# Define a relation

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

## The Attempt at a Solution

I said that the following set defines an equivalence class for the above problem:
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[(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.

Mentor
2022 Award
I thought about using the theorem of Thales.

Mathematicsresear
I thought about using the theorem of Thales.

Mentor
2022 Award
##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}?

Mentor
2022 Award
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.

Mentor
2022 Award
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.