Construction of Third Apotome
Theorem
In the words of Euclid:
- To find the third apotome.
(The Elements: Book $\text{X}$: Proposition $87$)
Proof
Let $A$ be a rational straight line.
Let $E$, $BC$ and $CD$ be numbers set out which do not have pairwise between them the ratio that a square number has to another square number.
But let $CB : BD$ be the ratio that a square number has to a square number.
Using Porism to Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable, let it be contrived that:
- $E : BC = A^2 : FG^2$
and:
- $BC : CD = FG^2 : GH^2$
We have that:
- $E : BC = A^2 : FG^2$.
From Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:
- $A^2$ is commensurable with $FG^2$.
But $A^2$ is rational.
Therefore $FG^2$ is rational.
Therefore $FG$ is rational.
We have that $E : BC$ is not the ratio that a square number has to another square number.
Therefore $A^2 : FG^2$ is not the ratio that a square number has to another square number.
Therefore from Proposition $9$ of Book $\text{X} $: Commensurability of Squares:
- $A$ is incommensurable in length with $FG$.
We have that:
- $BC : CD = FG^2 : GH^2$
From Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:
- $FG^2$ is commensurable with $GH^2$.
But $FG^2$ is rational.
Therefore $GH^2$ is rational.
Therefore $GH$ is rational.
We have that $BC : CD$ is not the ratio that a square number has to another square number.
Therefore $FG^2 : GH^2$ is not the ratio that a square number has to another square number.
Therefore from Proposition $9$ of Book $\text{X} $: Commensurability of Squares:
- $FG$ is incommensurable in length with $GH$.
Both $FG$ and $GH$ are rational.
Therefore $FG$ and $GH$ rational straight lines which are commensurable in square only.
Therefore $FH$ is an apotome.
It remains to be shown that $FH$ is a third apotome.
We have that:
- $E : BC = A^2 : FG^2$.
and:
- $BC : CD = FG^2 : GH^2$
Therefore from Proposition $22$ of Book $\text{V} $: Equality of Ratios Ex Aequali:
- $E : CD = A^2 : HG^2$
But $E : CD$ is not the ratio that a square number has to another square number.
Therefore neither is $A^2 : GH^2$ the ratio that a square number has to another square number.
Therefore from Proposition $9$ of Book $\text{X} $: Commensurability of Squares:
- $A$ is incommensurable in length with $GH$.
Therefore neither $FG$ and $GH$ is incommensurable in length with the rational straight line $A$.
Now let $K^2 = FG^2 - GH^2$.
We have that:
- $BC : CD = FG^2 GH^2$
Therefore by Porism to Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:
- $BC : BD = FG^2 : K^2$
But $BC$ has to $BD$ the ratio that a square number has to another square number.
Therefore $FG^2$ has to $K^2$ the ratio that a square number has to another square number.
Therefore by Proposition $9$ of Book $\text{X} $: Commensurability of Squares:
- $FG$ is incommensurable in length with $K$.
So:
- $FG^2$ is greater than $GH^2$ by the square on a straight line which is commensurable in length with the rational straight line $A$.
Therefore, by definition, $FH$ is a third apotome.
$\blacksquare$
Historical Note
This proof is Proposition $87$ 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