Commentary on Proposition X.60

Back to Proposition X.60.

In the hexad of propositions beginning with this we have the solution of the converse problem to that of X. 54—59. We find the squares of the irrational straight lines of X. 36—41 and prove that they are respectively equal to the rectangles contained by a rational straight line and the first, second, third, fourth, fifth and sixth binomials.

In X. 60 we prove that, $\rho + \sqrt{k} \,.\, \rho$ being a binomial straight line [X. 36],

$$\frac{(\rho + \sqrt{k} \,.\, \rho)^2}{\sigma}$$

is a first binomial straight line, and we find it geometrically.

The procedure may be represented thus. Take $x, y, z$ such that

$$\begin{aligned} \sigma x &= \rho^2,\\ \sigma y &= k\rho^2,\\ \sigma \cdot 2z &= 2\sqrt{k} \cdot \rho^2, \end{aligned}$$

$\rho^2, k\rho^2$ being of course the squares on the terms of the original binomial, and $2\sqrt{k} \cdot \rho^2$ twice the rectangle contained by them.

Then

$$(x + y) + 2z = \frac{(\rho + \sqrt{k} \cdot \rho)^2}{\sigma},$$

and we have to prove that $(x + y) + 2z$ is a first binomial straight line of which $(x + y), 2z$ are the terms and $(x + y)$ the greater.

Euclid divides the proof into two parts, showing first that $(x + y) + 2z$ is some binomial, and secondly that it is the first binomial.

($\alpha$) $\rho \sim \sqrt{k} \cdot \rho$, so that $\rho^2, k\rho^2$ are rational and commensurable; therefore $\rho^2 + k\rho^2$, or $\sigma (x + y)$, is a rational area, whence

$$(x + y)\text{ is rational and } \sim \sigma \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(1).$$

Next, $2\rho \cdot \sqrt{k} \cdot \rho$ is a medial area, so that $\sigma \cdot 2z$ is a medial area, whence

$$2z\text{ is rational but } \smile \sigma \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(2).$$

Hence [(1), (2)] $(x + y), 2z$ are rational and commensurable in square only $\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(3);$ thus $(x + y) + 2z$ is a binomial straight line. [X. 36]

($\beta$)

$$\rho^2 : \sqrt{k} \cdot \rho^2 = \sqrt{k} \cdot \rho^2 : k\rho^2,$$

so that

$$\sigma x : \sigma z = \sigma z : \sigma y,$$

and

$$x : z = z : y,$$

or

$$xy = z^2 = \frac{1}{4}(2z)^2 \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(4).$$

Now $\rho^2, k\rho^2$ are commensurable, so that $\sigma x, \sigma y$ are commensurable, and therefore

$$x \sim y \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots(5).$$

And, since [Lemma] $\rho^2 + k\rho^2 > 2\sqrt{k} \cdot \rho^2,$

$$x + y > 2z.$$

But $(x + y)$ is given, being equal to $\frac{\rho^2 + k\rho^2}{\sigma} \dots\dots\dots\dots\dots\dots\dots\dots(6).$

Therefore [(4), (5), (6), and X. 17] $\sqrt{(x + y)^2 - (2z)^2} \sim (x + y).$ And $(x + y), 2z$ are rational and $\smile$ [(3)], while $(x + y) \sim \sigma$ [(1)].

Hence $(x + y) + 2z$ is a first binomial.

The actual value of $(x + y) + 2z$ is, of course,

$$\frac{\rho^2}{\sigma}(1 + k + 2\sqrt{k}).$$