitely wrong. Therefore we are forced to conclude that the assumption (*) was incorrect, that is:
0.9999... = 1
One more perhaps?
Observe that the limit of the sequence
0.9
0.99
0.999
0.9999
0.99999
...
is
0.9999...
That is, the sequence gets closer and closer to 0.9999..., in fact, infinitely close.
But the sequence also gets closer and closer to 1.0000..., in fact, infinitely close. So 1.0000... is a limit of this sequence too.
But a sequence can only have one limit, so 0.9999... and 1.0000... must be the same.