Construction of Fifth Apotome
Theorem
In the words of Euclid:
- To find the fifth apotome.
(The Elements: Book $\text{X}$: Proposition $89$)
Proof
Let $A$ be a rational straight line.
Let $CG$ be commensurable in length with $A$.
Therefore, by definition, $CG$ is also rational.
Let $DF$ and $FE$ be square numbers such that $DE$ does not have to either $DF$ or $FE$ the ratio that a square number has to another square number.
Using Porism to Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable, let it be contrived that:
- $FE : ED = CG^2 : GB^2$
Therefore by Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:
- $GB^2$ is commensurable with $GC^2$.
Therefore $GB^2$ is rational.
Therefore $BG$ is rational.
We have that $DE : EF$ is not the ratio that a square number has to another square number.
Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number.
Therefore from Proposition $9$ of Book $\text{X} $: Commensurability of Squares:
- $BG$ is incommensurable in length with $GC$.
Both $BG$ and $GC$ are rational.
Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only.
Therefore $BC$ is an apotome.
It remains to be shown that $BC$ is a fifth apotome.
Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$.
We have that:
- $DE : EF = BG^2 : GC^2$
So by Porism to Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:
- $ED : EF = BG^2 : H^2$
So $ED$ does not have to $EF$ the ratio that a square number has to another square number.
Therefore neither does $GB^2$ have to $H^2$ the ratio that a square number has to another square number.
Therefore from Proposition $9$ of Book $\text{X} $: Commensurability of Squares:
- $BG$ is incommensurable in length with $H$.
We have that:
- $BG^2 = GC^2 + H^2$
Therefore $GB^2$ is greater than $GC^2$ by the square on a straight line which is incommensurable in length with $GB$.
Also, the annex $CG$ is commensurable in length with the rational straight line $A$.
Therefore, by definition, $BC$ is a fifth apotome.
$\blacksquare$
Historical Note
This proof is Proposition $89$ 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