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I've been looking at some equivalence relations and was wondering how to define an equivalence relation on $\mathbb{R}^2$ by $(w,y)\sim(x,z)$ if $(w,y)=(cx,cz)$.

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    @dls: Sorry; for some reason, I got my wires crossed and thought your comment was from the OP... And so it seemed in my head that he was almost there and was asking all the right questions, and I went ahead and wrote the answer. Sigh.2011-12-02

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So, we define $\sim$ as a relation on $\mathbb{R}^2$ by $(w,y)\sim (x,z)\iff \text{there exists a real number }c\text{ such that }(w,y)=(cx,cz).$

As stated, the relation is not an equivalence relation, because it is not symmetric.

For example, $(0,0)\sim(x,y)$ for every $(x,y)$, since we can take $c=0$, and then $(0,0)=(0x,0y)$.

However, if $(x,y)\sim(0,0)$, then $(x,y) = (c0,c0) = (0,0)$. So $(0,0)$ is related to everything, but the only thing related to $(0,0)$ is itself. E.g., $(0,0)\sim(1,1)$, but $(1,1)\not\sim(0,0)$. Thus, the relation is not symmetric, so it is not an equivalence relation.


There are two possible "fixes". One is exclude $c=0$. That is, define $(w,y)\sim (x,z)\iff \text{there exists a real number }c\neq 0\text{ such that }(w,y)=(cx,cz).$

We want to verify that this is an equivalence relation on $\mathbb{R}^2$. That means, verifying that the relation is Reflexive, Symmetric, and Transitive.

Reflexive. Let $(x,y)\in\mathbb{R}^2$. We need to show that $(x,y)\sim(x,y)$. This holds, because we can take $c=1$, and then $(x,y) = (1x,1y)$, so $(x,y)\sim(x,y)$. Thus, $\sim$ is a reflexive relation.

Symmetric. Let $(x,y),(z,w)\in\mathbb{R}^2$ and assume that $(x,y)\sim(z,w)$. Then there exists $c\neq 0$ such that $(x,y)=(cz,cw)$. Therefore, $x=cz$, $y=cw$. Hence, $z = \frac{1}{c}x$, $w=\frac{1}{c}w$. So, letting c'=\frac{1}{c} (possible since $c\neq 0$), we have that there exists c'\in\mathbb{R} such that (z,w)=(c'x,c'y). Hence $(z,w)\sim(x,y)$. So the relation is Symmetric.

Transitive. Assume that $(x,y)\sim (z,w)$ and $(z,w)\sim(r,s)$. Then there exist $c$ and $d$ in $\mathbb{R}$ such that $(x,y)=(cz,cw)$ and $(z,w)=(dr,ds)$. Let $e=cd$. Then $(er,es) = (c(dr),c(ds)) = (cz,cw) = (x,y)$; hence, $(x,y)\sim (r,s)$, so $\sim$ is transitive.

Since $\sim$ is reflexive, symmetric, and transitive, it follows that $\sim$ is an equivalence relation on $\mathbb{R}^2$.


A second possible fix is to change the domain from $\mathbb{R}^2$ to $\mathbb{R}^2-\{(0,0)\}$. Note that the proof of reflexivity and of transitivity above does not require $c$ to be nonzero in the definition: it's only symmetry where that comes into play. So let us prove that if $(x,y)\sim(z,w)$ and neither $(x,y)$ nor $(z,w)$ is $(0,0)$, then $(z,w)\sim(x,y)$.

Indeed, we know there is a $c$ such that $(x,y)=(cz,cw)$. Since $(x,y)$ is not $(0,0)$, either $x\neq 0$ or $y\neq 0$. Say $x\neq 0$; then $x=cz$, so $c\neq 0$ and $z\neq 0$. Since $c\neq 0$, we can now use the same argument as above to show that $(z,w)\sim(x,y)$. Symmetrically, if $y\neq 0$, then from $y=cw$ we conclude $c\neq 0$, so the argument above goes through.


Under the second fix, the partition induced on $\mathbb{R}$ consists of the lines through the origin with the origin removed.

Under the first fix, the partition induced on $\mathbb{R}$ is the same as above, but with a further equivalence class that contains only the origin.

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As indicated in the comments and Arturo's answer, this is almost an equivalence relation on the points of $\mathbb{R}^2$ whose equivalence classes are the lines through the origin; but the lines through the origin do not quite partition $\mathbb{R}^2$, because the origin belongs to every such line.

The most elegant fix (in my opinion) is to restrict $c$ to be positive. That is, let ${\bf x} ∼ {\bf y}$ if and only if ${\bf x} = c{\bf y}$ for some $c>0$. This does make $(\mathbb{R}^2, ∼)$ an equivalence relation: it is easily seen to be reflexive and transitive, and is symmetric because ${\bf x} = c{\bf y}$ implies ${\bf y} = c^{-1}{\bf x}$. Its equivalence classes are the open rays starting at the origin and the singleton set containing just the origin.