Straight Line Commensurable with Apotome
Theorem
In the words of Euclid:
- A straight line commensurable in length with an apotome is an apotome and the same in order.
(The Elements: Book $\text{X}$: Proposition $103$)
Proof
Let $AB$ be an apotome.
Let $CD$ be commensurable in length with $AB$.
It is to be demonstrated that:
- $CD$ is an apotome
and:
Let $BE$ be the annex of $CD$.
This can be constructed uniquely by Construction of Apotome is Unique.
By definition, $AE$ and $EB$ are rational straight lines which are commensurable in square only.
From Proposition $12$ of Book $\text{VI} $: Construction of Fourth Proportional Straight Line, let it be contrived that:
- $BE : DF = AB : CD$
From Proposition $12$ of Book $\text{V} $: Sum of Components of Equal Ratios:
- $AE : CF = AB : CD$
and so from Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:
- $AE : CF = BE : DF$
But $AB$ is commensurable in length with $CD$.
Therefore from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $AE$ is commensurable in length with $CF$
and:
- $BE$ is commensurable in length with $DF$.
We have that $AE$ and $EB$ are rational straight lines which are commensurable in square only.
Therefore from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:
- $CF$ and $FD$ are rational straight lines which are commensurable in square only.
Thus $CD$ is an apotome.
$\Box$
It remains to be shown that the order of $CD$ is the same as the order of $AB$.
We have that:
- $AE : CF = BE : DF$
So from Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:
- $AE : EB = CF : FD$
We have that:
- $AE^2 = EB^2 + \lambda^2$
where either:
- $\lambda$ is commensurable in length with $AE$
or:
- $\lambda$ is incommensurable in length with $AE$.
Suppose $\lambda$ is commensurable in length with $AE$.
Then by Proposition $14$ of Book $\text{X} $: Commensurability of Squares on Proportional Straight Lines:
- $CF^2 > FD^2 + \mu^2$
where $\mu$ is commensurable in length with $CF$.
Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out.
Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:
- $CD$ is commensurable in length with $\alpha$.
Similarly, let $BE$ be commensurable in length with $\alpha$.
Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:
- $DF$ is commensurable in length with $\alpha$.
Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$.
- neither $CF$ nor $FD$ will be commensurable in length with $\alpha$.
Let $\lambda$ be incommensurable in length with $AE$.
Then by Proposition $14$ of Book $\text{X} $: Commensurability of Squares on Proportional Straight Lines:
- $CF^2 > FD^2 + \mu^2$
where $\mu$ is incommensurable in length with $CF$.
Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out.
Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:
- $CD$ is commensurable in length with $\alpha$.
Similarly, let $BE$ be commensurable in length with $\alpha$.
Then by Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:
- $DF$ is commensurable in length with $\alpha$.
Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$.
- neither $CF$ nor $FD$ will be commensurable in length with $\alpha$.
It follows that $CD$ is an apotome of the same as the order as $AB$.
$\blacksquare$
Historical Note
This proof is Proposition $103$ 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