Why does an argument similiar to 0.999...=1 show 999...=-1? I accept that two numbers can have the same supremum depending on how you generate a decimal representation. So $2.4999\ldots = 2.5$ etc.
Can anyone point me to resources that would explain what the below argument that shows $999\ldots = -1$ is about?
Here is the most usual proof I see that $0.999\ldots = 1$:

$x=0.999\ldots$
$10x=9.999\ldots$
$10x - x = 9$
$x=1$

Using this same argument template I can show $999\ldots=-1$:

$x= \ldots9999.0 $
$0.1x= \ldots9999.9$
$0.1x - x = 0.9$
$x=-1$

What might this mean?
Edit from one of the comments:
$$\sum_{k=0}^{\infty}{9 \cdot 10^k}=-1$$
If you want to understand the mathematics behind these things, it is all based upon the notions of 'convergence' and of 'limits'. If you read any first course in analysis textbook you will find the concept rigorously treated there. 

Basically this is the point: Whenever you write 0.999... you are writing down a numeral that represents the 'limit' obtained when an infinite summation $\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...$ is performed. Since we can prove that this sum 'converges' to some real number (namely 1), we are justified in treating the numeral 0.999... as representing some real number. 

However, whenever you write down 999... I presume you are writing a numeral to represent the limit obtained when an infinite summation $9+90+900+...$ is performed. Since this limit does not converge to any real number, (it 'diverges'), we are not justified in treating the numeral 999... as any real number. So it does not make sense to divide it by ten, or take it away from itself. 

We usually denote such divergent limits by the numeral $\infty$, but this does not denote a real number, and there is no consistent way to define operations such as $\infty - \frac{1}{10}\infty$.

I hope this helps and I hope you are motivated to think more about these things :)
Folks have put forth the $p$-adic interpretation, but another way of understanding this is to consider it in the context of divergent series. One of the ideas behind the study of divergent series is to view $\sum$ as a linear functional from a particular domain $C(\mathbb{R})$ to $R$, where $C(\mathbb{R})$ is the vector space consisting of all real sequences with convergent series. This operator has many useful properties in and of itself, evidenced by how much we care about series in general.

However, we may also pose the question, Can we extend $\sum$ to an operator, call it $S$, on a vector space $V \subseteq \mathbb{R}^{\mathbb{N}}$ containing $C(\mathbb{R})$? In a very trivial way, the answer is yes, we can. To see this, recall that the axiom of choice implies that every vector space has a basis, so let $\{ e_{i} \in C(\mathbb{R}) : i \in I \} \subseteq \{ e_{j} \in V : j \in J \}$ be bases for $C(\mathbb{R}), V$ respectively. Then we can set
$$ S (e_{j}) = \begin{cases}
\sum (e_{j}) &amp; e_{j} \in C(\mathbb{R}) , \\
0 &amp; e_{j} \in V \setminus C(\mathbb{R}) .
\end{cases} $$
However, this is not very satisfying. Linearity is not the only thing we like about summation. For example, consider the shift operator $\sigma : (a_{k})_{k \in \mathbb{N}} \mapsto (a_{k + 1})_{k \in \mathbb{N}}$. A property of $\sum$ is that $\sum ((a_{k})_{k \in \mathbb{N}}) = a_{1} + \sum(\sigma((a_{k})_{k \in \mathbb{N}}))$. We like this, for example, because it means that when doing convergence tests we know that no matter how $(a_{k})_{k \in \mathbb{N}}$ "starts", if it eventually "looks like" a sequence whose sum converges, then we still have convergence; that is, changing any finite number of terms in a sequence won't change whether or not its series converges. I will call this the Front-End Property (FEP).

It is important to note that the proof you listed in the OP that $0. \overline{9} = 1$, though very common, is as written incomplete. In particular, it assumes (rightfully, of course) that $0. \overline{9}$ is defined. You can show this, for instance, by appealing to the fact the sequence $(0.9, 0.99, 0.999, \ldots )$ is monotonic and bounded by $1$, and thus convergent. From this, we consider that $0. \overline{9} \equiv \sum (9 \cdot 10^{-k})_{k \in \mathbb{N}}$, i.e. $0. \overline{9}$ is defined as the summation of the sequence $(0.9, 0.09, 0.009, 0.0009, \ldots)$. We then use the front-end property. Let $x = \sum (9 \cdot 10^{-k})_{k \in \mathbb{N}}$. Then
\begin{align*}
10 x &amp; = \sum(10 \cdot (0.9 \cdot 10^{-k}))_{k \in \mathbb{N}} \\
&amp; = (10 * 0.9) + \sum(10 \cdot (0.9 \cdot 10^{-(k + 1)}))_{k \in \mathbb{N}} \\
&amp; = 9 + \sum(0.9 \cdot 10^{-k})_{k \in \mathbb{N}} \\
&amp; = 9 + x \\
\Rightarrow 9x &amp; = 9 \\
\Rightarrow x = 1 .
\end{align*}
But line one to line two was just the FEP! What you actually proved in the OP is that $S(10 \cdot (0.9 \cdot 10^{-k}))_{k \in \mathbb{N}} = 1$ for any operator $S$ with the FEP. You're mimicking the same argument to show that $\overline{9} = -1$, i.e. you're showing that IF $S$ has the FEP, THEN $S(9, 90, 900, 9000, \ldots) = - 1$ for a linear map $S : \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$. It just happens that $\sum$ is not defined at $(9 \cdots 10^{k})_{k \in \mathbb{N}}$.

So in essence we're saying, "So we know we have this thing called summation that we can do to sequences that does neat stuff, and its defined in terms of these other sequences of partial sums converging. But we can also talk about this summation as a linear transformation, which is just the same as saying I can multiply by a scalar or add the sequences and get what I expect. But summation has other properties as an operator than linearity. We mentioned the FEP, but it's also what we may call positive, i.e. if $a_{k} \geq 0$ for all $k$, then $\sum (a_{k})_{k \in \mathbb{N}} \geq 0$. So we may ask, Can we extend this to all of $\mathbb{R}^{\mathbb{N}}$ and keep FEP and positivity? As you've shown, no.

Note also that if you apply your standard geometric series formula for when the ratio is $10$, you'll get $\overline{9} = - 1$.