# Construction of Components of Major

## Theorem

In the words of Euclid:

To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.

### Lemma

In the words of Euclid:

Let $ABC$ be a right-angled triangle having the angle $A$ right, and let the perpendicular $AD$ be drawn;

I say that the rectangle $CB, BD$ is equal to the square on $BA$, the rectangle $BC, CD$ equal to the square on $CA$, the rectangle $BD, DC$ equal to the square on $AD$, and, further, the rectangle $BC, AD$ equal to the rectangle $BA, AC$.

## Proof

Let $AB$ and $BC$ be rational straight lines which are commensurable in square only such that:

$AB^2 = BC^2 + \rho^2$

such that $\rho$ is incommensurable in length with $AB$.

Let $BC$ be bisected at $D$.

Let a parallelogram be applied to $AB$ equal to the square on either of $BD$ or $DC$, and deficient by a square.

Let this parallelogram be the rectangle contained by $AE$ and $EB$.

Let the semicircle $AFB$ be drawn with $AB$ as the diameter.

Let $EF$ be drawn perpendicular to $AB$.

Join $AF$ and $FB$.

We have that $AB > BC$ such that $AB^2 = BC^2 + \rho^2$ such that $\rho$ is incommensurable in length with $AB$.

We also have that the rectangle contained by $AE$ and $EB$ equals the parallelogram on $AB$ equal to $\dfrac {BC} 4$.

$AE$ is incommensurable in length with $EB$.

We have that:

$AE : BE = AB \cdot AE : AB \cdot BE$
$AB \cdot AE = AF^2$

and:

$AB \cdot BE = BF^2$

Therefore $AF^2$ and $BF^2$ are incommensurable.

By definition, $AF$ and $BF$ are therefore incommensurable in square.

As $AB$ is a rational straight line, it follows by definition that $AB^2$ is a rational area.

From Pythagoras's Theorem:

$AB^2 = \left({AF + FB}\right)^2$

Thus $\left({AF + FB}\right)^2$ is also a rational area.

Therefore $AF + FB$ is rational.

$AE \cdot EB = EF^2$

and by hypothesis:

$AE \cdot EB = BD^2$

then:

$FE = BD$

Therefore:

$BC = 2 FE$

Thus $AB \cdot BC$ is commensurable with $AB \cdot EF$.

But from Medial is Irrational:

$AB \cdot BC$ is medial.
$AB \cdot EF$ is also medial.
$AB \cdot EF = AF \cdot FB$

Therefore $AF \cdot FB$ is also medial.

But it has been proved that $AF + FB$ is rational.

Therefore we have found two rational straight lines which are incommensurable in square whose sum of squares is rational, but such that the rectangle contained by them is medial.

$\blacksquare$

## Historical Note

This proof is Proposition $33$ of Book $\text{X}$ of Euclid's The Elements.