That's exactly the same problem - you need to show that 0.9* = 9x 0.1*. I'd expect something like the following:
I'll use bold
N,
R for natural numbers, real numbers, etc
Axiom 1.1: Every non-empty bounded set of real numbers has a least upper bound
Theorem 1.2: Given a r in
R, there exists a n in
N such that n>r
Proof: If it is false then r is an upper bound for
N. By Axiom 1.1
N has least upper bound s. Then s-1 is not an upper bound for
N so there is n in
N with n>s-1. But then n+1>s, s in
N contradicting s = least upper bound for
N.
QED
Corollary 1.3: greatest lower bound { 1/n : n in
N } = 0
Proof: The set is non-empty and bounded below by 0, so t = greatest lower bound = inf {1/n : n in
N} exists. Certainly t >= 0.
If t > 0 then 1/t < n for all n in
N, contradicting Theorem 1.2. Hence t=0
QED
Corollary 1.4: 0.999... = 1
Proof: Let s = least upper bound = sup {a_n : n in
N} where a_n = 0.99...9=1-10^-n (n nines)
Then a_n =< 1 for all n in
N, so s exists by Axiom 1.1, and s =< 1. But if s<1 then by Corollary 1.3. 1-s>1-1/n for some n, so s<a_n. This is a contradiction, and hence s=1
QED
[Lecture notes for Mathematical Analysis by Prof Johnstone for the Mathematics Tripos at the University of Cambridge]
