Continued fraction expansion of $1-x$ for $x\in [0,1[$.

Question

If $\alpha$ is a real number in $[0,1[$ that can be written as $[a_0;a_1,a_2,\ldots]$, what can I say about the continued fraction expansion of $1-\alpha$?

Answer 1

If $\alpha$ is irrational, then $$1 - \alpha = \begin{cases} [0; 1, a_1 - 1, a_2, a_3, \ldots] & \text{if $a_1 > 1$}\\ [0; a_2 + 1, a_3, a_4, \ldots] & \text{if $a_1 = 1$} \end{cases}$$ To verify, first figure out which $\alpha$ have $a_1 > 1$ and which have $a_1 = 1$. Then find each partial quotient one at time in order using the recursive formula for $a_i$ and use what you have to get the next partial quotient. I worked backwards from the known answer to discover the algebraic manipulations. You will also find that once you "hit" the partial quotient $a_2$ in the first case and $a_3$ in the second, the remaining partial quotients follow in order.

If $\alpha$ is rational, there are more cases to work out.