Construction of Minor is Unique

Theorem

In the words of Euclid:

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

(The Elements: Book $\text{X}$: Proposition $82$)

Proof

Let $AB$ be a minor.

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

$AC$ and $CB$ are incommensurable in square

$AC^2 + CB^2$ is rational

$2 \cdot AC \cdot CB$ is a medial 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 rational

$2 \cdot AD \cdot DB$ is a medial 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 :$AD^2 + DB^2$ and $AC^2 + CB^2$ are both rational.

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

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

But by definition of medial area, this is impossible.

The result follows.

$\blacksquare$

Historical Note

This proof is Proposition $82$ 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