# Prove that $R\subseteq X\times X$ is an equivalence relation and construct its equivalence class

Prove that relation $$R\subseteq X\times X$$, where $$X= \mathbb{R}\times\mathbb{R}$$, is an equivalence relation and construct its equivalence class. $$R$$ is defined as: $$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \Longleftrightarrow x_1^2+y_1^2 = x_2^2+y_2^2$$

The way I did prove that it is an equivalence relation was based on equality relationship between two or more quantities. To make it easier to read, we can also substitute $$a = x_1^2+y_1^2$$, $$b = x_2^2+y_2^2$$, $$c = x_3^2+y_3^2$$, at least I think we can.

The first property is reflexivity:
$$\langle x_1, y_1\rangle R \langle x_1, y_1\rangle \Longleftrightarrow x_1^2+y_1^2 = x_1^2+y_1^2$$, which is always true, as $$a=a$$.

Then we check for symmetry:
$$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \Longrightarrow \langle x_2, y_2\rangle R \langle x_1, y_1\rangle$$, so if $$x_1^2+y_1^2 = x_2^2+y_2^2$$, then $$x_2^2+y_2^2 = x_1^2+y_1^2$$. Using the substitution we can say that $$(a=b) \Rightarrow (b=a)$$, which is also true.

The last one is transitivity:
$$\langle x_1, y_1\rangle R \langle x_2, y_2\rangle \wedge \langle x_2, y_2\rangle R \langle x_3, y_3\rangle \Longrightarrow \langle x_1, y_1\rangle R \langle x_3, y_3\rangle$$, which means $$x_1^2+y_1^2 = x_2^2+y_2^2 \wedge x_2^2+y_2^2 = x_3^2+y_3^2 \Longrightarrow x_1^2+y_1^2 = x_3^2+y_3^2$$, using substitutions it is $$(a=b \wedge b=c) \Longrightarrow (a=c)$$.

Up to this moment I think that my line of thinking is quite correct, but I am not sure whether that proof is good enough and it is what I would like to know.

When it comes to equivalence class of $$R$$, I would say that the relation tells us about points on the common circle, but I have to write it up with the definition, which is:

Equivalence class of element $$a\in A$$ in regard to equivalence relation $$R\subseteq A\times A$$ is a set $$[a]_R = \{b\in A\: |\: a R b\}$$.

$$[\langle a,b \rangle]_R = \{\langle a, b\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$$ I would like you to tell me whether my proof (also usage of substitutions) and construction of equivalence class is right, also how could I get grasp of it, as lectures I attend are not good enough.

• You first part(the relation part) looks good, the second part it should be $[\langle a,b \rangle]_R = \{\langle x, y\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$: this is the set of all elements of $A\times A$ that are in relation with $\langle a,b\rangle$(here $A$ is indeed $\Bbb R\times \Bbb R$) – Holo Dec 16 '18 at 21:14

I like your proof, but I'd write the equivalence classes as follows:

$$[(a,b)]_R=\{(x,y)\in \mathbb{R}^2|a^2+b^2=x^2+y^2\}$$

which translates to: "The equivalence class of $$(a,b)$$ is the set of all points $$(x,y)$$ having the same distance from 0 as $$(a,b)$$.

Your idea is quite good and can be better formalized.

Consider a set $$X$$ and a map $$f\colon X\to Z$$, $$Z$$ any set. Then you can define a relation $$\sim_f$$ on $$X$$ by decreeing that $$x\sim_f y\quad\text{if and only if}\quad f(x)=f(y)$$ Then $$\sim_f$$ is clearly an equivalence relation: the proof is easily based on the properties of equality, which is essentially what you did just with more complicated symbols.

For $$a\in X$$, its equivalence class is $$[a]_{\sim_f}=\{x\in X:f(x)=f(a)\}$$.

In your case, $$X=\mathbb{R}\times\mathbb{R}$$, $$Z=\mathbb{R}$$ and $$f(x,y)=x^2+y^2$$ (square of the distance of $$(x,y)$$ from the origin).

Thus the equivalence class of $$(a,b)\in X$$ is the set of all points in $$X=\mathbb{R}\times\mathbb{R}$$ that share the same distance from the origin as $$(a,b)$$.

• Is it more correct to say "and a map $f\colon X\to Z$, where $Z$ is any set"? – manooooh May 23 at 19:14
• @manooooh As correct. – egreg May 23 at 19:32