Ytukyg

Prove that relation $Rsubseteq Xtimes X$, where $X= mathbb{R}timesmathbb{R}$, is an equivalence relation and construct its equivalence class. $R$ is defined as:
$$langle x_1, y_1rangle R langle x_2, y_2rangle Longleftrightarrow x_1^2+y_1^2 = x_2^2+y_2^2$$

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:
$langle x_1, y_1rangle R langle x_1, y_1rangle Longleftrightarrow x_1^2+y_1^2 = x_1^2+y_1^2$, which is always true, as $a=a$.

Then we check for symmetry:
$langle x_1, y_1rangle R langle x_2, y_2rangle Longrightarrow langle x_2, y_2rangle R langle x_1, y_1rangle$, 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:
$langle x_1, y_1rangle R langle x_2, y_2rangle wedge langle x_2, y_2rangle R langle x_3, y_3rangle Longrightarrow langle x_1, y_1rangle R langle x_3, y_3rangle$, 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 $ain A$ in regard to equivalence
relation $Rsubseteq Atimes A$ is a set $[a]_R = {bin A: |: a R b}$.

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

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 $fcolon Xto Z$, $Z$ any set. Then you can define a relation $sim_f$ on $X$ by decreeing that
$$
xsim_f yquadtext{if and only if}quad f(x)=f(y)
$$
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 $ain X$, its equivalence class is $[a]_{sim_f}={xin X:f(x)=f(a)}$.

In your case, $X=mathbb{R}timesmathbb{R}$, $Z=mathbb{R}$ 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 $X=mathbb{R}timesmathbb{R}$ that share the same distance from the origin as $(a,b)$.