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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

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

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