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Commentary on Proposition X.57¶

The problem here is to find the square root of the expression [cf. X. 51]

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

The procedure is the same. Find $u$, $v$ from the equations

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

and, if

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

$(x+y)$ is the required square root.

To prove that $(x+y)$ is the major irrational straight line Euclid argues thus. By X. 18, $u \smile v$, therefore $\rho u \smile \rho v$, or $x^2 \smile y^2$, so that $x \bumpeq y \quad\quad\quad\quad\dots\dots(3).$

Now, since $(u+v) \sim \rho$, $(u+v)\rho$, or $(x^2+y^2)$, is a rational area $\dots\dots(4).$

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

Thus [(3), (4), (5)] $(x+y)$ is a major irrational straight line. [X. 39]

Actual solution gives

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

and

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

The corresponding square root found in X. 94 is the minor irrational straight line, the terms being separated by a minus sign, and the two straight lines are the roots (cf. note on X. 51) of the biquadratic equation

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