Greatest Common Measure of Commensurable Magnitudes
Theorem
In the words of Euclid:
- Given two commensurable magnitudes, to find their greatest common measure.
(The Elements: Book $\text{X}$: Proposition $3$)
Porism
In the words of Euclid:
- From this it is manifest that, if a magnitude measure two magnitudes, it will also measure their greatest common measure.
(The Elements: Book $\text{X}$: Proposition $3$ : Porism)
Proof
Let $AB$ and $CD$ be two commensurable magnitudes such that $AB < CD$.
$AB$ either measures $CD$ or it does not.
Let $AB$ measure $CD$.
As $AB$ also measures itself, it follows that $AB$ is a common measure for $AB$ and $CD$.
As no magnitude can measure a smaller magnitude, it follows that $AB$ is the greatest common measure of $AB$ and $CD$.
Let $AB$ not measure $CD$.
Let the less be continually subtracted in turn from the greater.
We are given that $AB$ and $CD$ are commensurable.
From Incommensurable Magnitudes do not Terminate in Euclid's Algorithm, that which is left over will eventually measure the one before it.
Let $AB$ measure $ED$ and leave $EC$ less than $AB$.
Let $EC$ measure $FB$ and leave $AF$ less than $EC$.
Let $AF$ measure $CE$.
Since:
- $AF$ measure $CE$
and:
- $CE$ measure $FB$
it follows that:
- $AF$ measure $FB$.
But $AF$ also measures itself.
Therefore $AF$ measures the whole of $AB$.
But $AB$ also measures $DE$.
Therefore $AF$ also measures $DE$.
But $AB$ also measures $CE$.
Therefore $AF$ also measures the whole of $CD$.
Therefore $AF$ is a common measure for $AB$ and $CD$.
Suppose there were some magnitude $G$ which is greater than $AF$ which is also a common measure for $AB$ and $CD$.
Since:
- $G$ measures $AB$
and:
- $AB$ measures $ED$
it follows that:
- $G$ measures $ED$.
But $G$ also measures the whole of $CD$.
Therefore $G$ also measures the remainder $CE$.
But $CE$ measures $FB$.
Therefore $G$ also measures $FB$.
But $G$ also measures the whole of $AB$.
Therefore it also measures the remainder $AF$.
But $G$ is greater than $AF$, and so cannot measure it.
From this contradiction it follows that $G$ cannot be a common measure for $AB$ and $CD$.
That is, $AF$ is the greatest common measure of $AB$ and $CD$.
$\blacksquare$
Historical Note
This proof is Proposition $3$ 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