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Let $r$ be a relation on $A \times A$ such that $(a,b) r (c,d) \iff ad = bc.$ How can I show that this relation is transitive, ie. $(a, b)r(c,d)$ and $(c,d)r(e, f) \implies (a,b)r(e,f)$?

I tried to say that $(a,b) r (c,d)$ means that $c=ka$ and $d=kb,$ $k$ a coefficient, such that $ad = bc \iff akb = bka,$ and going on from there, but I'm not sure this is valid for all values of $a, b, c,$ and $d$. How can I show the transitivity of the relation?

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If $ad=bc$ and $cf=de$, then $adcf=bcde$. If $dc\ne0$, $af=be$. This works for complex numbers, for example. Otherwise, use the relevant properties of the composition law on $A$. – Did Oct 30 '11 at 17:38
If $A$ is any subset of complex (including reals) containing $0$, then this relation is not transitive. Otherwise it is transitive. So, the answer depends on $A$. – Tapu Oct 30 '11 at 17:49
Both this and another question are special cases of a more general result:… – Martin Sleziak Oct 30 '11 at 17:59
up vote 1 down vote accepted

I think that the simplest method is straight-forward: $ad=bc$ and $cf = de$ implies $$ a(dc)f = b(cd)e \quad(1) $$ from which we need to obtain that $af = be$ which is iff $(a,b)r(e,f)$. The only case which is unclear is $dc=0$, otherwise we obtain $af=be$ just by dividing both sides in $(1)$ by $(dc)$.

It happens though that this case is crucial: indeed, if $c=d=0$ you can easily check that $(a,b)r(c,d)$ and $(c,d)r(e,f)$ for any $a,b,e,f$. So to find a counterexample you just take $$(a,b,e,f) = (1,2,3,4).$$

Indeed: $(1,2)r(0,0)$ since $0=0$ and $(3,4)r(0,0)$ by the same reason. On the other hand, $1\cdot4\neq 2\cdot3$ so it does not hold that $(1,2)r(3,4)$.

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I am writing the comment as a answer:

You have not specified $A$. The answer (and your reasoning) depends on $A$.
If A is any subset of complex numbers (including reals) containing $0$, then this relation is not transitive. Because in this case the element $(0,0)\in A\times A$ will spoil the transitivity. Otherwise it is transitive.

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