Commentary on Proposition IX.10¶
Back to Proposition IX.10.
If $1, a, a_2, a_3, a_4, \dots$ be a geometrical progression, then (1), if $a$ is not a square, none of the terms will be square except $a_2, a_4, a_6, \dots$ ; and (2), if $a$ is not a cube, none of the terms will be cube except $a_3, a_6, a_9, \dots$.
With reference to the first part of the proof, viz. that which proves that, if $a_3$ is a square, $a$ must be a square, Heiberg remarks that the words which I have bracketed are perhaps spurious ; for it is easier to use VIII. 24 than the converse of VIII. 26, and a use of VIII. 24 would correspond better to the use of VIII. 25 in the second part relating to cubes. I agree in this view and have bracketed the words accordingly. (See however note, p. 383, on converses of VIII. 26, 27 given by Heron.) If this change be made, the proof runs as follows.
If possible, let $a_3$ be square.
Now
$$a_2 : a_3 = a : a_2.$$
But $a_2$ is a square. [ix. 8]
Therefore $a$ is to $a_2$ in the ratio of a square to a square. And $a_2$ is square ; therefore $a$ is square [VIII. 24]: which is impossible.
If possible, let $a_4$ be a cube.
Now
$$a_3 : a_4 = a_2 : a_3.$$
And $a_3$ is a cube. [ix. 8]
Therefore $a_2$ is to $a_3$ in the ratio of a cube to a cube. And $a_3$ is a cube : therefore $a_2$ is a cube. [VIII. 25]
But, since
$$\begin{aligned} 1 : a &= a : a_2, \\ a_2 &= a^2. \end{aligned}$$
And, since $a^2$ is a cube, $a$ must be a cube [ix. 6]: which is impossible.
The propositions VIII. 24, 25 are here not quoted in their exact form in that the first and second squares, or cubes, change places. But there is no difficulty, since the method by which the theorems are proved shows that either inference is equally correct.