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Commentary on Proposition X.25¶

Two medial straight lines commensurable in square only are of the form $k^{\frac{1}{4}}\rho$, $\sqrt{\lambda} \,.\, k^{\frac{1}{4}}\rho$.

The rectangle contained by them is $\sqrt{\lambda} \,.\, k^{\frac{1}{2}}\rho^2$. Now this is in general medial; but, if $\sqrt{\lambda} = k' \sqrt{k}$, the rectangle is $k k'\rho^2$, which is rational.

Euclid’s argument is as follows. Let us, for convenience, put $x$ for $k^{\frac{1}{4}}\rho$, so that the medial straight lines are $x$, $\sqrt{\lambda} \,.\, x$.

Form the areas $x^2$, $x \,.\, \sqrt{\lambda} \,.\, x$, $\lambda x^2$, and let these be respectively equal to $\sigma u$, $\sigma v$, $\sigma w$, where $\sigma$ is a rational straight line.

Since $x^2$, $\lambda x^2$ are medial areas, so are $\sigma u$, $\sigma w$, whence $u$, $w$ are respectively rational and $\smile \sigma$.

[Folio 0062 | p. 58]

But

$$x^2 \frown \lambda x^2,$$

so that

$$\sigma u \frown \sigma w,$$

$$u \frown w \ \text{................................................(1).}$$

Therefore, $u$, $w$ being both rational, $u w$ is rational $\text{..........................(2).}$

Now

$$x^2 : \sqrt{\lambda} \,.\, x^2 = \sqrt{\lambda} \,.\, x^2 : \lambda x^2$$

$$\sigma u : \sigma v = \sigma v : \sigma w,$$

so that

$$u : v = v : w,$$

and

$$u w = v^2.$$

Hence, by (2), $v^2$, and therefore $v$, is rational $\text{.............................(3).}$

Now ($\alpha$) if $v \frown \sigma$, $\sigma v$ or $\sqrt{\lambda} \,.\, x^2$ is rational; ($\beta$) if $v \smile \sigma$, so that $v \sim \sigma$, $\sigma v$ or $\sqrt{\lambda} \,.\, x^2$ is medial.