# 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.*

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

## 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:

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

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

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

and

- $AB \cdot BC$ 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 $BC^2$.

- $2 AB \cdot BC$ 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**.

$\blacksquare$

## Historical Note

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

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions