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I have created a comprehensive set of notes recording my study of the chapter I am referring to in this question; with the exception of the last two sections, which I don't believe will be nearly as difficult as the rest of the chapter. There are still a couple of topics I haven't understood to my satisfaction. This is one of them.

This question is similar to Is the exclusion of infinite decimal expressions of the form $a_0.a_1\dots{a_n}\bar{9}$ logically necessary?

It might even be reasonable to conclude, based on the answer I accepted, the authors of my book were wrong. But I believe I am supposed to conclude something from their proof which has evaded me. At least now I can provide a link to an online copy of the book for others to examine.

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, pages 129 to 132 authors say that excluding expressions of the form $a_0.a_1\dots{a_n}\bar{9}$ is required when defining the real numbers as infinite decimal expressions.

For expedience I have written 9 in place of $g-1$, where $g$ is the base.

The only way I have found to interpret their argument as a "proof" that expressions terminating with $\bar{9}$ must be excluded is 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 $\dots a_n\bar{9}$ when working sequentially from $n=0$.

The proofs they give show that adding $g^{-n}$ to the forbidden $r=a_0.a_1\dots\bar{9}$ for any $n$ will result in a number greater than $r.$ But, to me that only says that $r+g^{-n}>r$ for all $n$. Which is pretty much the definition of $>$. So my question is: what is actually proved in the discussion to which I refer?

The authors were considered to be among the best in their respective fields, so when I find myself disagreeing with them, I am hard pressed to believe that I am correct.

For the sake of irony: enter image description here

The following is my rendering of their development: At the point in the development where this is discussed 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.

NB: For the next inequality, this is what is actually written in the English translation linked above:

Mathematical gibberish

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

$$ r_{k}-r=r_{k}-\left(r_{k}+0.0\dots0_{k}9_{k+1}\bar{9}\right)=-0.0\dots0_{k}9_{k+1}\bar{9}. $$

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    $\begingroup$ If you allow decimal expansions to end in an infinite block of 9's and if you then define real numbers to be decimal expansions, then you'll have that $1.5\bar0$ and $1.4\bar9$ are different real numbers, because they are different decimals. That means your real numbers won't be what the rest of the world calls real numbers, because for the rest of the world $1.5\bar0=1.4\bar9$. $\endgroup$ – Andreas Blass May 27 at 23:28
  • $\begingroup$ I don't understand. $1.5\bar{0}=1.4\bar{9}$ appears to be exactly what the authors are telling us is excluded. One might argue that their lexicographical definition of order will fail for such expressions. But that doesn't seem to be difficult to remedy. Just add the exception that $0.\bar{9}=1$. $\endgroup$ – Steven Thomas Hatton May 27 at 23:41
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    $\begingroup$ Once you add that exception, you're no longer "defining the real numbers as infinite decimal expressions." You're defining them as equivalence classes under the equivalence relation that incorporates all those exceptions. (That's why I said in my previous comment "if you then define real numbers to be decimal expansions"; if you define them some other way, for instance as equivalence classes, then the exclusion of $\bar9$ is no longer needed.) $\endgroup$ – Andreas Blass May 27 at 23:45
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    $\begingroup$ I have not read through all the details but it appears they are saying the specific definitions they use do not work if you allow decimals that end with all $9$'s. This in no way means that it's not possible to accommodate such decimals using different definitions. $\endgroup$ – Eric Wofsey May 28 at 0:09
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    $\begingroup$ @EricWofsey They sure went the long way around the point. If either you or Andreas wish to post that simple explanation, I will accept it. I guess it does follow pedantically from their definition of ordering for infinite decimal expressions. But, either way, there needs to be an exception made. Either exclude $0.\bar{9}$, or assert that it is synonymous with 1. $\endgroup$ – Steven Thomas Hatton May 28 at 0:27
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"The first thing we do, let's kill all the language lawyers - Henry VI, Part II" - Bjarne Stroustrup, The C++ Programming Language.

The issue here is one of definitions. And, I concede that circumlocutious demonstration presented in the text has merit.

In terms of Dedekind cuts, the exclusion of $\overline{.9}$ is a case of the third axiom: If a rational number least upper bound of a set $\mathcal{M}$ exists, it is an element of the set of upper bounds $S\left(\mathcal{M}\right)$. And by definition $\mathcal{M}\cap S\left(\mathcal{M}\right)=\emptyset$.

In the post of the original question, there is a quote from the Dedakind cuts section of BBFSK which I find ironically amusing.

One problem with treating $\overline{.9}$ as $\overline{.9}\equiv1$ is that it requires an exception be added to the definition of order for infinite decimal expressions. Also, defining two equivalent forms of infinite decimal representations of the same numbers requires the definition to be of equivalence classes, and not of the expressions themselves.

In so much as infinite decimal representation constitutes statements about numbers, the proposition that $.\bar{9}$ must be excluded from the representation set is a statement about statements, and not a statement about numbers, per se.

What is proved is that the framework, as given, will not support the ambiguous naming $\overline{.9}\equiv1$.

Two obvious problems arise if the ambiguity is accepted. The definition of equality fails because it requires one to one equality between elements of the place-value sets (sequences) $\left\{a_n\right\}_1^\infty$. Similarly, ordering, as defined, fails because it compares the sequences element-by-element.

My objection to the argument in the book is that the exclusion of $\overline{.9}$ is itself a modification of the proposed framework. That is, infinite decimal representation is initially proposed as the set of all possible sequences $\left\{a_n\right\}_1^\infty$, for a given base. The proposed exclusion may be the simplest remedy, but it is a modification intended to avoid the exposed problem. Since there are other conceivable remedies which can be shown to work, the proposition in the text is too vague to be sustained by the arguments presented.

I will add that I have profited much by struggling with this material, so if the goal of the authors is primarily to educate. They have achieved it.

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