Incommensurable Magnitudes do not Terminate in Euclid's Algorithm

Proof

 * Euclid-X-2.png

Let $AB$ and $CD$ be unequal magnitudes such that $AB < CD$ which fulfil the condition of the statement.

Suppose $AB$ and $CD$ are commensurable magnitudes.

Then by definition some magnitude will measure them both.

Let $E$ be such a magnitude that measures both $AB$ and $CD$.

Let $AB$ measure $FD$ and leave $CF$ from $CD$.

Let $CF$ measure $BG$ and leave $AG$ from $AB$.

By hypothesis, this process can be repeated indefinitely.

Let it be repeated until some magnitude be left that is less than $E$.

Since:
 * $E$ measures $AB$

and:
 * $AB$ measures $DF$

then:
 * $E$ measures $FD$.

But $E$ also measures $CD$.

Therefore $E$ also measures $CF$.

But $CF$ measures $BG$.

Therefore $E$ also measures $BG$.

But $E$ also measures the whole of $AB$.

Therefore $E$ will also measures $AG$.

But $AG$ is less than $E$.

That means the greater magnitude measures the lesser magnitude.

From this contradiction it follows that no magnitude can measure both $AB$ and $CD$.

Therefore, by definition, $AB$ and $CD$ are incommensurable.