If two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples of them (respectively). * For let two magnitudes $AB$ and $CD$ be equal multiples of two magnitudes $E$ and $F$ (respectively). * And let the (parts) taken away (from the former) $AG$ and $CH$ be equal multiples of $E$ and $F$ (respectively). * I say that the remainders $GB$ and $HD$ are also either equal to $E$ and $F$ (respectively), or (are) equal multiples of them.
If we are given two positive real numbers \(\alpha\), \(\beta\),^{1} and the following multiples of aliquot parts \(m\ge 1\), \(n\ge 1\): \[m\alpha=m\beta,\quad n\alpha=n\beta\quad\quad( * )\] then adding both equations gives us \[(m-n)\alpha=m\beta-n\beta.\]
See distributivity law for real numbers.
Proofs: 1
Sections: 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. ↩