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Commentary on Proposition IX.13¶

If $1, a, a_2, \dots a_n$ be a geometrical progression, and if $a$ is prime, $a_n$ will not be measured by any numbers except the preceding terms of the series.

If possible, let $a_n$ be measured by $b$, a number different from all the preceding terms.

Now $b$ cannot be prime, for, if it were, it would measure $a$. [IX. 12]

Therefore $b$ is composite, and hence will be measured by some prime number [VII. 31], say $p$.

Thus $p$ must measure $a_n$ and therefore $a$ [IX. 12]; so that $p$ cannot be different from $a$, and $b$ is not measured by any prime number except $a$.

Suppose that

$$a_n = b \cdot c.$$

Now $c$ cannot be identical with any of the terms $a, a_2, \dots a_{n-1}$; for, if it were, $b$ would be identical with another of them : [IX. 11]

which is contrary to the hypothesis.

We can now prove (just as for $b$) that $c$ cannot be prime and cannot be measured by any prime number except $a$.

Since

$$b \cdot c = a_n = a \cdot a_{n-1}, \quad \text{[IX. 11]}$$

$$a : b = c : a_{n-1},$$

whence, since $a$ measures $b$,

$$c \text{ measures } a_{n-1}.$$

Let

$$a_{n-1} = c \cdot d.$$

We now prove in the same way that $d$ is not identical with any of the terms $a, a_2, \dots a_{n-2}$, is not prime, and is not measured by any prime except $a$, and also that

$$d \text{ measures } a_{n-2}.$$

Proceeding in this way, we get a last factor, say $k$, which measures $a$ though different from it : which is absurd, since $a$ is prime.

Thus the original supposition that $a_n$ can be measured by a number $b$ different from all the terms $a, a_2, \dots a_{n-1}$ must be incorrect.

Therefore etc.