# Commensurability of Squares

## Theorem

In the words of Euclid:

*The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either.*

(*The Elements*: Book $\text{X}$: Proposition $9$)

### Porism

In the words of Euclid:

*And it is manifest from what has been proven that straight lines commensurable in length are always commensurable in square also, but those commensurable in square are not always commensurable in length also.*

(*The Elements*: Book $\text{X}$: Proposition $9$ : Porism)

## Proof

Let $A$ and $B$ be commensurable in length.

Then $A$ has to $B$ the ratio which a number has to a number.

Let $\dfrac A B = \dfrac C D$.

From Similar Polygons are composed of Similar Triangles: Porism:

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

$\Box$

Let:

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

Then it follows that:

- $\dfrac A B = \dfrac C D$

and from Magnitudes with Rational Ratio are Commensurable it follows that $A$ and $B$ are commensurable in length.

$\Box$

Let $A$ and $B$ be incommensurable in length.

Suppose:

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

where $C$ and $D$ are integers.

Then $A$ and $B$ would be commensurable in length, which they are not.

So the square on $A$ and the square on $B$ do not have to one another the ratio which a square number has to a square number.

$\Box$

Let the square on $A$ not have to the square on $B$ not have the ratio which a square number has to a square number.

Suppose $A$ was commensurable in length with $B$.

Then the square on $A$ would have to the square on $B$ the ratio which a square number has to a square number, which it does not.

So $A$ is not commensurable in length with $B$.

$\blacksquare$

## Historical Note

This proof is Proposition $9$ 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