First Bimedial is Irrational

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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.

(The Elements: Book $\text{X}$: Proposition $37$)


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:

$AB : BC = AB \cdot BC : BC^2$
$AB : BC = AB^2 : AB \cdot BC$

Therefore from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:

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


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

But by Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:

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

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

So from Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:

$AB^2 + BC^2$ is commensurable with $BC^2$.

So from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:

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

Thus from Proposition $16$ of Book $\text{X} $: Incommensurability of Sum of Incommensurable Magnitudes:

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

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

Thus $AB \cdot BC$ is rational.

From Proposition $4$ of Book $\text{II} $: Square of Sum:

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

Thus from Book $\text{X}$ Definition $4$: Rational Area:

$AC$ is irrational.

Such a straight line is called first bimedial.


Historical Note

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