← euclid / The Thirteen Books of Euclid’s Elements

Commentary on Proposition X.58¶

We have here to find the square root of the expression [cf. X. 52]

$$\rho(k\rho\sqrt{1+\lambda} + k\rho).$$

As usual, we put

$$\left.\begin{aligned} u+v &= k\rho\sqrt{1+\lambda} \\ uv &= \frac{1}{4}k^2\rho^2 \end{aligned}\right\}\quad\quad\quad\quad\dots\dots(1).$$

Then, $u$, $v$ being found, we take

$$\left.\begin{aligned} x^2 &= \rho u \\ y^2 &= \rho v \end{aligned}\right\}\quad\quad\quad\quad\dots\dots(2),$$

and $(x+y)$, so found, is our required square root.

Euclid’s proof of the class of $(x+y)$ is as follows: By X. 18, $u \smile v$; therefore $\rho u \smile \rho v$, so that $x^2 \smile y^2$, and $x \bumpeq y \quad\quad\quad\quad\dots\dots(3).$

Next $u+v \smile k\rho$ $\phantom{\text{Next }u+v}\smile \rho,$ whence $\rho(u+v)$, or $(x^2+y^2)$, is a medial area $\dots\dots(4).$

Lastly, $xy = \frac{1}{2}k\rho^2$, which is a rational area $\dots\dots(5).$

Hence [(3), (4), (5)] $(x+y)$ is the side of a rational plus a medial area. [X. 40]

If we solve algebraically, we obtain

$$\begin{aligned} u &= \frac{k\rho}{2}(\sqrt{1+\lambda} + \sqrt{\lambda}), \\ v &= \frac{k\rho}{2}(\sqrt{1+\lambda} - \sqrt{\lambda}), \end{aligned}$$

and

$$x+y = \rho\sqrt{\frac{k}{2}(\sqrt{1+\lambda}+\sqrt{\lambda})} + \rho\sqrt{\frac{k}{2}(\sqrt{1+\lambda}-\sqrt{\lambda})}.$$

The corresponding “side” found in X. 95 is a straight line which produces with a rational area a medial whole, being of the form $(x-y)$, where $x$, $y$ have the same values as above.

The two square roots are (cf. note on X. 52) the roots of the biquadratic equation

$$x^4 - 2k\rho^2\sqrt{1+\lambda}\cdot x^2 + \lambda k^2\rho^4 = 0.$$