# Simple Continued Fractions Question (I Think My Book Has a Mistake)

1. Apr 14, 2012

### tylerc1991

1. The problem statement, all variables and given/known data

Show that if the simple continued fraction expression of the rational number $\alpha$, $\alpha > 1$, is $[a_0; a_1, a_2, \dotsc, a_k]$, then the simple continued fraction expression of $\frac{1}{\alpha}$ is $[0; a_1, a_2, \dotsc, a_k]$.

2. Relevant equations

3. The attempt at a solution

Suppose $\alpha$ is a rational number greater than $1$ with simple continued fraction expression
$\alpha = [a_0; a_1, a_2, \dotsc, a_k] = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots a_{k - 1} + \cfrac{1}{a_k}}}}.$
Then we see that $\frac{1}{\alpha}$ is also a rational number with simple continued fraction expression
$\frac{1}{\alpha} = \cfrac{1}{a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{\ddots a_{k - 1} + \cfrac{1}{a_k}}}}} = [0; a_0, a_1, \dotsc, a_k].$

This is, of course, different than what the book says the solution should be. However, I don't see where I went wrong, if I did go wrong. Any help would be greatly appreciated!