Commentary on Proposition X.31¶
Back to Proposition X.31.
Take the rational straight lines commensurable in square only found in x. 29, i.e. $\rho$, $\rho\sqrt{1 - k^2}$.
Take the mean proportional $\rho(1 - k^2)^\frac{1}{4}$ and $x$ such that
$$\rho(1 - k^2)^\frac{1}{4} : \rho\sqrt{1 - k^2} = \rho\sqrt{1 - k^2} : x.$$
Then $\rho(1 - k^2)^\frac{1}{4}$, $x$, or $\rho(1 - k^2)^\frac{1}{4}$, $\rho(1 - k^2)^\frac{3}{4}$ are straight lines satisfying the given conditions.
For ($\alpha$) $\rho^2\sqrt{1-k^2}$ is a medial area, and therefore $\rho(1-k^2)^{\frac{1}{4}}$ is a medial straight line ………………………………………….(1); and $x\ .\ \rho(1-k^2)^{\frac{1}{4}} = \rho^2(1-k^2)$ and is therefore a rational area.
($\beta$) $\rho$, $\rho(1-k^2)^{\frac{1}{4}}$, $\rho\sqrt{1-k^2}$, $x$ are straight lines in continued proportion, by construction.
Therefore
$$\rho : \rho\sqrt{1-k^2} = \rho(1-k^2)^{\frac{1}{4}} : x \text{ ....................(2).}$$
(This Euclid has to prove in a somewhat roundabout way by means of the lemma after x. 21 to the effect that $a : b = ab : b^2$.)
From (2) it follows [x. 11] that $x \mathbin{\sim} \rho(1-k^2)^{\frac{1}{4}}$; whence, since $\rho(1-k^2)^{\frac{1}{4}}$ is medial, $x$ or $\rho(1-k^2)^{\frac{3}{4}}$ is medial also.
($\gamma$) From (2), since $\rho$, $\rho\sqrt{1-k^2}$ satisfy the remaining condition of the problem, $\rho(1-k^2)^{\frac{1}{4}}$, $\rho(1-k^2)^{\frac{3}{4}}$ do so also [x. 14].
According as $\rho$ is of the form $a$ or $\sqrt{A}$, the straight lines take the forms
$$\begin{array}{lcc} (1) & \sqrt{a\sqrt{a^2-b^2}}, & \dfrac{a^2-b^2}{\sqrt{a\sqrt{a^2-b^2}}}, \\ \text{or}\quad (2) & \sqrt[4]{A(A-k^2A)}, & \dfrac{A-k^2A}{\sqrt[4]{A(A-k^2A)}}. \end{array}$$
To find medial straight lines commensurable in square only containing a rational rectangle, and such that the square on one exceeds the square on the other by the square on a straight line incommensurable with the former, we simply begin with the rational straight lines having the corresponding property [x. 30], viz. $\rho$, $\dfrac{\rho}{\sqrt{1+k^2}}$, and we arrive at the straight lines
$$\dfrac{\rho}{(1+k^2)^{\frac{1}{4}}},\quad \dfrac{\rho}{(1+k^2)^{\frac{3}{4}}}.$$
According as $\rho$ is of the form $a$ or $\sqrt{A}$, these (if we use the same transformation as at the end of the note on x. 30) may take any of the forms
$$\begin{array}{lcc} (1) & \sqrt{a\sqrt{a^2-B}}, & \dfrac{a^2-B}{\sqrt{a\sqrt{a^2-B}}}, \\ \text{or}\quad (2) & \sqrt[4]{A(A-B)}, & \dfrac{A-B}{\sqrt[4]{A(A-B)}}, \\ \text{or} & \sqrt[4]{A(A-b^2)}, & \dfrac{A-b^2}{\sqrt[4]{A(A-b^2)}}. \end{array}$$