# 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?

-
Pick $a = d = 1$, $b = c = 0$. – Thomas Apr 18 '12 at 13:53
Note, in particular, that your usage of the quadratic formula fails in the case $c=0$. The $11$th commandment: Thou shalt not divide by zero. – Thomas Andrews Apr 18 '12 at 13:57

$$\alpha = \frac{ 1 \cdot \alpha + 0}{0\cdot \alpha + 1} .$$