Commentary on Proposition IX.11

Back to Proposition IX.11.

[Folio 0402 | p. 396]

The proposition and the porism together assert that, if $1, a, a_2, \dots a_n$ be a geometrical progression, $a_r$ measures $a_n$ and gives the quotient $a_{n-r}\ (r < n)$.

Euclid only proves that $a_n = a \cdot a_{n-1}$, as follows.

$$1 : a = a_{n-1} : a_n.$$

Therefore $1$ measures $a$ the same number of times as $a_{n-1}$ measures $a_n.$ Hence $1$ measures $a_{n-1}$ the same number of times as $a$ measures $a_n;$ [VII. 15]

that is,

$$a_n = a \cdot a_{n-1}.$$

We can supply the proof of the porism as follows.

$$\begin{aligned} 1 : a &= a_r : a_{r+1}, \\ a : a_2 &= a_{r+1} : a_{r+2}, \\ &\dots\dots\dots\dots\dots\dots \\ a_{n-r-1} : a_{n-r} &= a_{n-1} : a_n, \end{aligned}$$

whence, ex aequali,

$$1 : a_{n-r} = a_r : a_n.\text{ \hfill [VII. 14]}$$

It follows, by the same argument as before, that

$$a_n = a_r \cdot a_{n-r}.$$

With our notation, we have the theorem of indices that

$$a^{m+n} = a^m \cdot a^n.$$