I am attempting to prove that if $x$ has a finite-length base $p$ decimal expansion, that it has precisely two base $p$ expansions. ($x = \frac{a_1}{p} + \dots +\frac{a_n}{p^n}$)
My attempt: x can be re-written as $x = 0.a_1 a_2 a_3 a_4\dots a_n$ (i.e. $0.12345$ where $a_1 = 1$ and so on). Let $y = 0.b_1 b_2 b_3.... b_n$, where $x \neq y$.
There must exist a $n$ s.t. $x_n \neq y_n$. Let $n$ be the smallest value at which $x$ and $y$ differ.
$$d(x,y) \ge \frac{1}{p^n}$$
Break up the decimal expansion of $x$ and $y$ into the first $n$ digits ($f$) and the remaining digits ($t$) s.t. $x = f_x t_x$ and $y= f_y t_y$.
Then $d(x,y) = (f_x - f_y) + (t_x - t_y)$
Therefore, $d(f_x, f_y) \ge \frac{1}{p^n}$.
That's where I'm stuck. Did I just prove that because $n$ is contained in the $f_x$ and $f_y$, that it is $\ge \frac{1}{p^n}$, which means that $x_n \neq y_n$?
Is there anything else I need to do?
