That which produces Medial Whole with Medial Area is Irrational

Theorem

In the words of Euclid:

If from a straight line there be subtracted a straight line which is incommensurable in square with the whole and which with the whole makes the sum of the squares on them medial, twice the rectangle contained by them medial, and further the squares on them incommensurable with twice the rectangle contained by them, the remainder is irrational; and let it be called that which produces with a medial area a medial whole.

(The Elements: Book $\text{X}$: Proposition $78$)

Proof

Let $AB$ be a straight line.

Let a straight line $BC$ such that:

$BC$ is incommensurable in square with $AB$

$AB^2 + BC^2$ is medial

twice the rectangle contained by $AB$ and $BC$ is medial

$AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$

be cut off from $AB$.

Let $DI$ be a rational straight line.

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

Let $DE$ be a parallelogram set out on $DI$ equal to $AB^2 + BC^2$.

Let its breadth be $DG$.

Similarly:

Let $DH$ be a parallelogram set out on $DI$ equal to $2 \cdot AB \cdot BC$.

From Proposition $7$ of Book $\text{II} $: Square of Difference:

$FE = AC^2$

We have that $AB^2 + BC^2$ is medial and equal to $DE$.

Therefore $DE$ is medial.

We have that $DE$ has been applied to the rational straight line $DI$ producing $DG$ as breadth.

So from Proposition $22$ of Book $\text{X} $: Square on Medial Straight Line:

$DG$ is rational and incommensurable in length with $DI$.

We have that $2 \cdot AB \cdot BC$ is medial.

But $2 \cdot AB \cdot BC = DH$.

Therefore $DH$ is medial.

We have that $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.

Therefore $DE$ is incommensurable with $DH$.

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

$DE : DH = DG : DF$

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

$DG$ is incommensurable with $DF$.

But both $GD$ and $DF$ are rational.

Therefore $GD$ and $DF$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $FG$ is an apotome.

From Proposition $20$ of Book $\text{X} $: Quotient of Rationally Expressible Numbers is Rational:

a rectangle contained by a rational and an irrational straight line is irrational.

Hence its "side" is irrational.

But $AC$ is the "side" of $FE$.

Therefore $AC$ is irrational.

Such a straight line is known as that which produces with a medial area a medial whole.

$\blacksquare$

Historical Note

This proof is Proposition $78$ 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