Show that every rational number has exactly two finite simple continued fraction expansions.
(Does this have something to do with how you handle the end of the continued fraction?)
$\displaystyle \displaystyle \frac{225}{157}$
This can be represented as $\displaystyle <b_0;b_1,b_2,\cdots, b_k> \ \mbox{and} \ <b_0;b_1,b_2,\cdots, b_k-1,1>$
Therefore, our fraction can be represented like
$\displaystyle \displaystyle <1; 2,3,4,5>$
$\displaystyle \displaystyle 1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{5}}}}$
or
$\displaystyle \displaystyle <1;2,3,4,4,1>$
$\displaystyle \displaystyle 1+\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{4+\fra c{1}{1}}}}}}$