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 :)

In the $10$-adic numbers it is true that $\dots 9999 = -1$.

More precisely, the series $\sum_{n=0}^{\infty} 9 \cdot 10^n$ converges in $\mathbb{Q}_{10}$ and its limit there is $-1$.

As other users have noted, it doesn't make sense to have infinitely many nines to the left of the decimal point. This is because the sequence $9, 99, 999, \ldots$ doesn't converge to anything, unlike the sequence $0.9, 0.99, 0.999, \ldots$, which converges to $1$.

However, you have touched on something interesting. Are there number systems where this does converge? And if so, does it converge to $-1$? The answer to both is yes.

For an integer $k$, there's a ring called the $k$-adic numbers*. Let's use $k = 10$. It's similar to the regular real numbers in base $k$, but a few things are backwards.

$$ \small{\begin{array}{c|c|c} & \textrm{Reals (base }k\textrm{)} & k\textrm{-adic numbers} \\ \hline \textrm{number of digits left of the point} & \textrm{finitely many} & \textrm{infinitely many} \\ \textrm{number of digits right of the point} & \textrm{infinitely many} & \textrm{finitely many} \\ \textrm{two numbers are close if} & \textrm{they match in the leftmost digits} & \textrm{they match in the rightmost digits} \end{array}} $$

That's not really the way that people like to construct the $k$-adics, but it's easier than all that "completion wrt a metric" or "inverse limit" stuff.

Anyways, in this system, $9, 99, 99, \ldots$ does have a limit, and it's $\ldots 999.0$. And if you add $1$ to that, you have the sequence $10, 100, 1000, \ldots$, which converges to $0$. So in that system, $\ldots 999 = -1$.

*Usually people like $k$ to be prime, so they call it $p$. So if you wanna find out more, look up "$p$-adic numbers", not "$k$-adic numbers".

One thing to think about is how this plays with modular arithmetic, if you're familiar. Basically, arithmetic mod $10^n$ is arithmetic where we only care about the last $n$ digits of a number and is defined by writing $a\equiv b\pmod{c}$ if and only if $a-b$ is a multiple of $c$. So, if $c=10^n$, this is the same as saying the last $n$ digits of $a$ and $b$ coincide. We have a series of equalities expressed as follows: $$9\equiv -1\pmod{10}$$ $$99\equiv -1\pmod{100}$$ $$999\equiv -1\pmod{1000}$$ $$9999\equiv -1\pmod{1000}$$ $$99999\equiv -1\pmod{10000}$$ $$\underbrace{99\ldots 99}_{n\text{ times}}\equiv -1\pmod{10^n}$$ These are all quite easy to prove: clearly, if you add $1$ to $\underbrace{99\ldots 99}_{n\text{ times}}$, you get $1\underbrace{00\ldots 00}_{n\text{ times}}=10^n$, where the last $n$ digits are $0$ meaning, mod $10^n$, this sum is $0$. Since $-1$ is more or less defined as the number whose sum with $1$ is $0$, the equality is proven.

The identity you have is, more or less, what happens when we take all the above identities and send $n$ to infinity. One precise way to do this is to say that two numbers $a$ and $b$ are close to each other whenever $a\equiv b\pmod{10^n}$ for large $n$. This takes us into the $10$-adic numbers, as others have suggested.

Another precise way not involving analysis would be to consider that we can consider a "number" $x$ to be something where we can always ask for the value of $x\pmod {10^n}$ in a consistent way - basically, it is just a string of digits. Then, we can define addition and multiplication of "numbers" in a way consistent with their truncations mod $10^n$. This again gives us that the infinite string of $9$'s equals $-1$, but this time in an algebraic way. (This give us the $10$-adic integers, which is a subset of the $10$-adic numbers. To be precise, the construction one can use for this is called an inverse limit, which is a scary sounding name for a scary looking definition)

It's worth noting that your proof, though not a proof that $\ldots 999$ is a sensible thing to think about, is a proof that if it is defined in any reasonable way (i.e. multiplying by $10$ shifts the digits and subtraction works digitwise when no carrying is at play), it equals $-1$. So, this is going to hold of any "reasonable" notion of summation, as well as in any "reasonable" extension of our algebraic system. For instance, one other answer used a method where we take the sum as a power series $$9x+90x^2+900x^3+9000x^4+\ldots$$ and equated this with $\frac{9}{1-10x}$ near $x=0$, which is a rational function. Without even checking, your proof tells us that this function had better equal $-1$ at $x=1$, since summing by this method allows you to do all the manipulations that you used.