Apotome not same with Binomial Straight Line

Proof

 * Euclid-X-111.png

Let $AB$ be an apotome.

Suppose $AB$ were the same with a binomial.

Let $DC$ be a rational straight line.

Let the rectangle $CE$ be applied to $CD$ equal to the square on $AB$ and producing $DE$ as breadth.

We have that $AB$ is an apotome.

Therefore from :
 * $DE$ is a first apotome.

Let $EF$ be the annex of $DE$.

Therefore by :
 * $DF$ and $EF$ are rational straight lines which are commensurable in square only
 * $DF^2 = FE^2 + \lambda^2$ where $\lambda$ is a straight line which is commensurable in length with $DF$
 * $DF$ is commensurable in length with $DC$.

We have that $AB$ is a binomial.

Therefore by :
 * $DE$ is a first binomial.

Let $DE$ be divided into its terms at $G$.

Let $DG$ be the greater term.

Then by :
 * $DG$ and $GE$ are rational straight lines which are commensurable in square only
 * $DG^2 = GE^2 + \mu^2$ where $\mu$ is a straight line which is commensurable in length with $DF$
 * $DG$ is commensurable in length with $DC$.

Therefore by :
 * $DG$ is commensurable in length with $DF$.

Therefore by :
 * $GF$ is commensurable in length with $DG$.

But $DF$ and $EF$ are rational straight lines which are commensurable in square only.

So $DF$ is incommensurable in length with $EF$.

Therefore by :
 * $FG$ is incommensurable in length with $EF$.

Therefore $GF$ and $EF$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $EG$ is an apotome.

But $EG$ is also rational, which is impossible.

Hence the result.