Commentary on Proposition X.44¶
Back to Proposition X.44.
As the irrationality of the second bimedial straight line [x. 38] is proved by means of the irrationality of the binomial straight line [x. 36], so the present theorem is reduced to that of x. 42.
Suppose, if possible, that the second bimedial straight line can be divided into its terms as such in two ways, i.e. that
$$x + y = x' + y',$$
where $x', y'$ are nearer equality than $x, y$.
Apply $x^2 + y^2$, $2xy$ to a rational straight line $\sigma$, i.e. let
$$\begin{aligned} x^2 + y^2 &= \sigma u, \\ 2xy &= \sigma v. \end{aligned}$$
Then, as in x. 38, the areas $x^2 + y^2$, $2xy$ are medial, so that $\sigma u, \sigma v$ are medial ; therefore $u, v$ are both rational and $\smile \sigma$ ……………………………………………..(1).
Again, by hypothesis, $x, y$ are medial straight lines commensurable in square only; therefore
$$x \smile y.$$
Hence
$$x^2 \smile xy.$$
And $x^2 \sim (x^2 + y^2)$, while $xy \sim 2xy$; therefore
$$(x^2 + y^2) \smile 2xy,$$
or
$$\sigma u \smile \sigma v,$$
and hence
$$u \smile v$$
………………………………………………………..(2).
Therefore, by (1) and (2), $u, v$ are rational straight lines commensurable in square only; therefore $u + v$ is a binomial straight line.
Similarly, if $x'^2 + y'^2 = \sigma u'$ and $2x'y' = \sigma v'$, $u' + v'$ will be proved to be a binomial straight line.
And, since $(x + y)^2 = (x' + y')^2$, and therefore $(u + v) = (u' + v')$, it follows that a binomial straight line is divided as such in two ways : which is impossible. [x. 42]
Therefore $x + y$, the given second bimedial straight line, can only be so divided in one way.
In order to prove that $u + v$, $u' + v'$ represent a different division of the same straight line, Euclid assumes that $x^2 + y^2 > 2xy$. This is of course an easy inference from II. 7; but the assumption of it here renders it probable that the Lemma after x. 59 is interpolated.