If there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second). * Let there be any number of magnitudes whatsoever, $AB$, $CD$, (which are) equal multiples, respectively, of some (other) magnitudes, $E$, $F$, of equal number (to them). * I say that as many times as $AB$ is (divisible) by $E$, so many times will $AB$, $CD$ also be (divisible) by $E$, $F$.
If we are given the multiples of aliquot parts. \[k\alpha_1,k\alpha_2,\ldots,k\alpha_n\] i.e. where \(k\ge 2\) is a natural number and where \(\alpha_1,\alpha_2,\ldots,\alpha_n\) are some^{1} positive real numbers, then the sum of all multiples equals the multiple of the sum of all these magnitudes:
\[k\alpha_1+k\alpha_2+\ldots+k\alpha_n=k(\alpha_1+\alpha_2+\ldots+\alpha_n).\]
See general law of distributivity of real numbers.
Proofs: 1
From a geometrical point of view, \(\alpha_1,\alpha_2,\ldots,\alpha_n\) could e.g. mean the lengths of some segments, the areas of some plane figures or the volumes of some solids. ↩