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
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 $(x,z)\in R.$