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Commentary on Proposition VII.20¶

If $a$, $b$ are the least numbers among those which have the same ratio (i.e. if $a/b$ is a fraction in its lowest terms), and $c$, $d$ are any others in the same ratio, i.e. if

$$a : b = c : d,$$

then $a = \frac{1}{n} c$ and $b = \frac{1}{n} d$, where $n$ is some integer.

The proof is by reductio ad absurdum, thus.

[Since $a < c$, $a$ is some proper fraction (“part” or “parts”) of $c$, by VII. 4.]

Now $a$ cannot be equal to $\frac{m}{n} c$, where $m$ is an integer less than $n$ but greater than 1.

For, if $a = \frac{m}{n} c$, $b = \frac{m}{n} d$ also. [VII. 13 and Def. 20]

Take each of the $m$ parts of $a$ with each of the $m$ parts of $b$, two and two; the ratio of the members of all pairs is the same ratio $\frac{1}{m} a : \frac{1}{m} b$.

Therefore

$$\frac{1}{m} a : \frac{1}{m} b = a : b. \qquad \text{[VII. 12]}$$

But $\frac{1}{m} a$ and $\frac{1}{m} b$ are respectively less than $a$, $b$ and they are in the same ratio : which contradicts the hypothesis.

[Folio 0328 | p. 322]

Hence $a$ can only be “a part” of $c$, or

$$a \text{ is of the form } \frac{1}{n} c,$$

and therefore

$$b \text{ is of the form } \frac{1}{n} d.$$

Here also Heiberg omits a proposition which was no doubt interpolated by Theon (B, V, p, $\phi$ have it as VII. 22, but P only has it in the margin and in a later hand; Campanus also omits it) proving for numbers the ex aequali proposition when “the proportion is perturbed,” i.e. (cf. enunciation of V. 22) if

$$\begin{aligned} a : b &= e : f, \dots\dots\dots\dots\dots\dots\dots\dots\dots(1) \\ \text{and}\qquad b : c &= d : e, \dots\dots\dots\dots\dots\dots\dots\dots\dots(2) \\ \text{then}\qquad a : c &= d : f. \end{aligned}$$

The proof (see Heiberg’s Appendix) depends on VII. 19. From (1) we have

$$af = be,$$

and from (2)

$$be = cd. \qquad \text{[VII. 19]}$$

Therefore

$$af = cd,$$

and accordingly

$$a : c = d : f. \qquad \text{[VII. 19]}$$