Re: .999... = 1

MoeBlee wrote: 
.999... = 1. As famous as this theorem is, I haven't been able to find
a rigorous proof (one that does not use infinite strings in additive
columns as if the strings were finite) and I haven't been able to prove
it for myself. I'd like to have a proof that:

lim of SUM[k = 1 to n]9/(10^k) = 1. 

It's a geometric series, with a = 9/10, r = 1/10.
Since |r| < 1; it's convergent to a / (1 - r).

I think that a more rigorous approach is to
start with the definition of real numbers as
equivalence classes of Cauchy sequences.  Then,
showing that the Cauchy sequences for 1 and .999...
are in the same equivalence class is easy.

I tried the following (since this is such a familiar subject, such
things as that 'e' ranges over reals and 'n' over naturals are
implicit):

Let f(n) = SUM[k = 1 to n]9/(10^k)

Show that, for all e, there exists n, such that, for all k > n,
|1 - e| > |1 - f(k)|.


Let e be 1.  There isn't any n such that 0 > | f(n) - 1 |.

Suppose |1 - e| > 1/(10^j) for some j > 0. Then let n = j. Let k > n.
Show, by induction on k, that 1/(10^j) > |1 - f(k)|.

For the basis step, k = 0, the result is vacuously true, since k > n >
0. Or, the basis step could have k = 2 (since k > n > 0) and, in this
case, the basis step holds since if k = 2, then j = 1, and 99/100 is
closer to 1 than is 9/10. But at the inductive step I just get tangled
in calculations that don't seem to be leading to the result, as well as
I do not see how to use the crucial fact that 9 is the numerator. Maybe
I'm approaching this wrong from the start.

This is not an exercise, as I just would like to have this proof as a
matter of record for a conversation about the subject. So, while hints
are appreciated, a finished proof would be even more appreciated. I
hope the proof doesn't require much more than the little I know, but if
the proof requires some more advanced theorems, then so be it, as I'll
need to learn these theorems anyway.

Thanks in advance for any help you would provide.

MoeBlee

P.S. This is posted to sci.logic rather than sci.math, since I just
happen to be familiar with the postings of enough people in sci.logic
that I know there are some who I can trust their expertise.


Seems off topic.  There are plenty of sharp people in sci.math,
including the ones who also follow sci.logic.
.