Construction of Apotome is Unique

Theorem
Let $D$ be the domain $\left\{{x \in \R_{>0} : x^2 \in \Q}\right\}$, the rationally expressible numbers.

Let $a, b \in D$ be two rationally expressible numbers such that $a - b$ is an apotome.

Then, there exists only one $x \in D$ such that $a - b + x$ and $a$ are commensurable in square only.

Proof

 * Euclid-X-79.png

Let $AB$ be an apotome.

Let $BC$ be added to $AB$ so that $AC$ and $CB$ are rational straight lines which are commensurable in square only.

It is to be proved that no other rational straight line can be added to $AB$ which is commensurable in square only with the whole.

Suppose $BD$ can be added to $AB$ so as to fulfil the conditions stated.

Then $AD$ and $DB$ are rational straight lines which are commensurable in square only.

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

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 from :
 * $2 \cdot AD \cdot DB$ and $2 \cdot AC \cdot CB$ are both medial.

From :
 * a medial area does not exceed a medial area by a rational area.

Therefore no other rational straight line can be added to $AB$ which is commensurable in square only with the whole.

Therefore only one rational straight line can be added to $AB$ which is commensurable in square only with the whole.