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Commentary on Proposition X.50¶

Let $\rho$ be a rational straight line.

Take the numbers $q(m^2 - n^2)$, $qn^2$, and let $p$ be a third number which is not a square and which has not to $qm^2$ or $q(m^2 - n^2)$ the ratio of square to square.

Take $x$ such that

$$p : qm^2 = \rho^2 : x^2 \dots\dots\dots\dots\dots\dots\dots(1).$$

Thus $x$ is rational and $\sim \rho$ $\dots\dots\dots\dots\dots\dots\dots(2)$.

Next suppose that

$$qm^2 : q(m^2 - n^2) = x^2 : y^2 \dots\dots\dots\dots\dots\dots(3).$$

It follows that $y$ is rational and $\sim x$ $\dots\dots\dots\dots\dots\dots(4)$.

Thus $(x + y)$ is a binomial straight line.

Again, from (1) and (3), ex aequali,

$$p : q(m^2 - n^2) = \rho^2 : y^2 \dots\dots\dots\dots\dots\dots(5),$$

whence

$$y \cup \rho \dots\dots\dots\dots\dots\dots(6).$$

Suppose that

$$x^2 - y^2 = z^2.$$

Then, from (3), convertendo,

$$qm^2 : qn^2 = x^2 : z^2,$$

whence

$$z \sim x.$$

Thus

$$\sqrt{x^2 - y^2} \sim x,$$

and $x, y$ are both $\cup\ \rho$; therefore $x + y$ is a third binomial straight line.

Now, from (1),

$$x = \rho \cdot \frac{m\sqrt{q}}{\sqrt{p}},$$

and, by (5),

$$y = \rho \cdot \frac{\sqrt{m^2 - n^2} \cdot \sqrt{q}}{\sqrt{p}}.$$

Thus the third binomial is

$$\sqrt{\frac{q}{p}} \cdot \rho(m + \sqrt{m^2 - n^2}),$$

which we may write in the form

$$m\sqrt{k} \cdot \rho + m\sqrt{k} \cdot \rho\sqrt{1 - \lambda^2}.$$

[Folio 0113 | p. 109]

The corresponding third apotome [x. 87] is

$$m\sqrt{k}\cdot p - m\sqrt{k}\cdot p\sqrt{1 - \lambda^2}.$$

The two expressions are accordingly the roots of the equation

$$x^2 - 2m\sqrt{k}\cdot px + \lambda^2 m^2 kp^2 = 0,$$

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

where

$$a = m\sqrt{k}\cdot p.$$