First Bimedial is Irrational

Theorem

In the words of Euclid:

If two medial straight lines commensurable in square only and containing a rational rectangle be added together, the whole is irrational; and let it be called a first bimedial straight line.

Proof

Let $AB$ and $BC$ be medial straight lines which are commensurable in square only.

Let $AB$ and $BC$ contain a rational rectangle.

By definition, $AB$ and $BC$ are incommensurable in length.

We have:

$AB : BC = AB \cdot BC : BC^2$
$AB : BC = AB^2 : AB \cdot BC$
$AB \cdot BC$ is incommensurable with $AB^2$

and

$AB \cdot BC$ is incommensurable with $BC^2$.
$2 AB \cdot BC$ is commensurable with $AB \cdot BC$.

We have that $AB$ and $BC$ are commensurable in square.

$AB^2 + BC^2$ is commensurable with $BC^2$.
$2 AB \cdot BC$ is incommensurable with $AB^2 + BC^2$.
$2 AB \cdot BC + AB^2 + BC^2$ is incommensurable with $AB \cdot BC$.

We have that $AB$ and $BC$ contain a rational rectangle.

Thus $AB \cdot BC$ is rational.

$AC^2 = \left({AB + BC}\right)^2 = 2 AB \cdot BC + AB^2 + BC^2$
$AC$ is irrational.

Such a straight line is called first bimedial.

$\blacksquare$

Historical Note

This proof is Proposition $37$ of Book $\text{X}$ of Euclid's The Elements.