Commentary on Proposition X.88¶
Back to Proposition X.88.
Beginning with $\rho$, $k\rho$, as in x. 85, 86, we take numbers $m$, $n$ such that $(m+n)$ has not to either of the numbers $m$, $n$ the ratio of a square number to a square number.
Take $x$ such that
$$(m+n) : n = k^2\rho^2 : x^2 \text{ ..............................(1),}$$
whence
$$x = k\rho \sqrt{\frac{n}{m+n}}$$
$$= \frac{k\rho}{\sqrt{1+\lambda}}\text{, say.}$$
Then shall $(k\rho - x)$, or $\left(k\rho - \frac{k\rho}{\sqrt{1+\lambda}}\right)$, be a fourth apotome.
For, by (1), $x$ is rational and $\sim k\rho$. Also $\sqrt{k^2\rho^2 - x^2}$ is incommensurable with $k\rho$, since
$$(m+n) : m = k^2\rho^2 : (k^2\rho^2 - x^2)\text{,}$$
and the ratio $(m+n) : m$ is not that of a square number to a square number. And $k\rho \sim \rho$. As explained in the note on x. 51, the fourth apotome
$$k\rho - \frac{k\rho}{\sqrt{1+\lambda}}$$
is the lesser root of the quadratic equation
$$x^2 - 2k\rho \cdot x + \frac{\lambda}{1+\lambda} k^2\rho^2 = 0.$$