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Commentary on Proposition IX.12¶

If $1, a, a_2, \dots a_n$ be a geometrical progression, and $a_n$ be measured by any prime number $p$, $a$ will also be measured by $p$.

[Folio 0405 | p. 399]

For, if possible, suppose that $p$ does not measure $a$; then, $p$ being prime, $p, a$ are prime to one another. [VII. 29]

Suppose

$$a_n = m \cdot p.$$

Now

$$a_n = a \cdot a_{n-1}. \qquad \text{[IX. 11]}$$

Therefore

$$a \cdot a_{n-1} = m \cdot p,$$

and

$$a : p = m : a_{n-1}. \qquad \text{[VII. 19]}$$

Hence, $a, p$ being prime to one another,

$$p \text{ measures } a_{n-1}. \qquad \text{[VII. 20, 21]}$$

By a repetition of the same process, we can prove that $p$ measures $a_{n-2}$ and therefore $a_{n-3}$, and so on, and finally that $p$ measures $a$.

But, by hypothesis, $p$ does not measure $a$: which is impossible.

Hence $p, a$ are not prime to one another: therefore they have some common factor. [VII. Def. 14]

But $p$ is the only number which measures $p$; therefore $p$ measures $a$.

Heiberg remarks that, as, in the ἔκθεσις, Euclid sets himself to prove that E measures A, the words bracketed above are unnecessary and therefore perhaps interpolated.