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Let n $\in \mathbb{N}$ and $\sim_n \subseteq \mathbb{R} \times \mathbb{R}$ be defined by:
$x \sim_n y\Leftrightarrow_{df} x^n - y^n = nx - ny$
Show that $\sim_n$ is an equivalence relation.

And here is my take on it (I apologize in advance for any notation mistakes, I'm still not quite good at it.):

Proof Reflexivity:
Let $x \in \mathbb{R}$ and $n \in \mathbb{N}$; to show: $x \sim_n x$
$\begin{align}x \sim_n x &\Leftrightarrow_{df} x^n - x^n = nx - nx\\ &\Leftrightarrow 0 = 0\end{align} \implies \sim_n$ is reflexive

Proof Symmetry:
Let $x \in \mathbb{R}$ and $n \in \mathbb{N}$; to show: $x \sim_n y \implies y \sim_n x$
$\begin{align}x \sim_n y &\Leftrightarrow_{df} x^n - y^n = nx - ny\\ y \sim_n x &\Leftrightarrow -(x^n - y^n) = -nx + ny\end{align} \implies y\nsim x$
So, the assumption made is wrong, that is, the relation is not symmetric, and therefore not an equivalence relation.

However, I'm unsure whether my proof is valid or not, as the wording of the question, and the following question, that requires ~ to be an equivalence relation, makes me believe it is in fact an equivalence relation.

pportela
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    You have to show transitivity as well. Moreover, your proof of symmetry is also incomplete. – Aryaman Maithani Nov 10 '21 at 21:56
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    @AryamanMaithani Well, I didn't show transitivity because it must meet all three requirements, so I thought if it didn't meet one, then the proposition does not stand. And thanks for the heads up, I didn't realize that. – pportela Nov 11 '21 at 02:32
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    Follows immediately via the kernel criterion in the linked dupe, with $,f(x) = x^n-nx\ \ $ – Bill Dubuque Dec 06 '21 at 12:53
  • That's not an exact duplicate. In fact, @Bill, you are presuming the user has covered "the kernal criterion"? In discrete math?? – amWhy Dec 06 '21 at 15:31
  • @amWhy I see no claim that it is an "exact" dupe. But it is an abstract dupe. The linked answer contains a full proof and exposition of this criterion. This proof is quite elementary and no more difficult than any direct proof (it essentially just organizes the common structure behind direct proofs of this form). This is what the answer below by A.M. is also hinting at and the OP says he will try doing it that way - so it does not seem beyond their reach. – Bill Dubuque Dec 06 '21 at 15:47

1 Answers1

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Here's a general fact you can show:

Let $f : A \to B$ be a function and define $\sim_f$ on $A$ by $$x \sim_f y \Leftrightarrow f(x) = f(y).$$

I encourage you to prove this is always an equivalence relation on your own and let me know if you face any issues with this.

Once you have done that, note that in your question, the relation is defined by $$x \sim_n y \Leftrightarrow x^n - y^n = nx - ny.$$ Note that the last equation can be rewritten as $$x^n - nx = y^n - ny.$$ Thus, defining $f_n : \Bbb R \to \Bbb R$ by $f_n(x) = x^n - nx$ and using the earlier result finishes the job.

Berci
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Aryaman Maithani
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