Commentary on Proposition X.85

Back to Proposition X.85.

[Folio 0183 | p. 179]

Take $k\rho$ commensurable in length with $\rho$, the given rational straight line. Let $m^2, n^2$ be square numbers such that $(m^2 - n^2)$ is not square. Take $x$ such that

$$m^2 : (m^2 - n^2) = k^2\rho^2 : x^2 \tag{1},$$

so that

$$x = k\rho \frac{\sqrt{m^2 - n^2}}{m}$$

$$= k\rho \sqrt{1 - \lambda^2}, \text{ say.}$$

Then shall $k\rho - x$, or $k\rho - k\rho\sqrt{1 - \lambda^2}$, be a first apotome.

For ($\alpha$) it follows from (1) that $x$ is rational but incommensurable with $k\rho$, whence $k\rho, x$ are rational and $\sim$, so that $(k\rho - x)$ is an apotome.

($\beta$) If $y^2 = k^2\rho^2 - x^2$, then, by (1), convertendo,

$$m^2 : n^2 = k^2\rho^2 : y^2,$$

whence $y$, that is, $\sqrt{k^2\rho^2 - x^2}$, is commensurable in length with $k\rho$.

And $k\rho \frown \rho$ ; therefore $k\rho - x$ is a first apotome.

As explained in the note to x. 48, the first apotome

$$k\rho - k\rho\sqrt{1 - \lambda^2}$$

is one of the roots of the equation

$$x^2 - 2k\rho \cdot x + \lambda^2 k^2\rho^2 = 0.$$