I am having trouble parsing a proof of the theorem stated in the title. For clarity here is the complete proof from Martin Liebeck's A Concise Introduction to Pure Mathematics:
Suppose that $a_0.a_1a_2a_3...$ and $b_0.b_1b_2b_3...$ are two different decimal expressions for the same real number. Then one of these expressions ends in $9999...$ and the other ends in $0000...$
Proof: Suppose first that $a_0 = b_0 = 0$. Call the real number with these two expresssion x, so that $$x = 0.a_1a_2a_3... = 0.b_1b_2b_3...$$ Let the first place where the two expressions disagreee be the $k^{th}$ place. Thus $x = 0.a_1...a_{k-1}a_k... = 0.a_1...a_{k-1}b_k...,$ where $a_k \neq b_k$. Assume $a_k > b_k$, hence $a_k \geq b_k+1$. Then $$x \geq 0.a_1...a_{k-1}a_k000...$$ and $$x \leq 0.a_1...a_{k-1}b_k999... = 0.a_1...a_{k-1}(b_{k+1} + 1)000...$$ It follows that $a_k = b_k + 1$ and that the two expresssions for x are $0.a_1...a_k000...$ and $0.a_1...a_{k-1}(a_k-1)9999...
My trouble comes when trying to deduce that $x \leq 0.a_1...a_{k-1}b_k999...$. Since if we assume (WLOG) that $a_k > b _k$ how is it possible that $x$ could less then $0.a_1...a_{k-1}b_k999...$? I suppose another answer to this question would be an alternative proof.
We're looking at the biggest number possible that starts with $0.a_1...a_{k-1}b_k$ and that number is $0.a_1...a_{k-1}b_k999...$ With the other representation, we're looking at the smallest possible number that starts with $0.a_1...a_{k-1}a_k$, and that number is $0.a_1...a_{k-1}a_k000...$ These two ranges do touch at $0.a_1...a_{k-1}a_k$ so that is the value.