Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 6 down vote accepted

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

share|cite|improve this answer
I cannot believe I just asked such a silly question... thanks. – EsX_Raptor Apr 18 '12 at 13:55
@JosuéMolina Not silly at all, everyone stumbles on small things every now and then. – Ragib Zaman Apr 18 '12 at 13:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.