If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and equal multiples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal multiples of the second (magnitude) and the fourth, respectively. * For let a first (magnitude) $A$ and a third $C$ be equal multiples of a second $B$ and a fourth $D$ (respectively), and let the equal multiples $EF$ and $GH$ have been taken of $A$ and $C$ (respectively). * I say that $EF$ and $GH$ are equal multiples of $B$ and $D$ (respectively).
If we are given two^{1} positive real numbers \(\alpha\), \(\beta\), and the following multiples of aliquot parts \(m\ge 1\), \(n\ge 1\): \[m\alpha=m\beta,\quad n(m\alpha)=n(m\beta)\quad\quad( * )\] then it follows \[(nm)\alpha=(nm)\beta.\]
See multiplication of real numbers is associative.
Proofs: 1
From a geometrical point of view, \(\alpha,\beta\) could e.g. mean the lengths of some segments, the areas of some plane figures or the volumes of some solids. ↩