Question 2
Let $a$ and $b$ be the bases of two number systems and let
$$A_n=\overline{x_1x_2\ldots x_n}^{(a)},\quad A_{n+1}=\overline{x_0x_1x_2\ldots x_n}^{(a)},$$
$$B_n=\overline{x_1x_2\ldots x_n}^{(b)},\quad B_{n+1}=\overline{x_0x_1x_2\ldots x_n}^{(b)},$$
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 $a\gt b$ if and only if
$$\frac{A_n}{A_{n+1}}\lt \frac{B_n}{B_{n+1}}.$$
$$A_n=\overline{x_1x_2\ldots x_n}^{(a)},\quad A_{n+1}=\overline{x_0x_1x_2\ldots x_n}^{(a)},$$
$$B_n=\overline{x_1x_2\ldots x_n}^{(b)},\quad B_{n+1}=\overline{x_0x_1x_2\ldots x_n}^{(b)},$$
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 $a\gt b$ if and only if
$$\frac{A_n}{A_{n+1}}\lt \frac{B_n}{B_{n+1}}.$$