Commentary on Proposition X.40¶

Here $x$, $y$ have [X. 34] the forms

$$\frac{\rho}{\sqrt{2(1+k^2)}}\sqrt{\sqrt{1+k^2}+k},\quad \frac{\rho}{\sqrt{2(1+k^2)}}\sqrt{\sqrt{1+k^2}-k}.$$

In this case $(x^2+y^2)$ is a medial, and $2xy$ a rational, area; thus

$$x^2+y^2 \cup 2xy.$$

Therefore

$$(x+y)^2 \cup 2xy,$$

whence, since $2xy$ is rational, $(x+y)^2$, and consequently $(x+y)$, is irrational.

The irrational straight line

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

is called (for an obvious reason) the “side” of a rational plus a medial (area).

This and the corresponding irrational with a minus sign found in x. 77 are the positive roots of the equation

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