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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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Where $c$ is...? –  Zev Chonoles Dec 2 '11 at 3:05
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(a) Is $c$ supposed to be nonzero? (b) This already defines an equivalence relation. What exactly is your question at this stage? –  Srivatsan Dec 2 '11 at 3:06
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It's late but I don't think that's an equivalence relation - for instance does (1,1) ~ (1,1); that is (1,1) = (c,c)? Does (2,2) ~ (2,2) , ... ? If you say; two ordered pairs are equivalent if there exists such a c then this is good (the classes are lines of the plane). –  Adam Dec 2 '11 at 3:08
    
@Zev: $c$ is a real number. –  johnnymath Dec 2 '11 at 3:09
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@dls: The relation given is not an equivalence relation. –  Arturo Magidin Dec 2 '11 at 3:32
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2 Answers

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.

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