Second Bimedial Straight Line is Divisible Uniquely
Theorem
In the words of Euclid:
- A second bimedial straight line is divided at one point only.
(The Elements: Book $\text{X}$: Proposition $44$)
Proof
Let $AB$ be a second bimedial straight line.
Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:
- $AC$ and $CB$ are medial straight lines
- $AC$ and $CB$ are commensurable in square only
- $AC$ and $CB$ contain a medial rectangle.
Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.
WLOG let $AC > DB$.
- $AD^2 + DB^2 < AC^2 + CB^2$
From Proposition $4$ of Book $\text{II} $: Square of Sum:
- $AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
and:
- $AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$
Let $EF$ be a rational straight line.
Let $EG = EK - AC^2 - CB^2$.
Then $HK = 2 \cdot AC \cdot CB$.
Similarly, let $EL = AD^2 + DB^2$.
Then $HL = 2 \cdot AD \cdot DB$.
Since the squares on $AC$ and $CB$ are by definition medial, it follows that $EG$ is medial.
But $EG$ is applied to the rational straight line $EF$.
Therefore from Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:
- $EH$ is a rational straight line which is incommensurable in length with $EF$.
Using the same reasoning:
- $HN$ is a rational straight line which is incommensurable in length with $EF$.
We have that $AC$ and $CB$ are medial straight lines which are commensurable in square only.
Therefore $AC$ and $CB$ are incommensurable in length.
But:
- $AC : CB = AC^2 : AC \cdot CB$
So from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $AC^2$ is incommensurable with $AC \cdot CB$.
But we have that $AC$ and $CB$ are commensurable in square.
So from Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:
- $AC^2 + CB^2$ is commensurable in square with $AC^2$.
From Proposition $6$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:
- $2 \cdot AC \cdot CB$ is commensurable with $AC \cdot CB$.
- $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.
We have:
- $EG = AC^2 + CB^2$
and:
- $HK = 2 \cdot AC \cdot CB$
Therefore $EG$ is incommensurable with $HK$.
So by:
and:
it follows that:
- $EH$ is incommensurable in length with $HN$.
We also have that $EH$ and $HN$ are rational.
Therefore $EH$ and $HN$ are rational straight lines which are commensurable in square only.
But from Proposition $36$ of Book $\text{X} $: Binomial is Irrational:
- Two rational straight lines commensurable in square only, when added together, form an irrational straight line which is a binomial.
Therefore $EN$ is a binomial straight line which is divided at $H$.
In the same way $EM$ and $MN$ can be shown to be rational straight lines commensurable in square only.
Thus $EN$ is a binomial straight line which is divided into terms in two places: $H$ and $M$.
We have that:
- $AC^2 + CB^2 > AD^2 + DB^2$
Also:
- $AD^2 + DB^2 > 2 \cdot AD \cdot DB$
Thus $EG = AC^2 + CB^2 > 2 \cdot AD \cdot DB = MK$
Thus $EH > MN$.
But from Proposition $42$ of Book $\text{X} $: Binomial Straight Line is Divisible into Terms Uniquely:
- $EN$ can be divided into its terms in only one way.
This contradicts the conclusion that has been drawn: that $H$ and $M$ divide $EH$ into terms in two different ways.
So the supposition that $AB$ can be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$ must be false.
$\blacksquare$
Historical Note
This proof is Proposition $44$ 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