Commentary on Proposition VII.8

Back to Proposition VII.8.

[Folio 0314 | p. 308]

If

$$a = \frac{m}{n}b \text{ and } c = \frac{m}{n}d, \quad (m < n)$$

then

$$a - c = \frac{m}{n}(b - d).$$

Euclid’s proof amounts to the following.

Take $e$ equal to $\frac{1}{n}b$, and $f$ equal to $\frac{1}{n}d$.

Then since, by hypothesis, $b > d$,

$$e > f,$$

and, by VII. 7,

$$e - f = \frac{1}{n}(b - d).$$

[Folio 0315 | p. 309]

Repeat this for all the parts equal to $e$ and $f$ that there are in $a$, $b$ respectively, and we have, by addition ($a$, $b$ containing $m$ of such parts respectively),

$$m(e - f) = \frac{m}{n}(b - d).$$

But

$$m(e - f) = a - c.$$

Therefore

$$a - c = \frac{m}{n}(b - d).$$

The propositions VII. 7, 8 are complementary to V. 5 which gives the corresponding result with multiple in the place of “part” or “parts.”