How can I show that $x_1=[a_0;a_1,a_2,\dots]$ is equivalent to $x_2=[a_1;a_2,a_3,\dots]$?

Question

anonymous · Answer

This is elementary number theory: continued fractions.

anonymous · Answer

With equivalence, I mean that there are integers $a$, $b$, $c$, and $d$ such that $ad-bc=\pm1$ and $x_1=\frac{ax_2+b}{cx_2+d}$.

anonymous · Answer

By the way, the notation here means$$x_1=a_0+\frac{1}{a_1+\frac{1}{\begin{matrix}
a_2+ &  & \ 
 & \ddots & \ 
 &  & 
\end{matrix}}}$$

experimentx · Answer

seems like x1 = a0 + 1/x2

across · Answer

@experimentX gave you a good clue:$$x_1=a_0+\frac{1}{x_2}=\frac{a_0\cdot x_2+1}{1\cdot x_2+0}.$$This implies that$$ad-bc=a_0\cdot0-1\cdot1=-1,$$and thus $x_1$ is equivalent to $x_2$.

experimentx · Answer

Oh! that wasn't a clue at all ... just an observation.