If a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever. * For let a first (magnitude) $A$ have the same ratio to a second $B$ that a third $C$ (has) to a fourth $D$. * And let equal multiples $E$ and $F$ have been taken of $A$ and $C$ (respectively), and other random equal multiples $G$ and $H$ of $B$ and $D$ (respectively). * I say that as $E$ (is) to $G$, so $F$ (is) to $H$.
Given four positive real numbers \(\alpha,\beta,\gamma,\delta\) with the equation of ratios. \[\frac \alpha\beta=\frac \gamma\delta,\quad\quad( * )\]
the ratios of the corresponding multiples of aliquot parts \(m\ge 1\), \(n\ge 1\) will also be equal: \[\frac{m\alpha}{n\beta}=\frac{m\gamma}{n\delta}.\]
In other words, the equation \((*)\) remains valid if we multiply it by any positive rational number \(\frac mn\).
See multiplication of real numbers is cancellative.
Proofs: 1