# Proof: By Euclid

• Let $AB$ and $CD$ be the two given magnitudes, of which (let) $AB$ (be) the lesser.
• So, it is required to find the greatest common measure of $AB$ and $CD$. • For the magnitude $AB$ either measures, or (does) not (measure), $CD$.
• Therefore, if it measures ($CD$), and (since) it also measures itself, $AB$ is thus a common measure of $AB$ and $CD$.
• And (it is) clear that (it is) also (the) greatest.
• For a (magnitude) greater than magnitude $AB$ cannot measure $AB$.
• So let $AB$ not measure $CD$.
• And continually subtracting in turn the lesser (magnitude) from the greater, the remaining (magnitude) will (at) some time measure the (magnitude) before it, on account of $AB$ and $CD$ not being incommensurable [Prop. 10.2].
• And let $AB$ leave $EC$ less than itself (in) measuring $ED$, and let $EC$ leave $AF$ less than itself (in) measuring $FB$, and let $AF$ measure $CE$.
• Therefore, since $AF$ measures $CE$, but $CE$ measures $FB$, $AF$ will thus also measure $FB$.
• And it also measures itself.
• Thus, $AF$ will also measure the whole (of) $AB$.
• But, $AB$ measures $DE$.
• Thus, $AF$ will also measure $ED$.
• And it also measures $CE$.
• Thus, it also measures the whole of $CD$.
• Thus, $AF$ is a common measure of $AB$ and $CD$.
• So I say that (it is) also (the) greatest (common measure).
• For, if not, there will be some magnitude, greater than $AF$, which will measure (both) $AB$ and $CD$.
• Let it be $G$.
• Therefore, since $G$ measures $AB$, but $AB$ measures $ED$, $G$ will thus also measure $ED$.
• And it also measures the whole of $CD$.
• Thus, $G$ will also measure the remainder $CE$.
• But $CE$ measures $FB$.
• Thus, $G$ will also measure $FB$.
• And it also measures the whole (of) $AB$.
• And (so) it will measure the remainder $AF$, the greater (measuring) the lesser.
• The very thing is impossible.
• Thus, some magnitude greater than $AF$ cannot measure (both) $AB$ and $CD$.
• Thus, $AF$ is the greatest common measure of $AB$ and $CD$.
• Thus, the greatest common measure of two given commensurable magnitudes, $AB$ and $CD$, has been found.
• (Which is) the very thing it was required to show.

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