<s>[INST] Equivalence of a Real Number to Itself 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 a real number $\alpha$ is equivalent to itself?
What I have in mind is to assume that $\alpha=\frac{a\alpha+b}{c\alpha+d}$, and then show that it must be the case that $ad-bc=\pm1$:
$$\begin{align}
c\alpha^2+(d-a)\alpha-b&=0\\
\Longrightarrow\alpha&=\frac{a-d\pm\sqrt{(d-a)^2+4bc}}{2c}\\
&=\frac{a-d\pm\sqrt{d^2-2ad+a^2+4bc}}{2c}\\
&=\frac{a-d\pm\sqrt{-2(ad-2bc)+a^2+d^2}}{2c}\\
&=\frac{a-d\pm\sqrt{a^2+d^2+2bc\pm2}}{2c}
\end{align}$$
I do not know how to proceed. Am I on the correct track?
 [/INST] $$ \alpha = \frac{ 1 \cdot \alpha + 0}{0\cdot \alpha + 1} .$$
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