# Equivalence Relation

1. Apr 2, 2009

### latentcorpse

I'm having a bit of difficulty in showing the following is an equivalence relation:

$z~w iff \frac{z}{w} \in \mathbb{R} \forall z,w \in \mathbb{C}*=\mathbb{C} \backslash \{0\]$

clearly z~z is ok as 1 is real

then i considered $\frac{z}{y} \in \mathbb{R}$ and tried to show y divided by z is real. i decided to multiply through by $\frac{\bar{y}}{\bar{y}}$ to get $\frac{z \bar{y}}{|y|^2} \in \mathbb{R}$
then clearly $z \bar{y} \in \mathbb{R}$
and dividing by $|z|^2$ we get $\frac{\bar{y}}{\bar{z}} \in \mathbb{R}$
but now what?

2. Apr 3, 2009

### xepma

1) If a number $$w$$ is real, what does that tell you about the inverse $$w^{-1}$$? (remember: $$w * w^{-1} = 1$$)

2) For the final step, use the fact that you can write a fraction as: $$\frac{a}{b} = \frac{a}{c}\frac{c}{b}$$

3. Apr 3, 2009

### HallsofIvy

Staff Emeritus
(z/w)(w/z)= 1. Therefore, w/z is the multiplicative inverse of z/w. But z/w is real so its multiplicative inverse is ....

4. Apr 3, 2009

### latentcorpse

erm.. just w/z???

5. Apr 3, 2009

### HallsofIvy

Staff Emeritus
Yes, but what I was looking for is "since z/w is real, its multiplicative inverse, w/z, is real". That was what xepema meant.