# 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 $$2.4999\ldots = 2.52.4999\ldots = 2.5$$ etc.

Can anyone point me to resources that would explain what the below argument that shows $$999\ldots = -1999\ldots = -1$$ is about?

Here is the most usual proof I see that $$0.999\ldots = 10.999\ldots = 1$$:

$$x=0.999\ldotsx=0.999\ldots$$

$$10x=9.999\ldots10x=9.999\ldots$$

$$10x – x = 910x - x = 9$$

$$x=1x=1$$

Using this same argument template I can show $$999\ldots=-1999\ldots=-1$$:

$$x= \ldots9999.0 x= \ldots9999.0$$

$$0.1x= \ldots9999.90.1x= \ldots9999.9$$

$$0.1x – x = 0.90.1x - x = 0.9$$

$$x=-1x=-1$$

What might this mean?

Edit from one of the comments:

$$\sum_{k=0}^{\infty}{9 \cdot 10^k}=-1\sum_{k=0}^{\infty}{9 \cdot 10^k}=-1$$

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 $\infty - \frac{1}{10}\infty$.