Cantor's diagonalization- why we must add $2 \pmod {10}$ to each digit rather than $1 \pmod {10}$?

I am reading this following explanation of why in Cantor's diagonalization to show the uncountability of the reals, the digits of the real number are created by adding $$2 \pmod {10}$$ to the digit we are on in the diagonalization. I have a few questions about this explanation, which reads as follows:

Let us remark that the reason that we modified each digit by adding $$2(\bmod 10)$$ as opposed to adding 1 is that the same real number can have two decimal expansions; for example $$0.999 \ldots=1.000 \ldots$$ But if two real numbers differ by more than 1 in any digit they cannot be equal. Thus we are completely safe in our assertion. (An alternative way of avoiding this potential pitfall is to replace each digit by some different digit chosen from the range $$\{1,2, \ldots, 8\}$$.)

I am confused with the statement "if two real numbers differ by more than 1 in any digit they cannot be equal"- $$0.9999 \ldots$$ and $$1.0000$$ differ by $$0.9$$ in digits, which is more than one, but are equal; are they treating $$0$$ as $$10$$? This didn't make sense to me though when talking about $$\pmod {10}$$, when we don't consider the number $$10$$ but only go up to $$9$$. How should I compute this difference between digits; is it merely the absolute value of their difference? Also, how can I justify why real numbers differing by more than $$1$$ in a digit are different, from the perspective of someone who doesn't know much about the reals formally?

In the case of the decimal expansions $$0.9999\dots$$ and $$1.0000\dots$$, the first digit after the decimal dot are $$9$$ and $$0$$, which are consecutive digits (modulo $$10$$). So, yes, $$0$$ is being treated as $$10$$.
And the difference between digits is this: the difference between $$3$$ and $$9$$ is $$4$$ because you can go from $$9$$ to $$3$$ in $$4$$ steps: from $$9$$ to $$0$$, from $$0$$ to $$1$$, from $$1$$ to $$2$$, and from $$2$$ to $$3$$. Of course, you can go from $$3$$ to $$9$$ upwards, but then it takes $$6$$ steps, rather than $$4$$. In particular, the difference between $$9$$ and $$0$$ is $$1$$, as I wrote in the first paragraph.
• Yes, it's the minimum if you go in both directions. And one real number has two distinct decimal expansions only one one such expansion is like $0.432$ and the other one is $0.43199999\ldots$, in which case, the difference between one digit of the first expansion and the corresponding digit of the second one is always $1$. Commented Sep 22, 2023 at 7:25