Equivalence of Two Different Irrational Numbers 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.
 A: 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.
A: Proceed by induction on the number of terms after which they agree. 
Base Case: Agree after $0$ terms. Then $\alpha = \beta = [c_1, c_2, \cdots ] $ and so $\alpha $ is equivalent to $ \beta$ (by your previous question). 
Induction hypothesis: Suppose that if $ \alpha = [a_1, a_2, \cdots, a_n, c_1, c_2, \cdots] $ and $ \beta = [b_1, b_2, \cdots, b_n, c_1, c_2, \cdots ]$ then $\beta = \frac{ a \alpha + b}{c\alpha + d}$ for some integers $a,b,c,d$ such that $ad-bc = \pm 1.$
Induction step: 
Let $ \alpha = [a_0, a_1, a_2, \cdots, a_n, c_1, c_2, \cdots] $ and $ \beta = [b_0, b_1, b_2, \cdots, b_n, c_1, c_2, \cdots ].$
Then $\alpha = a_0 + \frac{1}{\alpha'} $ and $\beta = b_0 + \frac{1}{\beta'}$ where (by the Induction hypothesis) $\alpha'$ is equivalent to $\beta'.$ 
Thus $ \frac{1}{\alpha - a_0} $ is equivalent to $ \frac{1}{\beta - b_0} $ so there exists $a,b,c,d$ with $ad-bc=\pm 1$ such that $$ \frac{1}{\alpha -a_0} = \frac{ \frac{a}{\beta - b_0} + b}{ \frac{c}{\beta- b_0} + d} .$$
Multiply top and bottom of the fraction by $\beta - b_0$: $$ \frac{1}{\alpha - a_0} = \frac{a+b\beta - bb_0}{c+d\beta-db_0} $$ so solving for $\alpha$: $$\alpha = \frac{c+d\beta-db_0}{a+b\beta-bb_0} + a_0 =  \frac{c+d\beta-db_0+ aa_0+a_0b\beta-a_0bb_0}{a+b\beta-bb_0} $$ 
$$= \frac{ (a_0b+d)\beta + (aa_0+a_0bb_0-db_0+c) }{b\beta + (a-bb_0)} .$$  
Now since $$(a-bb_0)(a_0b+d) - b(aa_0+a_0bb_0-db_0+c) = ad   - bc =\pm 1$$
we have that $\alpha$ is equivalent to $\beta$ as desired. 
A: I did what Thomas Andrews suggested and obtained this:
Let $x_1=[a_0;a_1,a_2,\dots]$ and $x_2=[a_1;a_2,a_3,\dots]$. Then
$$x_1=a_0+\frac{1}{x_2}=\frac{a_0x_2+1}{x_2}=\frac{a_0\cdot x_2+1}{1\cdot x_2+0}.$$
Therefore, $a=a_0$, $b=1$, $c=1$, and $d=0$;
$$ad-bc=a_0\cdot0-1\cdot1=-1;$$
and $x_1$ is equivalent to $x_2$.
Now it is just a matter of using this to finish up the proof.
