# Construction of that which produces Medial Whole with Rational Area is Unique

## Theorem

In the words of Euclid:

*To a straight line which produces with a rational area a medial whole only one straight line can be annexed which is incommensurable in square with the whole straight line and which with the whole straight line makes the sum of the squares on them medial, but twice the rectangle contained by them rational.*

(*The Elements*: Book $\text{X}$: Proposition $83$)

## Proof

Let $AB$ be a straight line which produces with a rational area a medial whole.

Let $BC$ be added to $AB$ such that:

- $AC$ and $CB$ are incommensurable in square
- $AC^2 + CB^2$ is medial
- $2 \cdot AC \cdot CB$ is a rational] rectangle.

It is to be proved that no other straight line can be added to $AB$ which fulfils these conditions.

Suppose $BD$, different from $BC$, can be added to $AB$ such that:

- $AD$ and $DB$ are incommensurable in square
- $AD^2 + DB^2$ is [[Definition:Medial Area|medial]
- $2 \cdot AD \cdot DB$ is a rational] rectangle.

From Proposition $7$ of Book $\text{II} $: Square of Difference:

- $AD^2 + DB^2 - 2 \cdot AD \cdot DB = AC^2 + CB^2 - 2 \cdot AC \cdot CB = AB^2$

Therefore:

- $AD^2 + DB^2 - AC^2 + CB^2 = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$

But $2 \cdot AD \cdot DB$ and $2 \cdot AC \cdot CB$ are both rational.

Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational.

Therefore $AD^2 + DB^2 - AC^2 + CB^2$ is rational.

But from:

and:

both $AD^2 + DB^2$ and $AC^2 + CB^2$ is medial.

By Proposition $26$ of Book $\text{X} $: Medial Area not greater than Medial Area by Rational Area this cannot happen.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $83$ 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