Equivalence relation of complex x,y, where rx = y I came up with an exercise telling to check if a relation is an equivalence relation. I would appreciate help with determining if my thoughts are correct.
The relation is:
$$\forall (x,y) \in \mathbb{C}(xRy \iff \exists r \in \mathbb{R} - \{0\}: rx=y)$$
The things that define equivalence relation are:


*

*Reflexivity.

*Symmetry.

*Transitivity.


So, to decide what type or relation is it:


*

*$$xRx ?: rx = x \rightarrow r = 1, r \in \mathbb{R} - \{0\} $$
Which seems all right - the reflexivity is preserved.

*$$xRy \rightarrow yRx?$$
Here I took two random complex numbers, let's say:
$$x=1+2i, y=2+3i.$$
Then I attempted to do the standard calculations putting these values in 2.
$$r_1(1+2i) = (2+3i) \rightarrow r_2(2+3i) = (1+2i)$$
Then I get the r1 and r2 values:
$$r_1=\frac{8-i}{9}, r_2=\frac{8+i}{13}$$
$$r_1 \ne r_2; r_1, r_2 \notin \mathbb{R}-\{0\}$$
Hence the relation is not an equivalence relation.(EDIT) Please verify if it makes sense or not. Thanks for all help.


EDIT: According to your comments, picking "two random complex numbers" couldn't work. I thought about other explanation; can I go this way?
$$xRy \rightarrow yRx?$$
I can go with:
$$rx=y \rightarrow ry=x$$
I take x from the first equation and put in in the second one:
$$ry=\frac{y}{r} \rightarrow r^2 = 1 \rightarrow r = -1 \vee 1$$
These numbers are in the given domain, so it is preserved as well.
$$xRy \wedge yRz \rightarrow xRz$$
We can take r from the first relation and z from the second and try putting it in the third, getting:
$$\frac{y}{x}x = ry \rightarrow r = 1$$
It is in the R without 0, so this is also preserved.
 A: You misunderstood the symmetry. It say: If $xRy$, then $yRx$. But in your example $1+2i$ and $2+3i$ have no relation! So you start:
Let be $x,y\in\mathbb C$ such that $xRy$. By definition there exists $r\in\mathbb R\setminus \{0\}$ such that $rx=y$.
Now you have to deduce that there exists $\tilde r\in\mathbb R\setminus\{0\}$ such that $\tilde ry=x$ and you get $yRx$.
A: Note that : $$ xRy \implies xr = y \equiv x = \frac{1}{r}y \equiv \alpha y = x$$
Hence, $yRx$. You have taken a $(x,y)$ which have no relation between them.
In a similar vein: $$x R y \, \text{ and } \, y R z \implies x = \frac{1}{r} y \, \text{ and } \, y = \frac{1}{r}z \implies x = \frac{1}{r^2} z \equiv \beta x = z $$ where $\beta = r^2$.
Hence, the relation is an equivalence relation.
A: Your $x$ and $y$ in the check for symmetry are not in relation, so the calculations you make are kind of pointless.
With word, symmetry means that is $x$ is related to $y$, $y$ has to be related to $x$. that is not hard to prove in your case.
You don't address transitivity at all, but that is also easy to verify for your relation, so your relation is in fact an equivalence relation, but I'm leaving the (easy) calculations to you.
A: Note. The symmetry condition say that IF $xRy$ then $yRx$. In your example with $x=1+2i$ and $y=2+3i$ we do not have $xRy$, thus this does not disprove symmetry. Instead, as Mauro comments, if you have $xRy$ then $rx=y$, thus $\frac{1}{r}y=x$ and hence $yRx$.
Now you just need to prove the transitivity, and then you are done.
