# Apotome not same with Binomial Straight Line

## Theorem

In the words of Euclid:

The apotome is not the same with the binomial straight line.

## Proof

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.

$DE$ is a first apotome.

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

Therefore by Book $\text{X (III)}$ Definition $1$: First Apotome:

$DF$ 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.

$DE$ is a first binomial.

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

Let $DG$ be the greater term.

$DG$ 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 $DC$.
$DG$ is commensurable in length with $DF$.
$GF$ 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 $EF$.

$FG$ 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.

$\blacksquare$

## Historical Note

This proof is Proposition $111$ of Book $\text{X}$ of Euclid's The Elements.