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.

Using Proposition $45$ of Book $\text{I} $: Construction of Parallelogram in Given Angle equal to Given Polygon:

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:

Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base

and:

Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes

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)