Commentary on Proposition V.17

Back to Proposition V.17.

$$\begin{aligned} \text{Algebraically, if} \qquad a : b &= c : d, \\ \text{then} \qquad (a - b) : b &= (c - d) : d. \end{aligned}$$

I have already noted the somewhat strange use of the participles of συγκεῖσθαι and διαιρεῖσθαι to convey the sense of the technical σύνθεσις and διαίρεσις λόγου, or what we denote by componendo and separando. ἐὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται is, literally, “if magnitudes compounded be proportional, they will also be proportional separated,” by which is meant “if one magnitude made up of two parts is to one of its parts as another magnitude made up of two parts is to one of its parts, the remainder of the first whole is to the part of it first taken as the remainder of the second whole is to the part of it first taken.” In the algebraical formula above $a, c$ are the wholes and $b, a - b$ and $d, c - d$ are the parts and remainders respectively. The formula might also be stated thus:

$$\begin{aligned} \text{If} \qquad a + b : b &= c + d : d, \\ \text{then} \qquad a : b &= c : d, \end{aligned}$$

in which case $a + b, c + d$ are the wholes and $b, a$ and $d, c$ the parts and remainders respectively. Looking at the last formula, we observe that “separated,” διαιρεθέντα, is used with reference not to the magnitudes $a, b, c, d$ but to the compounded magnitudes $a + b, b, c + d, d$.

As the proof is somewhat long, it will be useful to give a conspectus of it in the more symbolical form. To avoid minuses, we will take for the hypothesis the form

$$a + b \text{ is to } b \text{ as } c + d \text{ is to } d.$$

Take any equimultiples of the four magnitudes $a, b, c, d$, viz.

$$ma, \quad mb, \quad mc, \quad md,$$

and any other equimultiples of the consequents, viz.

$$nb \text{ and } nd.$$

Then, by V. 1, $m(a + b), m(c + d)$ are equimultiples of $a + b, c + d$, and, by V. 2, $(m + n)b, (m + n)d$ are equimultiples of $b, d$.

Therefore, by Def. 5, since $a + b$ is to $b$ as $c + d$ is to $d$, according as $m(a + b) > = < (m + n)b, \quad m(c + d) > = < (m + n)d$.

Subtract from $m(a + b), (m + n)b$ the common part $mb$, and from $m(c + d), (m + n)d$ the common part $md$; and we have, according as $ma > = < nb, \quad mc > = < nd$.

But $ma, mc$ are any equimultiples of $a, c$, and $nb, nd$ any equimultiples of $b, d$, therefore, by V. Def. 5,

$$a \text{ is to } b \text{ as } c \text{ is to } d.$$

Smith and Bryant’s proof follows, mutatis mutandis, their alternative proof of the next proposition (see pp. 173—4 below).