Proof: By Euclid

For, $AB$ and $CD$ being two unequal magnitudes, and $AB$ (being) the lesser, let the remainder never measure the (magnitude) before it, (when) the lesser (magnitude is) continually subtracted in turn from the greater.
I say that the magnitudes $AB$ and $CD$ are incommensurable.
For if they are commensurable then some magnitude will measure them (both).
If possible, let it (so) measure (them), and let it be $E$.
And let $AB$ leave $CF$ less than itself (in) measuring $FD$, and let $CF$ leave $AG$ less than itself (in) measuring $BG$, and let this happen continually, until some magnitude which is less than $E$ is left.
Let (this) have occurred,¹ and let $AG$, (which is) less than $E$, have been left.
Therefore, since $E$ measures $AB$, but $AB$ measures $DF$, $E$ will thus also measure $FD$.
And it also measures the whole (of) $CD$.
Thus, it will also measure the remainder $CF$.
But, $CF$ measures $BG$.
Thus, $E$ also measures $BG$.
And it also measures the whole (of) $AB$.
Thus, it will also measure the remainder $AG$, the greater (measuring) the lesser.
The very thing is impossible.
Thus, some magnitude cannot measure (both) the magnitudes $AB$ and $CD$.
Thus, the magnitudes $AB$ and $CD$ are incommensurable [Def. 10.1] .
Thus, if ... of two unequal magnitudes, and so on ....

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The fact that this will eventually occur is guaranteed by [Prop. 10.1] (translator's note). ↩