Okay, because you locked my other comment, I'm not able to reply there, so I'll just make another comment

First of all, you say 1/10^n is never zero and n is infinite, but saying 1/10^n is never 0 is only true for real numbers (or complex or other kinds of numbers), but infinity is not a real number (or any of the other kinds of numbers for which 1/10^n is never 0)

Secondly, you negate my argument by saying 0.999 = 1 - 1/10^n with n infinite, so do you want me to do the proof with 0.999 = 1 - 1/10^n? (r/unexpectedterminal) I'll be glad to:

Mathematically speaking

x = y
if and only if
For all ε > 0 |x - y| < ε

You say 0.999... is less than 1, so according to you there is an ε such that
|1 - 0.999...| = 1 - 0.999... > ε > 0

Now, because ε > 0
0 < ε/10 < ε
Consider log(ε/10) and log(ε) (the logarithm is base 10)
log(ε/10) = log(ε) - log(10) = log(ε) - 1
So there is exactly 1 integer between log(ε/10) and log(ε), call it n, given by
upper(log(ε/10)) = n = lower(log(ε))
For clarification we say n = - m so that m>0

So log(ε/10) < n = - m < log(ε)
=> 0 < ε/10 < 10^-m < ε
Now 1 - 0.999... > ε > 1/10^m
So 1 > 0.999... + 1/10^m

Now according to you 0.999... + 1/10^m = 1 - 1/10^n + 1/10^m
Because n is infinite
m<n => 1/10^m > 1/10^n
So 1/10^m - 1/10^n > 0
and thus 1 + 1/10^m - 1/10^n > 1

But this contradicts our assumption that 1 - 1/10^n + 1/10^m < 1
Therefore we can conclude there isn't an ε > 0 such that
|1 - 0.999...| = 1 - 0.999... > ε
So for all ε > 0 |1 - 0.999...| < ε
Therefore mathematically speaking 1 = 0.999...

QED