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How do i prove that this is a equivalence relation

I'm aware this may get removed as a dupe but I can't comment on that post since I'm below 50 reputation. I think the answer the post got just highlighted the thing I don't understand about relations (It's not that the answer isn't clear)

Symmetry: $(x,y)\in R\implies y=x\times e^n$ for some integer $n$. But then $x=y\times e^{-n}$ and, of course, $-n$ is still an integer. Thus $(y,x)\in R$ as desired.

  • How did they specifically come to the conclusion that $(y,x)\in R$ from their workings out, which I understood. It feels like they assumed the conclusion instead of working towards it (This sounds like a shot on them but it's just me not understanding)

Transitivity: $(x,y)\,\&\,(y,z)\in R$ means that there are integers $n,m$ with $y=x\times e^n$ and $z=y\times e^m$. But in that case we have $z=x\times e^n\times e^m=x\times e^{n+m}$. And of course $n+m$ is still an integer, so we have $(x,z)\in R$.

  • I have the same confusion with this one as the previous one. I get all the steps behind it but how do you get $(x,z)\in R$ from such steps? For example does $(x,xe^n),(xe^n,xe^{n+m})$ imply $(x,z)\in R$?

I hope I can clear this confusion before it gets shut. I'm just really puzzled with the last steps. Can anyone help me please

Bill Cogn
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1 Answers1

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In the first case, it is proved that for some integer $k$ ($k=-n$) $x=ye^k,$ i.e. $(y,x)=(y,ye^k),$ which is exactly the definition of $(y,x)\in R.$

Similarly, in the second case, it is proved that $z=xe^k$ for some integer $k$ ($k=m+n$), i.e. $(x,z)=(x,xe^k),$ hence $(x,z)\in R.$

Anne Bauval
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    Right, I think I understand. Thanks! – Bill Cogn Mar 03 '23 at 00:28
  • Quick question. Given $x=ye^{-n}$ and $y=xe^n$, can I substitute these expressions in the already known pair $(x,y)$ in R, let n = 0 and get $(ye^0,xe^0) = (y,x)$ as required? – Bill Cogn Mar 03 '23 at 00:49