I want to prove that the relation $\sim$ on fractions given by $\frac{a}{b} \sim \frac{c}{d}$ if $ad = cb$, where $a, c \in \mathbb Z$ and $b, d \in \mathbb Z_{> 0}$, is transitive.
(My last question was a similar sort of "is-this-a-valid-proof"-question; sorry if this is tiresome, but asking really, really turned out to help me, thanks to a very insightful user.) So is this a valid proof (:-))?
Assume that $\frac{a}{b} \sim \frac{c}{d}$ and that $\frac{c}{d} \sim \frac{e}{f}$. This means that $ad = cb$ and that $cf = de$. We want to show that $af = be$ from the two previous equalities: $\textbf{af = be} \iff a = \frac{be}{f} \iff \frac{cb}{d} = \frac{be}{f} \iff \frac{c}{d} = \frac{e}{f} \iff \textbf{cf = de} \iff \frac{c}{d} = \frac{e}{f} \iff \frac{cb}{d} = \frac{be}{f} \iff \frac{ad}{d} = \frac{be}{f} \iff ad = b(\frac{de}{f}) \iff \textbf{ad = bc}.$
I've seen this relation defined on pairs of integers $(a,b)$ when constructing the rational numbers from the integers. So, in this sense, maybe it's best to work only with properties of the integers (meaning not using fractions).
Supossing $a,c,e\neq 0$, since $ad=bc$ and $cf=de$ (and $b,d,f$ are already non-zero) we have
$acdf=bcde$
since we can cancel equal terms on both sides of the equation (making use of the cancellation law of the integers), we can cancel $cd$. Thus
$af=be$.