# Medial Area not greater than Medial Area by Rational Area

## Theorem

In the words of Euclid:

A medial area does not exceed a medial area by a rational area.

## Proof

Let $\rho \cdot \sqrt k \rho$ and $\rho \cdot \sqrt \lambda \rho$ be two medial areas which have been applied to the same rational number $\rho$.

Their difference is then:

$\paren {\sqrt k - \sqrt \lambda} \rho^2$

Let $x = \sqrt k \rho$ and $y = \sqrt \lambda \rho$.

Let:

$\rho \paren {x - y} = \rho z$

Suppose $\rho z$ is a rational area.

Then $z$ must be a rational straight line and:

$(1): \quad z \frown \rho$

where $\frown$ denotes commensurability in length.

Since $\rho x$ and $\rho y$ are medial areas, it follows that $x$ and $y$ are rational and:

$(2): \quad x \smile \rho$ and $y \smile \rho$

where $\smile$ denotes incommensurability in length.

Then from $(1)$ and $(2)$:

$y \smile z$

Then:

 $\ds y : z$ $=$ $\ds y^2 : y z$ $\ds y^2$ $\smile$ $\ds y z$

But:

 $\ds y^2 + z^2$ $\frown$ $\ds y^2$ $\ds 2 y z$ $\frown$ $\ds y z$ $\ds \leadsto \ \$ $\ds y^2 + z^2$ $\smile$ $\ds 2 y z$ $\ds \leadsto \ \$ $\ds \paren {y + z}^2$ $\smile$ $\ds \paren {y^2 + z^2}$ $\ds \leadsto \ \$ $\ds x^2$ $\smile$ $\ds \paren {y^2 + z^2}$

But $y^2 + z^2$ is rational.

Therefore $x^2$ must be irrational.

Therefore $x$ must be irrational.

But from $(2)$ $x$ is rational.

The result follows from this contradiction.

$\blacksquare$

## Historical Note

This proof is Proposition $26$ of Book $\text{X}$ of Euclid's The Elements.