Commentary on Proposition X.48¶

Let $kp$ be a straight line commensurable in length with $\rho$, a given rational straight line.

The two numbers taken may be written $p(m^2 - n^2)$, $pn^2$, where $(m^2 - n^2)$ is not a square.

Take $x$ such that

$$pm^2 : p(m^2 - n^2) = k^2 p^2 : x^2 \dots\dots\dots\dots\dots\dots\dots(1),$$

whence

$$x = kp \frac{\sqrt{m^2 - n^2}}{m}.$$

Then $kp + x$, or $kp + kp \frac{\sqrt{m^2 - n^2}}{m}$, is a first binomial straight line $\dots\dots(2)$.

To prove this we have, from (1),

$$x^2 \sim k^2 p^2,$$

and $x$ is rational, but $x \cup kp$; that is, $x$ is rational and $\sim kp$, so that $kp + x$ is a binomial straight line.

Also, $k^2 p^2$ being greater than $x^2$, suppose $k^2 p^2 - x^2 = y^2$.

Then, from (1),

$$pm^2 : pn^2 = k^2 p^2 : y^2,$$

whence $y$ is rational and $\sim kp$.

Therefore $kp + x$ is a first binomial straight line [x. Deff. II. 1].

This binomial straight line may be written thus,

$$kp + kp \sqrt{1 - \lambda^2}.$$

When we come to x. 85, we shall find that the corresponding straight line with a negative sign is the first apotome,

$$kp - kp \sqrt{1 - \lambda^2}.$$

Consider now the equation of which these two expressions are the roots. The equation is

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

In other words, the first binomial and the first apotome correspond to the roots of the equation

$$x^2 - 2ax + \lambda^2 a^2 = 0,$$

where $a = kp$.