# Incommensurability of Sum of Incommensurable Magnitudes

## Theorem

In the words of Euclid:

If two incommensurable magnitudes be added together, the whole will also be incommensurable with one of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.

## Proof

Let $AB$ and $BC$ be incommensurable magnitudes which are added together to make $AC$.

Suppose $AC$ and $AB$ are not incommensurable.

Then some magnitude $D$ will measure them both.

Since $D$ measures both $AC$ and $AB$, $D$ also measures the remainder $BC$.

That is, $D$ measures $AB$ and $BC$.

Therefore from Book $\text{X}$ Definition $1$: Commensurable, $AB$ and $BC$ are commensurable.

But by hypothesis $AB$ and $BC$ are incommensurable.

Therefore no magnitude $D$ will measure both $AC$ and $AB$.

Therefore $AC$ and $AB$ are incommensurable.

In the same way it is shown that $AC$ and $BC$ are also incommensurable.

$\Box$

Now let $AC$ be incommensurable with one of either $AB$ and $BC$.

Without loss of generality, let $AC$ be incommensurable $AB$.

Suppose $AB$ and $BC$ are not incommensurable.

Then some magnitude $D$ will measure them both.

Since $D$ measures both $AB$ and $BC$, $D$ also measures the whole $AC$.

But $D$ measures $AB$.

Therefore $D$ measures both $AC$ and $AB$.

Therefore $AC$ and $AB$ are commensurable.

But by hypothesis $AB$ and $AC$ are incommensurable.

Therefore no magnitude will measure both $AB$ and $BC$.

Therefore $AB$ and $BC$ are incommensurable.

$\blacksquare$

## Historical Note

This proof is Proposition $16$ of Book $\text{X}$ of Euclid's The Elements.