Commentary on Proposition X.42

Back to Proposition X.42.

[Folio 0097 | p. 93]

This proposition proves the equivalent of the well-known theorem in surds that,

$$\text{if } a + \sqrt{b} = x + \sqrt{y},$$

then

$$a = x,\quad b = y,$$

and if

$$\sqrt{a} + \sqrt{b} = \sqrt{x} + \sqrt{y},$$

then

$$a = x,\quad b = y\text{ (or } a = y,\quad b = x\text{)}.$$

[Folio 0098 | p. 94]

The proposition states that a binomial straight line cannot be split up into terms (ὀνόματα) in two ways. For, if possible, let

$$x + y = x' + y',$$

where $x$, $y$, and also $x'$, $y'$, are the terms of a binomial straight line, $x'$, $y'$ being different from $x$, $y$ (or $y$, $x$).

One pair is necessarily more nearly equal than the other. Let $x'$, $y'$ be more nearly equal than $x$, $y$.

Then

$$(x^2 + y^2) - (x'^2 + y'^2) = 2x'y' - 2xy.$$

Now by hypothesis $(x^2 + y^2)$, $(x'^2 + y'^2)$ are rational areas, being of the form $\rho^2 + k\rho^2$; but $2x'y'$, $2xy$ are medial areas, being of the form $\sqrt{k}\cdot\rho^2$; therefore the difference of two medial areas is rational: which is impossible. [x. 26]

Therefore $x'$, $y'$ cannot be different from $x$, $y$ (or $y$, $x$).