I have been working through an exercise from Pugh's 'Real Mathematical Analysis' and although there is an answer on this site discussing the exercise (Pugh exercise 1.18) there is a part of the explanation given there that I cannot figure out. It is in the answerer's explanation of part (a): Show that each $x_k$ in the decimal expansion is an integer between $0$ and $9$.
The issue I'm having is with how to show that $$10(x-N) < x_1 + 1, \text{ given that } 0\leq x_1 \leq 9$$ and also how to prove the related statement for the induction step (which I assume uses similar reasoning): $$10^n\left(x-\left(N+\sum_{k=1}^{n-1}\frac{x_k}{10^k}\right)\right) < x_n + 1.$$
Here is my attempt for the first inequality:
Suppose that $x_1 + 1 \leq 10(x-N)$. Then since $x-N < 1$ by definition of the floor function we have $x_1 + 1 \leq 10(x-N) < 10$. Therefore $x_1 < 9$. But we know that $x_1 \leq 9$ (but this isn't a contradiction is it? since we already know $x_1<9$? Or is this a contradiction because we know that $x_1$ can equal $9$.).
I'm not sure what else to do so any hints and/or explanations are welcome. Thanks.
Recall the definition of $x_1$. It is the largest integer less than or equal to $10(x-N)$. So any integer larger than $x_1$ must be greater than $10(x-N)$. In particular this is true for $x_1 + 1$, i.e., $$x_1 + 1 > 10(x-N).$$