Commentary on Proposition X.22

Back to Proposition X.22.

Our algebraical notation makes the result of this proposition almost self-evident. We have seen that the square of a medial straight line is of the form $\sqrt{k} \cdot \rho^2$. If we “apply” this area to another rational straight line $\sigma$, the breadth is $\frac{\sqrt{k} \cdot \rho^2}{\sigma}$.

This is equal to $\frac{\sqrt{k} \cdot \rho^2}{\sigma^2} \cdot \sigma = \sqrt{k} \cdot \frac{m}{n} \sigma$, where $m$, $n$ are integers. The latter straight line, which we may express, if we please, in the form $\sqrt{k'} \cdot \sigma$, is clearly commensurable with $\sigma$ in square only, and therefore rational but incommensurable in length with $\sigma$.

Euclid’s proof, necessarily longer, is in two parts.

Suppose that the rectangle $\sqrt{k} \cdot \rho^2 = \sigma \cdot x$.

Then (1)

$$\sigma : \rho = \sqrt{k} \cdot \rho : x, \qquad \text{[VI. 14]}$$

whence

$$\sigma^2 : \rho^2 = k\rho^2 : x^2. \qquad \text{[VI. 22]}$$

But $\sigma^2 \sim \rho^2$, and therefore $k\rho^2 \sim x^2$. [x. 11]

And $k\rho^2$ is rational, therefore $x^2$, and therefore $x$, is rational. [x. Def. 4]

  1. Since $\sqrt{k} \cdot \rho \sim \rho$, $\sqrt{k} \cdot \rho \smile \rho$.

But [Lemma]

$$\sqrt{k} \cdot \rho : \rho = k\rho^2 : \sqrt{k} \cdot \rho^2,$$

whence

$$k\rho^2 \smile \sqrt{k} \cdot \rho^2. \qquad \text{[x. 11]}$$

But $\sqrt{k} \cdot \rho^2 = \sigma x$, and $k\rho^2 \sim x^2$ (from above); therefore

$$x^2 \smile \sigma x; \qquad \text{[x. 13]}$$

and, since $x^2 : \sigma x = x : \sigma$, [Lemma]

$$x \smile \sigma.$$