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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:

  1. Reflexivity.
  2. Symmetry.
  3. Transitivity.

So, to decide what type or relation is it:

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

Sqoshu
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4 Answers4

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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 $yRx$.

Mundron Schmidt
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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.

  • Thanks for help, this seems reasonable. I've came with another way to prove the equivalence, I think. Can you see if things I did in the EDIT are correct? – Sqoshu Dec 21 '17 at 09:15
  • @Sqoshu What you are doing is that you are solving the equation. That is not required. Note that $ xr = y \implies x = \frac{1}{r}y$. Now, think of this $\frac{1}{r}$ as another constant, say, $\alpha$ because $r$ is also a constant. Thus, $\beta y = r$ holds which is exactly what we are supposed to prove. So, the relation is symmetric. –  Dec 21 '17 at 09:20
  • @Sqoshu Your working for the transitive part is correct! –  Dec 21 '17 at 09:20
  • Thank you for the contribution. If the relation is equivalence relation, then, can you give me any hints on how I define equivalence classes in this situation? – Sqoshu Dec 21 '17 at 09:28
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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.

Henrik supports the community
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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.

Ove Ahlman
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