Apotome not same with Binomial Straight Line

Theorem

In the words of Euclid:

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

(The Elements: Book $\text{X}$: Proposition $111$)

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.

Therefore from Proposition $97$ of Book $\text{X} $: Square on Apotome applied to Rational Straight Line:

$DE$ 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.

Therefore by Proposition $60$ of Book $\text{X} $: Square on Binomial Straight Line applied to Rational Straight Line:

$DE$ is a first binomial.

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

Let $DG$ be the greater term.

Then by Book $\text{X (II)}$ Definition $1$: First Binomial:

$DG$ 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 Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:

$DG$ is commensurable in length with $DF$.

Therefore by Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:

$GF$ 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 Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:

$FG$ 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.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions