Second Bimedial is Irrational
Theorem
In the words of Euclid:
- If two medial straight lines commensurable in square only and containing a medial rectangle be added together, the whole is irrational; and let it be called a second bimedial straight line.
(The Elements: Book $\text{X}$: Proposition $38$)
Proof
Let $AB$ and $BC$ be medial straight lines which are commensurable in square only.
Let $AB$ and $BC$ contain a medial rectangle.
Let $DE$ be a rational straight line.
- Let $DF$ be a parallelogram set out on $DE$ equal to the square on $AC$.
Let its breadth be $DG$.
From Proposition $4$ of Book $\text{II} $: Square of Sum:
- $AC^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC$
Let $EH$ be a rectangle applied to $DE$ whose area equals $AB^2 + BC^2$.
The rectangle $HF$ which remains from $DF$ having had $EH$ removed is therefore equal to $2 \cdot AB \cdot BC$.
Since each of $AB$ and $BC$ are medial, both of $AB^2$ and $BC^2$ are also medial.
By hypothesis, $2 \cdot AB \cdot BC$ is also medial.
As:
- $EH = AB^2 + BC^2$
and:
- $FH = 2 \cdot AB \cdot BC$
each of the rectangles $EH$ and $FH$ is medial.
Each of $EH$ and $FH$ is applied to the rational straight line $DE$.
From Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:
- Each of $DH$ and $HG$ is rational and incommensurable in length with $DE$.
We have that $AB$ is incommensurable in length with $BC$.
We also have that:
- $AB : BC = AB^2 : AB \cdot BC$
So by Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $AB^2$ is incommensurable with $AB \cdot BC$.
But from Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:
- $AB^2 + BC^2$ is commensurable with $AB^2$
and from Proposition $15$ of Book $\text{X} $: Magnitudes with Rational Ratio are Commensurable:
- $2 \cdot AB \cdot BC$ is commensurable with $AB \cdot BC$.
Therefore from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:
- $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.
But:
- $EH = AB^2 + BC^2$
and:
- $HF = 2 \cdot AB \cdot BC$
Therefore $EH$ is incommensurable with $HF$.
So from:
and:
it follows that $DH$ is incommensurable in length with $HG$.
Therefore $DH$ and $HG$ are rational straight lines which are commensurable in square only.
By Proposition $11$ of Book $\text{X} $: Binomial is Irrational:
- $DG$ is irrational.
But $DE$ is rational.
From Proposition $20$ of Book $\text{X} $: Quotient of Rationally Expressible Numbers is Rational:
- The rectangle contained by an irrational and a rational straight line is irrational.
Therefore $DF$ is irrational.
From Book $\text{X}$ Definition $4$: Rational Area:
- The side of a square equal to an irrational area is irrational.
But $AC$ is the side of a square equal to $DF$.
Therefore $AC$ is irrational.
Such a straight line is called second bimedial.
$\blacksquare$
Historical Note
This proof is Proposition $38$ 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
- (in which a mistake apppears)