TITLE: Show that for any $n\in\mathbb{N}$ we have $ |x-a_0.a_1a_2\cdots a_n|\leq \frac{1}{10^n}$
QUESTION [5 upvotes]: Let $a_0\in \mathbb{N}\cup\{0\}$ and $\{a_1,a_2,\ldots\}\subset\{0,1,2,\ldots, 9\}$. Define the infinite decimal exansion $x=a_0.a_1a_2\cdots$. Show that for any $n\in\mathbb{N}$ we have $ |x-a_0.a_1a_2\cdots a_n|\leq \frac{1}{10^n}$.
I was thinking of using the proof that $\sup(S+T)=\sup(S)+\sup(T)$ (I've done this in class already) and separating the infinite decimal sequence as $a_0 + a_1*(10^{-1}) + a_2*(10^{-2})+\cdots$ to prove this but I'm not exactly sure how.

REPLY [3 votes]: You have
$$
|x-a_0.a_1a_2\cdots a_n|=\sum_{k=n+1}^\infty a_k\,10^{-k}
\leq\sum_{k=n+1}^\infty9\times10^{-k}=\frac{9\times10^{-n-1}}{1-10^{-1}}=10^{-n}
$$