Decimal Digits and Unique Real Numbers I understand that every real number has an infinite decimal expansion. 
How can I use the Axiom of Completeness to prove that every string of decimal digits corresponds to a unique real number alpha?
 A: It's worth noting that, while every string of decimal digits signifies a unique real number, there are many real numbers with non-unique decimal representations, as Raskolnikov points out in the comments above.
It is readily seen that a finite string of decimal digits is a rational number. Given an infinite string, say $0.b_1b_2b_3b_4b_5...,$ consider the set $$\{0,0.b_1,0.b_1b_2,0.b_1b_2b_3,0.b_1b_2b_3b_4,...\}$$ What is the supremum of this set of rational numbers?
A: Given a decimal $d$ write $d = \sum_{i=0}^\infty\frac {a_i}{10^i} = a_0.a_1a_2a_3...$ and apply your Axiom of Completeness to this series to show that it is convergent. I don't know what formulation of completeness you're using, but let's say you're using the axiom that every Cauchy sequence of real numbers converges. Then all you have to do is prove that the sum is a Cauchy sequence.
That is, $$\forall \epsilon >0 \; \exists N : m > n > N \Rightarrow \left|\sum_{i=n+1}^m\frac{a_i}{10^i}\right| < \epsilon$$
Proof:
Choose $m = n+1$ and $N=\log_{10}\frac 1 \epsilon$ and note $a_i<10, \;\forall i$. Then:
$$\left|\sum_{i=n+1}^m\frac{a_i}{10^i}\right| < \left|\sum_{i=n+1}^m\frac{10}{10^i}\right| < \left|\sum_{i=n+1}^m\frac{1}{10^{i-1}}\right| < \left|\sum_{i=n+1}^m\frac{1}{10^{n+1}}\right|$$
$$= (m - (n+1)+1)\left(\frac{1}{10^{n+1}}\right) = (m - n)\left(\frac{1}{10^{n+1}}\right)$$
$$= (n + 1 - n)\left(\frac{1}{10^{n+1}}\right) = \left(\frac{1}{10^{n+1}}\right) < \left(\frac{1}{10^N}\right) = \epsilon$$
