# Construction of Incommensurable Lines

## Theorem

In the words of Euclid:

*To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line.*

(*The Elements*: Book $\text{X}$: Proposition $10$)

### Lemma

In the words of Euclid:

*It has been proved in the arithmetical books that similar plane numbers have to one another the ratio which a square number has to a square number,*

and that, if two numbers have to one another the ratio which a square number has to a square number, they are similar plane numbers.

And it is manifest from these propositions that numbers which have not to one another the ratio which a square number has to a square number, that is, those which have not their sides proportional, have not to one another the ratio which a square number has to a square number.

(*The Elements*: Book $\text{X}$: Proposition $10$ : Lemma)

## Proof

Let $A$ be the assigned straight line.

Let $B$ and $C$ be two numbers which do not have to one another the ratio which a square number has to a square number.

That is, from the lemma, that $B$ and $C$ are not similar plane numbers.

From Magnitudes with Rational Ratio are Commensurable: Porism, let $D$ be constructed such that:

- $\dfrac {A^2} {D^2} = \dfrac B C$

Therefore from Magnitudes with Rational Ratio are Commensurable, the square on $A$ is commensurable with the square on $D$.

Since:

- $B$ does not have to one $C$ the ratio which a square number has to a square number

then:

- $A^2$ does not have to one $D^2$ the ratio which a square number has to a square number.

Therefore from Commensurability of Squares $A$ is incommensurable in length with $D$.

Let a mean proportional $E$ be taken between $A$ and $D$.

From Book $\text{V}$ Definition $9$: Duplicate Ratio:

- $\dfrac A D = \dfrac {A^2} {E^2}$

But $A$ is incommensurable in length with $D$.

Therefore $A^2$ is incommensurable with $E^2$.

Therefore $A$ is incommensurable in square with $E$.

Therefore two straight lines $D$ and $E$ have been found, such that:

- $D$ is incommensurable in length with $A$
- $E$ is incommensurable in square with $A$.

$\blacksquare$

## Historical Note

This proof is Proposition $10$ of Book $\text{X}$ of Euclid's *The Elements*.

It was suggested by Heiberg, and is generally believed, that this theorem, along with the lemma that accompanies it, is spurious.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions