Square on Medial Straight Line

Theorem

In the words of Euclid:

The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.

(The Elements: Book $\text{X}$: Proposition $22$)

Lemma

In the words of Euclid:

If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.

(The Elements: Book $\text{X}$: Proposition $22$ : Lemma)

Proof

Let $A = \rho \sqrt [4] k$ be a medial straight line.

Let $BC = \sigma$ be a rational straight line.

The square on $A$ is $\rho^2 \sqrt k$.

The breadth $CD$ of the rectangle whose area is $\rho^2 \sqrt k$ and whose side is $BC$ is:

\(\ds CD\)

\(=\)

\(\ds \frac {\rho^2 \sqrt k} \sigma\)

\(\ds \)

\(=\)

\(\ds \dfrac {\rho^2 \sqrt k} {\sigma^2} \sigma\)

\(\ds \)

\(=\)

\(\ds \sqrt k \dfrac m n \sigma\)

where $m$ and $n$ are integers.

Thus $CD$ can be expressed in the form:

$CD = \sqrt {k'} \sigma$

which is commensurable in square only with $\sigma$.

By definition, therefore, $CD$ is incommensurable in length with $BC$.

$\blacksquare$

Historical Note

This proof is Proposition $22$ 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