If a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively). * For let the magnitude $AB$ be the same multiple of the magnitude $CD$ that the (part) taken away $AE$ (is) of the (part) taken away $CF$ (respectively). * I say that the remainder $EB$ will also be the same multiple of the remainder $FD$ as that which the whole $AB$ (is) of the whole $CD$ (respectively).
If we are given two positive real numbers \(\alpha\), \(\beta\),^{1} and the following multiples of aliquot parts \(m\ge 1\), \(n\ge 1\): \[\alpha m=\beta m,\quad \alpha n=\beta n\quad\quad( * )\] then adding both equations gives us \[\alpha(m-n)=\beta m-\beta n.\]
See distributivity law for real numbers.
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. ↩