Second Apotome of Medial is Irrational

Theorem

In the words of Euclid:

If from a medial straight line there be subtracted a medial straight line commensurable with the whole in square only, and which contains with the whole a medial rectangle, the remainder is irrational; and let it be called a second apotome of a medial straight line.

(The Elements: Book $\text{X}$: Proposition $75$)

Proof

Let $AB$ be a medial straight line.

Let a medial straight line $CB$ such that:

$CB$ is commensurable in square only with $AB$

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

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 $AB^2$ and $BC^2$ are medial areas which are commensurable.

Therefore from:

Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes

and:

Porism to Proposition $23$ of Book $\text{X} $: Straight Line Commensurable with Medial Straight Line is Medial:

it follows that:

$DE$ 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 $AB \cdot BC$ is medial.

So fromPorism to Proposition $23$ of Book $\text{X} $: Straight Line Commensurable with Medial Straight Line is Medial:

$2 \cdot AB \cdot BC$ is medial.

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

Therefore $DH$ is medial.

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

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

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

We have that $AB$ and $BC$ are commensurable in square only.

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

$AB$ is incommensurable in length with $BC$.

But from Proposition $15$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:

$AB^2 + BC^2$ are commensurable with $AB^2$

and from Proposition $6$ of Book $\text{X} $: Commensurability of Sum of Commensurable Magnitudes:

$2 \cdot AB \cdot BC$ is commensurable with $AB \cdot BC$

therefore from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:

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

But:

$DE = AB^2 + BC^2$

and:

$DH = 2 \cdot AB \cdot BC$

and so $DE$ is incommensurable with $DH$.

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

$DE : DH = GD : DF$

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

$GD$ 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.

But $DI$ is rational.

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 a second apotome of a medial.

$\blacksquare$

Historical Note

This proof is Proposition $75$ 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