Prove properties of a generalization of the decimal expansion of a number

We may generalize the decimal expansion of a number by replacing the integer 10 with any integer $b> 1$ . If $x > 0$ , let $a_0$ denote the greatest integer greater than $x$ . Assuming the integers $a_0, a_1, \ldots, a_{n-1}$ have been defined, let $a$ , denote the largest integer such that

$\sum_{k=0}^n \frac{a_k}{b^k} \leq x.$

Show that $0 \leq a_k \leq b-1$ for each $k \geq 1$ .

Proof. Suppose otherwise, that for some $n \geq 1$ we have $a_n \geq b$ . Hence,

$\begin{align*} \sum_{k=0}^n \frac{a_k}{b^k} \leq x && \implies && \sum_{k=0}^{n=1} \frac{a_k}{b^k} + \frac{a_n}{b^n} &\leq x \\[9pt] && \implies && \sum_{k=0}^{n-1} \frac{a_k}{b^k} + \frac{(b+r)}{b^n} &\leq x \\[9pt] && \implies && \sum_{k=0}^{n-1} \frac{(a_k + 1)}{b^k} + \frac{r}{b^n} &\leq x \\[9pt] && \implies && \sum_{k=0}^{n-1} \frac{(a_k+1)}{b^k} &\leq x. \end{align*}$

This contradicts that $a_{n-1}$ is the greatest integer such that

$\sum_{k=0}^{n-1} \frac{a_k}{b^k} \leq x. \qquad \blacksquare$

Stumbling Robot

Prove properties of a generalization of the decimal expansion of a number

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