Commentary on Proposition X.26

Back to Proposition X.26.

[Folio 0063 | p. 59]

“Apply” the two given medial areas to one and the same rational straight line $\rho$. They can then be written in the form $\rho \,.\, k^{\frac{1}{2}}\rho$, $\rho \,.\, \lambda^{\frac{1}{2}}\rho$.

The difference is then $(\sqrt{k} - \sqrt{\lambda})\rho^2$; and the proposition asserts that this cannot be rational, i.e. $(\sqrt{k} - \sqrt{\lambda})$ cannot be equal to $k'$. Cf. the proposition corresponding to this in algebraical text-books.

To make Euclid’s proof clear we will put $x$ for $k^{\frac{1}{2}}\rho$ and $y$ for $\lambda^{\frac{1}{2}}\rho$. Suppose

$$\rho(x - y) = \rho z,$$

and, if possible, let $\rho z$ be rational, so that $z$ must be rational and $\frown \rho \text{ ...(1).}$

Since $\rho x$, $\rho y$ are medial,

$$x \text{ and } y \text{ are respectively rational and } \smile \rho \text{ .................(2).}$$

From (1) and (2),

$$y \smile z.$$

Now

$$y : z = y^2 : yz,$$

so that

$$y^2 \smile yz.$$

[Folio 0064 | p. 60]

But

$$y^2 + z^2 \frown y^2,$$

and

$$2yz \frown yz.$$

Therefore

$$y^2 + z^2 \smile 2yz,$$

whence

$$(y + z)^2 \smile (y^2 + z^2),$$

or

$$x^2 \smile (y^2 + z^2).$$

And $(y^2 + z^2)$ is rational; therefore $x^2$, and consequently $x$, is irrational.

But, by (2), $x$ is rational: which is impossible.

Therefore $\rho z$ is not rational.