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