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