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If we have a relation $\sim$ on $\mathbb{Z}/6\mathbb{Z}\times (\mathbb{Z}/6\mathbb{Z}\setminus\{0\})$ so that $(w,x)\sim(y,z)$ if $wz=xy$, how is $\sim$ not an equivalence relation?

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Obviously $(w,x)\sim(w,x)$ because $wx=wx$. Next, if $(w,x)\sim(y,z)$, we see that $wx=yz$ and so $zy=xw$ and we see that $(y,z)\sim(w,x)$. Finally, if $(w,x)\sim (y,z)$ and $(y,z)\sim(a,b)$ then (after some work) $wzb=xza$ or $wb=xa$ which means that $(w,x)\sim(a,b)$. But I am told this is not an equivalence relation. Any help? – johnnymath Nov 17 '11 at 8:48
What do you mean by $wz=xy$? – user7530 Nov 17 '11 at 8:50
Think about involving divisors of zero – marwalix Nov 17 '11 at 8:54
To give a counterexample proving that the relation is not transitive – marwalix Nov 17 '11 at 8:57

In your comment (which you could preferably have added to the question), you cancel $z$ on both sides of an equation. This is not valid in $\mathbb Z/6\mathbb Z$, since it contains zero divisors. Thus, if you have $wb=2$ and $xa=4$, then with $z=3$ you get $wzb=xza$ despite $wb\ne xa$. Thus for instance $(1,1)\sim(3,3)$ and $(3,3)\sim(2,4)$, but $(1,1)\nsim(2,4)$.

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