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Commentary on Proposition X.84¶

With the usual notation, suppose that

$$x - y = x' - y'. \qquad (x > x')$$

Let

$$\left. \begin{array}{r} x^2 + y^2 = \sigma u \\ 2xy = \sigma v \end{array} \right\} \quad \text{and} \quad \left. \begin{array}{r} x'^2 + y'^2 = \sigma u' \\ 2x'y' = \sigma v' \end{array} \right\}.$$

Consider $(x - y)$ first ; it follows, since $(x^2 + y^2)$, $2xy$ are both medial areas, that $u, v$ are both rational and $\smile \sigma$ ………………………………………….(1).

But

$$x^2 + y^2 \smile 2xy,$$

that is,

$$\sigma u \smile \sigma v,$$

and therefore

$$u \smile v \text{ .................................................(2).}$$

Therefore [(1) and (2)] $u, v$ are rational and $\sim$ ; hence $(u - v)$ is an apotome.

Similarly $(u' - v')$ is proved to be the same apotome.

Thus the same apotome is formed as such in two ways : which is impossible [X. 79].

Therefore, etc.