If $\alpha$ and $\beta$ are two real numbers, we say that $\beta$ is equivalent to $\alpha$ if there are integers $a$, $b$, $c$, and $d$ such that $ad-bc=\pm1$ and $\beta=\frac{a\alpha+b}{c\alpha+d}$.
How can I show that two irrational numbers $\alpha$ and $\beta$ are equivalent if the tails of their simple continued fractions agree, that is, if
$$\begin{align}
\alpha&=[a_0;a_1,a_2,\dots,a_j,c_1,c_2,c_3,\cdots]\\
\beta&=[b_0;b_1,b_2,\dots,b_k,c_1,c_2,c_3,\dots]
\end{align}$$
where $a_i$, $i=0,1,2,\dots,j$; $b_i$, $i=0,1,2,\dots,k$; and $c_1$, $i=1,2,3,\dots$ are integers, all positive except perhaps $a_0$ and $b_0$?
This is what I tried: I assumed the above and tried to show that there exist $a$, $b$, $c$, and $d$ such that $ad-bc=\pm1$ and
$$\begin{align}
\beta&=\frac{a\alpha+b}{c\alpha+d}\\
\Longrightarrow b_0+\frac{1}{b_1+\frac{1}{b_2+\cdots}}&=\frac{a\left(a_0+\frac{1}{a_1+\frac{1}{a_2+\cdots}}\right)+b}{c\left(a_0+\frac{1}{a_1+\frac{1}{a_2+\cdots}}\right)+d}
\end{align}$$
However, the operations on the fractions are confusing, and I have the feeling that algebraic manipulations will not help or that there may be an easier way to tackle this problem.
Answer
Hint $\ $ This - like many properties of continued fractions - is clearer when continued fractions are viewed as sequences of Möbius maps. For an introduction to the Möbius viewpoint see the following exposition: A. Beardon, Continued Fractions, Discrete Groups and Complex
Dynamics.
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