If a number is that part of a number that a (part) taken away (is) of a (part) taken away then the remainder will also be the same part of the remainder that the whole (is) of the whole. * For let a number $AB$ be that part of a number $CD$ that a (part) taken away $AE$ (is) of a part taken away $CF$. * I say that the remainder $EB$ is also the same part of the remainder $FD$ that the whole $AB$ (is) of the whole $CD$.
This proposition states $$\begin{array}{rcl} \underbrace{n\cdot (AE+EB)}_{n\cdot AB}=\underbrace{n\cdot AB}_{CD}&=&\underbrace{n\cdot AE}_{CF}+\underbrace{n\cdot EB}_{FD}\\ &\Downarrow&\\ n\cdot EB&=&\underbrace{n\cdot (AB-AE)}_{=FD}. \end{array}$$ In particular, $$n\mid (CF+FD)\wedge n\mid CF\Rightarrow n\mid FD.$$
Proofs: 1
Proofs: 1
