If a number is those parts of a number that a (part) taken away (is) of a (part) taken away then the remainder will also be the same parts of the remainder that the whole (is) of the whole. * For let a number $AB$ be those parts 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 parts of the remainder $FD$ that the whole $AB$ (is) of the whole $CD$.
This proposition states (for integers $0 < r_1 < r_0 < AL$ and $0 < m < n$) $$\begin{array}{rclclc}CD&=&AB+r_0&=&n\cdot AL+r_0&\wedge\\ CF&=&AE+r_1&=&m\cdot AL+r_1\\ &\Downarrow&\\ CD-CF&=&(AB-AE)+(r_0-r_1)&=&(n-m)AL+(r_0-r_1) \end{array}$$
with $0 < (r_0-r_1) < AL.$ In particular,
$$AL\not\mid (CD-CF)\Rightarrow AL\not\mid CD\vee AL\not\mid CF.$$
Proofs: 1
Proofs: 1
