I have a question from my book. The question is $ \mathbb{R^2} - (0,0)$, where $(a,b) \sim(c,d)$ if $ad-bc=0$. The question is to prove that it is equivalence relation.
I get to the transition property part, then the answer key shows me that $(a,b)\sim(c,d)$ and $(c,d)\sim(e,f)$, then $ad-bc=0$ and $cf-de=0$, so $adf-bde=d(af-be)=0$. I want to know how the book got $adf-bde=d(af-be)=0$ - what happened to $c$? I would be nice if you could show me how to find the equivalence class of this.
Well since, $ad-bc=0$ and $cf-de=0$ then $c=de/f$. Substituting that into the first equation gives:
$ad-b(de/f)=0$ and multiplying through by f gives $daf-bde=d(af-be)$
and its "transitive" property, not "transition"