$$ \forall a \in \mathbb{N}: \quad a = a_{n-1}\times10^{n-1} + a_{n-2}\times10^{n-2} + \dots + a_1\times10 + a_0 \\ a_{n-i} \in \{0;1;2;3;4;5;6;7;8;9\}; \quad a_{n-1} \neq 0 $$
We say that $a$ has $n$ digits; it's important for proving something later but that's not what I'm having trouble with so it's not really relevant for this problem.
Although I've already proved $10^{n-1} \le a$, I'm struggling with the second part $a < 10^n$. I have all the elements for the proof but I'd like some help with putting them together:
- $\forall a \in \mathbb{N}: a_{n-1} \lt 10 \Rightarrow a_{n-1}\times10^{n-1} \lt 10\times10^{n-1} \Rightarrow a_{n-1} \lt 10^n$
- $a_{n-1} \le 9 \Rightarrow a_{n-2}\times10^{n-2} + \dots + a_1\times10 + a_0$ would have to equal $10^{n-1}$ in order for $$ \begin{align} a & = 9\times10^{n-1} + 10^{n-1} \\ & = 10^{n-1}\times(9+1) \\ & = 10^{n-1}\times10 \\ & = 10^n \end{align} $$ but thanks to part 1, we already know $\sum_{k=2}^n a_{n-k}\times10^{n-k} \lt 10^{n-1}$, as we can recursively apply the same argument to each group of terms until we're just left with $n_0 \lt 10$.
I'm rather certain that a nice inductive proof can be constructed from this, but I'm not sure how to start.
Given a decomposition of $a$ into digits as $a=\sum_{i=0}^{n-1}a_i10^i$ such that $a_{n-1}\ne0$ and each $a_i$ satisfies $0\le a_i<10$, we say that $a$ is an $n$-digit number (in base 10). Under these circumstances, we want to show:
For the lower inequality, use the lower bounds $0\le a_i$ and $1\le a_{n-1}$ to get $$a=a_{n-1}10^{n-1}+\sum_{i=0}^{n-2}a_i10^i\ge10^{n-1}+0.$$ For the upper bound, use the upper bound $a_i\le9$ to get $$a=\sum_{i=0}^{n-1}a_i10^i\le\sum_{i=0}^{n-1}9\cdot10^i=9\cdot\frac{10^n-1}9<10^n$$ using the formula for the sum of a geometric series $\sum_{i=0}^{n-1}r^i=\frac{r^n-1}{r-1}$ in the last step.