Commensurability of Sum of Commensurable Magnitudes

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Theorem

In the words of Euclid:

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

(The Elements: Book $\text{X}$: Proposition $15$)


Proof

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

Since $AB$ and $BC$ are commensurable, then some magnitude $D$ will measure them both.

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

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

Therefore from Book $\text{X}$ Definition $1$: Commensurable, $AC$ is commensurable with each of the magnitudes $AB$ and $BC$.

$\Box$


Let $AC$ be commensurable with $AB$.

Since $AC$ and $AB$ are commensurable, 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 $AB$, $BC$ and $AC$.

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

$\blacksquare$


Historical Note

This proof is Proposition $15$ of Book $\text{X}$ of Euclid's The Elements.


Sources