Greatest Common Measure of Three Commensurable Magnitudes

Theorem

In the words of Euclid:

Given three commensurable magnitudes, to find their greatest common measure.

(The Elements: Book $\text{X}$: Proposition $4$)

Porism

In the words of Euclid:

From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure.
Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended.

(The Elements: Book $\text{X}$: Proposition $4$ : Porism)

Proof

Let $A, B, C$ be three commensurable magnitudes.

From Greatest Common Measure of Commensurable Magnitudes, let the greatest common measure of $A$ and $B$ be $D$.

Either $D$ measures $C$ or it does not.

Let $D$ measure $C$.

Since:

$D$ measures $C$

and:

$D$ measures $A$ and $B$

then:

$D$ is a common measure for $A, B$ and $C$.

As no magnitude can measure a smaller magnitude, it follows that $D$ is the greatest common measure of $A, B$ and $C$.

Let $D$ not measure $C$.

As $A$, $B$ and $C$ are commensurable, then some magnitude $E$ will measure them by definition.

$E$ will measure both $A$ and $B$.

From the porism to Greatest Common Measure of Commensurable Magnitudes, $E$ also measures their greatest common measure $D$.

But $E$ also measures $C$.

Therefore $E$ will measure $C$ and $D$.

Therefore $C$ and $D$ are commensurable.

So from Greatest Common Measure of Commensurable Magnitudes, let $E$ be the greatest common measure of $C$ and $D$.

Since:

$E$ measures $D$

and:

$D$ measures $A$ and $B$

then:

$E$ measures $A$ and $B$.

But $E$ measures $C$ also.

So $E$ is a common measure for $A, B$ and $C$.

Suppose there were some magnitude $F$ which is greater than $E$ which is also a common measure for $A, B$ and $C$.

Since $F$ measures $A, B$ and $C$, it measures $A$ and $B$.

From the porism to Greatest Common Measure of Commensurable Magnitudes, $F$ also measures the greatest common measure of $A$ and $B$.

That is, $F$ measures $C$

But $F$ measures $D$.

From the porism to Greatest Common Measure of Commensurable Magnitudes, $F$ measures the greatest common measure of $D$ and $C$.

That is, $F$ measures $E$.

But $F$ is greater than $E$ and so cannot measure $E$.

From this contradiction it follows that no magnitude greater than $E$ can measure $A, B$ and $C$.

The result follows: $E$ is the greatest common measure of $A, B$ and $C$, unless $D$ measures $C$.

If $D$ measures $C$ then $D$ is the greatest common measure of $A, B$ and $C$.

$\blacksquare$

Historical Note

This proof is Proposition $4$ 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