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Let R be the relation defined on the set of integer pairs by $(x_1,y_1)R(x_2,y_2)$ when $x_{1}^{2}$ + $y_{1}^{2}$ = $x_{2}^{2}$ + $y_{2}^{2}$.

Prove that R is an equivalence relation and determine the equivalence classes.

So to prove it's an equivalence relation

I said

Symmetric - Yes, because $x_{2}^{2}$ + $y_{2}^{2}$ = $x_{1}^{2}$ + $y_{1}^{2}$.

Reflexive - Yes, because $x_{1}^{2}$ + $y_{1}^{2}$ = $x_{1}^{2}$ + $y_{1}^{2}$.

Transitive = Yes, because

$x_{1}^{2}$ + $y_{1}^{2}$ = $x_{2}^{2}$ + $y_{2}^{2}$.

$x_{2}^{2}$ + $y_{2}^{2}$ = $x_{3}^{2}$ + $y_{3}^{2}$.

$x_{1}^{2}$ + $y_{1}^{2}$ = $x_{3}^{2}$ + $y_{3}^{2}$.

I think this is enough to prove it's an equivalence relation.

Now I am new to this, and my understanding of equivalence classes is basically list all the things that satisfy this relation.

I'm not sure if that's correct, but here is what I got from that

$$[x,y] =\{(x,y),(x,-y),(-x,y),(x,y)\}$$

is this correct, are there alternate ways?

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    $\begingroup$ Note that e.g. $(4,3)\sim (5,0)$ $\endgroup$ – Hagen von Eitzen Apr 25 at 17:59
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    $\begingroup$ Hint: Consider the circle with center at origin. $\endgroup$ – Thomas Shelby Apr 25 at 18:05
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Your proof is rather sloppy. To prove, for example, that the relation is symmetric, you should prove that $$\forall (x_1,y_1),(x_2,y_2)\in\Bbb{Z}^2:\qquad (x_1,y_1)R(x_2,y_2)\quad\implies\quad(x_2,y_2)R(x_1,y_1).$$ Similarly, to prove reflexivity, you should prove that $$\forall (x_1,y_1)\in\Bbb{Z}^2:\qquad (x_1,y_1)R(x_1,y_1),$$ and something similar for transitivity.

Your characterization of the equivalence classes is incorrect. As noted in the comments $(4,3)R(5,0)$, for example. In stead, consider the fact that for every $(x_i,y_i)\in\Bbb{Z}^2$ we can set $c_i:=x_i^2+y_i^2$, and then for all $(x,y)\in[(x_i,y_i)]$ we have $$x^2+y^2=x_i^2+y_i^2=c_i.$$ Can you tell what these sets (these equivalence classes) look like geometrically?

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  • $\begingroup$ Sorry so I'm a real beginner at this, so starting with the proofs. I understand what your trying to say, but for the symmetric proof it seems obvious to me that if things equal on another flipping them would still keep them equal. How can I prove this more thoroughly? Should it be something like for all $\forall (x_1,y_1),(x_2,y_2)\in\Bbb{Z}^2$ if $x_1^2+y_1^2 =x_2^2+y_2^2$ then $x_2^2+y_2^2 = x_1^2+y_1^2$. I feel like this is just restating what I have in fancier terms, and not sure how to do it properly. $\endgroup$ – Brownie Apr 25 at 18:22
  • $\begingroup$ As for how the equivalence class will look like geometrically, going off of Thomas Shelby's answer, $x^{2} +y^{2}$ is used to graph a circle. So if I'm understanding this correctly, if every point on the edge of the circle has multiple other points its related to, then is the circumference of the circle my equivalence set? $\endgroup$ – Brownie Apr 25 at 18:36
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    $\begingroup$ @Brownie Your first comment is spot on; it is rather trivial to prove that the relation is symmetric, and your suggested proof here is perfectly fine. But just stating that $x_2^2+y_2^2=x_1^2+y_1^2$ without any context, as you did originally, is not a proof. $\endgroup$ – Servaes Apr 25 at 20:14
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    $\begingroup$ @Brownie Your second comment is unclear to me. The equation shows that two points are in the same equivalence class precisely when they are on the same circle centered at the origin. $\endgroup$ – Servaes Apr 25 at 20:16
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More generally, if $f\colon X\to Y$ is a map, then $$a\operatorname{R}b:\Leftrightarrow f(a)=f(b) $$ is always an equivalence relation. This is readily proved and after that comes in handy in mayn situations of life. Moreover, we can loosely identify the set of equivalence classes with the image of $f$.

Here $f\colon \Bbb Z^2\to \Bbb N_0$ is given by $(x_1,x_2)\mapsto x_1^2+x_2^2$, and with some non-trivial results from number theory, $\operatorname{im}f$ contains $0$ and all naturals $n$ such that for all primes $p$ with $p\equiv 3\pmod4$, the highest power of $p$ dividing $n$ is of even exponent.

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