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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 $[a]_R = \{b\in A\: |\: a R b\}$.

Sadly I can not get it completely, but I thought about
$$[\langle a,b \rangle]_R = \{\langle a, b\rangle \in \mathbb{R} \times \mathbb{R}\:|\: a^2+b^2=x^2+y^2\}$$ 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.

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  • $\begingroup$ 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 $\Bbb R\times \Bbb R$) $\endgroup$ – Holo Dec 16 '18 at 21:14
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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 $(a,b)$.

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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 $(a,b)$.

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

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