Find All Equivalence Classes for Relation R⊆ℝ2

[Mathematicsresear]

Homework Statement

a relation R⊆ℝ²
It is defined if and only if a²+b²=c²+d² where (a,b) ∧ (c,d)∈ℝ²
Find all equivalence classes

Homework Equations

The Attempt at a Solution

I said that the following set defines an equivalence class for the above problem:
[/B]
[(a,b)] = {(c,d)∈ℝ² : ((a,b),(c,d))∈R}⊆ℝ²

so I asked myself when is a²+b²=c²+d² such that (a,b) and (c,d) are elements of the relation and a subset of ℝ².

I said when d= sqrt(a²+b²-c²) and a²+b² > c²

I am not sure what to do next.

 

[fresh_42]

I thought about using the theorem of Thales.

 

[Mathematicsresear]

I thought about using the theorem of Thales.

Please elaborate

 

[fresh_42]

$a^2+b^2=h^2$ are the side lengths of a right triangle, and Thales together with the definition of your equivalence means: all points on the circle with diameter $h$ are equivalent. This reduces the problem to some signs and classes represented by a circle. So the solution has probably to do with concentric circles.

 

[Mathematicsresear]

$a^2+b^2=h^2$ are the side lengths of a right triangle, and Thales together with the definition of your equivalence means: all points on the circle with diameter $h$ are equivalent. This reduces the problem to some signs and classes represented by a circle. So the solution has probably to do with concentric circles.

Can't I say that the equivalence classes are:
[(a,b)] = {(c,sqrt(a²+b²-c²))∈ℝ² : a²+b² > c² }⊆ℝ²}?

 

[fresh_42]

Can't I say that the equivalence classes are:
[(a,b)] = {(c,sqrt(a²+b²-c²))∈ℝ² : a²+b² > c² }⊆ℝ²}?

If then $\pm \sqrt{}$, but are your classes distinct?

 

[Mathematicsresear]

If then $\pm \sqrt{}$, but are your classes distinct?

I want to describe all possible equivalence classes.

 

[fresh_42]

Given any equivalence relation, $\mathbb{R}^2$ can be written as a disjoint union of the classes. Can you prove this for your classes? In any case you will need the negative roots, too, as $(c,d) \sim (c,-d)$. Thus you will at least have $[(a,b)] = \{(c,\sqrt{\ldots}\,\vert \,\ldots\} \cup \{(c,-\sqrt{\ldots}\,\vert \,\ldots\}$. And what about $a^2+b^2 =c^2\,?$ You basically parameterized the classes by $h^2=a^2+b^2$ which is the same as my concentric circles. And in which class do you have $(0,0)\,?$ For this point there is no $c$.

In both cases, you will need a formal proof.