My question really is as simple as: Is the exclusion of infinite decimal expressions of the form $a_0.a_1\dots{a_n}\bar{9}$ logically necessary?
The obvious alternative would be $a_0.a_1\dots{a_n}\bar{9}\equiv{a_0.a_1\dots{a_n}}+10^{-n}.$
The rest of this post is really just background. The authors of the book I am using say that excluding such expressions is "the" solution to the problem which they expose. I am asking if this is the only reasonable solution.
The question is motivated by the discussion in Fundamentals of Mathematics, Volume 1 Foundations of Mathematics: The Real Number System and Algebra; Edited by H. Behnke, F. Bachmann, K. Fladt, W. Suess and H. Kunle
At this point in the development we have already established all of rational number arithmetic in terms of equivalence classes of quotients with the form $a/b;0\ne b,a\in\mathbb{Z}.$ Our extension of $\mathbb{Q}$ to $\mathbb{R}$ begins by considering only non-negative rational numbers: $0\le r\in\mathbb{Q}.$ We previously established the Archimedean ordering of the field $\left\langle \mathbb{Q},+,\times\right\rangle ,$ which means
$$ \forall_{n}0\le a<n^{-1}\iff a=0. $$
Using $n,g,a_{n}\in\mathbb{N}_{0};$$g>1$ we define infinite decimal representation as
$$ r\equiv\sum_{i=0}^{\infty}a_{i}g^{-i}\equiv a_{0}.a_{1}a_{2}a_{3}\ldots; $$
where the properties of the various components are determined by
$$ r_{n}\equiv\sum_{i=0}^{n}a_{i}g^{-i}\equiv a_{0}.a_{1}a_{2}\dots a_{n}\text{ such that } $$
$$ r_{n}\le r<r_{n}+g^{-n}. $$
This requires
$$ 0\le r_{n+1}-r_{n}=a_{n+1}g^{-\left(n+1\right)}=0.00\dots a_{n+1}, $$
and therefore
$$ a_{n+1}=\left(r_{n+1}-r_{n}\right)g^{n+1}. $$
Since these conditions require $r_{n+1}\le r$ we establish $$ 0\le r_{n+1}-r_{n}\le r-r_{n}<g^{-n}. $$
So by
$$ \left(0\le\left(a_{n+1}\right)g^{-\left(n+1\right)}\le r-r_{n}<g^{-n}\right)g^{n+1} $$
we show
$$ 0\le a_{n+1}<g. $$
Order is defined lexicographically. That is, if for $n<k$ we have $a_{n}=b_{n},$ and $a_{k}<b_{k}$ then
$a_{0}.a_{1}a_{2}\dots<b_{0}.b_{1}b_{2}\dots.$
Addition of $g^{-k}$ to $a_{0}.a_{1}a_{2}\dots a_{n}$ is defined as follows:
If $k=0$ or $a_{k}\ne g-1$ then $a_{0}.a_{1}a_{2}\dots+g^{-k}=b_{0}.b_{1}b_{2}\dots$ is given by
$$ n\ne k\implies b_{n}=a_{n}, $$
$$ b_{k}=a_{k}+1. $$
That is
$$ a_{0}.a_{1}a_{2}\dots+g^{-k}=a_{0}.a_{1}a_{2}\dots\left(a_{k}+1\right)a_{k+1}\dots. $$
If $a_{n}=g-1$ when $h<n\le k$ and $a_{h\ne0}\ne g-1,$
$$ n<h\lor n>k\implies b_{n}=a_{n}, $$
$$ b_{h}=a_{h}+1, $$
$$ h<n\le k\implies b_{n}=0. $$
That is
$$ a_{0}.a_{1}a_{2}\dots+g^{-k}=a_{0}.a_{1}a_{2}\dots\left(a_{h}+1\right)0\dots0a_{k+1}\dots. $$
Note that we have not fully defined addition of infinite decimal expressions.
Now consider the case of $a_{n}=g-1$ for all $n>k$. This results in the following inequality
$$ a_{0}.a_{1}a_{2}a_{3}\ldots+g^{-n}>\sum_{i=0}^{k}a_{i}g^{-i}+g^{-k}=a_{0}.a_{1}\dots a_{k-1}\left(a_{k}+1\right). $$
That is $$ r+g^{-n}>r_{k}+g^{-k}=a_{0}.a_{1}\dots a_{k-1}\left(a_{k}+1\right). $$
That is because $a_{m>n}=g-1$ in
$$ r+g^{-n}=r_{k}+g^{-k}+0.00\dots a_{n+1}\dots $$
$$ =r_{k}+g^{-k}+\sum_{i=n+1}^{\infty}\left(g-1\right)g^{-i}. $$
If the monotonic law is to hold for addition and if subtraction is to be possible (for the case when the subtrahend is smaller than the minuend), we have the following inequality
[for all $n>k$]
$$ 0<d=\sum_{i=0}^{k}a_{i}g^{-i}+g^{-k}-a_{0}.a_{1}a_{2}a_{3}\ldots<g^{-n}, $$
$$ 0<d=r_{k}+g^{-k}-r<g^{-n}. $$
Since $g-1\ge1$ we have $g^{n}=\left(1+\left(g-1\right)\right)^{n}>n\left(g-1\right),$ so $g^{-n}<\left(g-1\right)^{-1}n^{-1}<n^{-1}.$ Our expression therefore appears to contradict the established Archimedean ordering of the rational numbers. The solution the authors provide is to exclude infinite decimal expressions with $a_{n}=g-1$ for all $n>k.$ They then assert:
In fact, such sequences do not occur in the decimal expansion of rational numbers.
Their argument is as follows: with the assumption $a_{n}=g-1$ for all $n>k$ we get
$$ r_{n}-r_{k}=\left(g-1\right)\sum_{i=k+1}^{n}g^{-i}=\left(g-1\right)g^{-n}\sum_{i=0}^{n-\left(k+1\right)}g^{-i} $$
$$ =\left(g^{n-k}-1\right)g^{-n}=g^{-k}-g^{-n}. $$
So $r_{n}+g^{-n}=r_{k}+g^{-k}.$ By the requirement $r_{n}\le r<r_{n}+g^{-n}$ we get
$$ r<r_{n}+g^{-n}\le r+g^{-n}, $$
$$ 0<r_{k}+g^{-k}-r\le g^{-n}. $$
This again appears to contradict the Archimedean ordering of $\mathbb{Q}.$ In both cases the problem appears when we subtract $r$. But we don't really have a definition for the subtraction of an infinite decimal expression at this point.
Neither of these situations are quite as mysterious as the notation indicates. To put this in grade school language, in the first case
$$ r+g^{-n}>r_{k}+g^{-k}, $$
the right-hand expression is rounding $r$ up on the $k^{th}$ decimal place. The right hand side is rounding up on the $n^{th}$ decimal place, and then some. The second case
$$ r_{n}+g^{-n}=r_{k}+g^{-k} $$ is just a fancy way of restating our rule for addition. We ``carry a one from the $n^{th}$ decimal place to the $k^{th}$ decimal place''. We got that from the more mysterious equation
$$ r_{n}-r_{k}=\left(g-1\right)\sum_{i=k+1}^{n}g^{-i}=g^{-k}-g^{-n}. $$
Which in base 10 really just means, $a_{k}\ne9,$
$$ r_{k}=a_{0}.a_{1}\dots a_{k}, $$
$$ r=a_{0}.a_{1}\dots a_{k}\bar{9}=r_{k}+0.0\dots0_{k}9_{k+1}\bar{9}, $$
$$ r_{n}=r_{k}+0.0\dots0_{k}9_{k+1}\dots9_{n}, $$
$$ r_{n}-r_{k}=0.0\dots0_{k}9_{k+1}\dots9_{n}. $$
Then
$$ g^{-k}-g^{-n}=0.0\dots1_{k}0\dots0_{n}-0.0\dots1_{n}, $$
where we have to ``borrow a one from the $n-1^{th}$ decimal place, etc.'' Using the above notation we write
$$ r_{k}-r=r_{k}-\left(r_{k}+0.0\dots0_{k}9_{k+1}\bar{9}\right)=-0.0\dots0_{k}9_{k+1}\bar{9}. $$
Now we try
$$ r_{k}+g^{-k}-r=0.0\dots1_{k}-0.0\dots0_{k}9_{k+1}\bar{9}. $$
So it all really boils down to the meaning of $1-0.\bar{9}.$ Lets start more modestly with $1-0.9=0.1,$ then $1-0.99=0.01,$etc. The typical way we approach this is to start with the right-most digits and borrow from the left. But in the case of an infinite decimal expression, there is no right-most digit. So why not argue that we are "borrowing one at infinity"?
Upon further reflection, I must concede that, from a purely constructive point of view, we would never produce $a_0.a_1\dots\bar{9}$ as an expansion of a rational number. This is because the "algorithm" determined by $r_n\le{r}<r_n+g^{-n}$ will always produce $\dots \left(a_n+1\right)$ instead of $\dots a_n\bar{9}.$
It is indeed not the only one way. If we want to define real numbers using decimal expansion, we will encounter the problem with having both $1.000\ldots$ and $0.999\ldots$. There are essentially three ways to deal with it: exclude $0.999\ldots$, exclude $1.000\ldots$, or exclude both $1.000\ldots$ and $0.999\ldots$, but include $\{0.999\ldots, 1.000\ldots\}$ (ie declare that this number is denoted by set of two different decimal fractions).
All three ways have some small advantages and some small disadvantages, but they all are the same in the end as soon as we end with construction of $\mathbb R$ and prove it's basic properties.