Nobody else has had the poor taste to muddy the waters further by introducing non-standard arithmetic. So I shan't either. WFL On 1/28/13, Dan Asimov <dasimov@earthlink.net> wrote:
To Eric Angelini:
Besides what Whit and Michael said,
The only way to understand an infinite decimal representation N.abcde... is to understand
a) That it represents the limit of the sequence S = N.a, N.ab, N.abc, N.abcd, N.abcde, ... .
b) What it means for a sequence of real numbers to have a limit, and that when this is true, the sequence has only one limit.
c) That the sequence S must in fact have a limit.
So e.g. 0.99999... means the limit of the sequence of real numbers
S_1 = 0.9, 0.99, 0.999, 0.9999, 0.99999, ... .
This must be 1, because no matter how close you may want the sequence to get to the limit (say within the number e > 0 of the limit), then after a certain number of terms of the sequence, the rest of the terms all lie within e of 1. (In other words, they are all between 1-e and 1+e.)
Let's *check* this claim. If we pick a sequence of closenesses e_1 > e_2 > e_3 > ... > e_n > ... > 0 such that this sequence contains arbitrarily small numbers, and we verify the above sentence for e = e_n no matter what n we choose, then we have verified that 0.99999... = 1.
So, let's choose e_n = 1/10^n for each n = 1,2,3,... . Clearly these get arbitrarily small.
Then for any e_n = 1/10^n, we can look at all the terms of the sequence S_1 *after the nth term*. Now, the nth term has n 9's after the decimal point, so the terms *after that* have at least n+1 9's after the decimal point. It's easy to check that this means they all lie between 1-(1/10^n) and 1+(1/10^n). This proves that the limit of the sequence S_1 is 1. This means that 0.99999... = 1.
--Dan