1970 IMO 第 2 题

(ROM 1) ${ }^{\mathrm{IMO} 2}$ Let $a$ and $b$ be the bases of two number systems and let $$\begin{array}{ll} A_{n}={\overline{x_{1} x_{2} \ldots x_{n}}}^{(a)}, & A_{n+1}={\overline{x_{0} x_{1} x_{2} \ldots x_{n}}}^{(a)}, \\ B_{n}={\overline{x_{1} x_{2} \ldots x_{n}}}^{(b)}, & B_{n+1}={\overline{x_{0} x_{1} x_{2} \ldots x_{n}}}^{(b)}, \end{array}$$ be numbers in the number systems with respective bases $a$ and $b$, so that $x_{0}, x_{1}, x_{2}, \ldots, x_{n}$ denote digits in the number system with base $a$ as well as in the number system with base $b$. Suppose that neither $x_{0}$ nor $x_{1}$ is zero. Prove that $a>b$ if and only if $$\frac{A_{n}}{A_{n+1}}<\frac{B_{n}}{B_{n+1}} .$$

(ROM 1) ${ }^{\mathrm{IMO} 2}$ 设 $a$ 和 $b$ 为两个数字系统的基数，并设 $$\begin{array}{ll} A_{n}={\overline{x_{1} x_{2} \ldots x_{n}}}^{(a)}, & A_{n+1}={\overline{x_{0} x_{1} x_{2} \ldots x_{n}}}^{(a)}, \\ B_{n}={\overline{x_{1} x_{2} \ldots x_{n}}}^{(b)}, & B_{n+1}={\overline{x_{0} x_{1} x_{2} \ldots x_{n}}}^{(b)}, \end{array}$$分别为基数为 $a$ 和 $b$ 的数系中的数字，因此 $x_{0}、x_{1}、x_{2}、\ldots、x_{n}$ 表示基数为 $a$ 的数系以及基数为 $b$ 的数系中的数字。假设 $x_{0}$ 和 $x_{1}$ 都不为零。证明 $a>b$ 当且仅当 $$\frac{A_{n}}{A_{n+1}}<\frac{B_{n}}{B_{n+1}} 。$$