# Proof: By Euclid

• For $A$ and $B$ are either prime to one another, or not.
• Let them, first of all, be prime to one another.
• And let $A$ make $C$ (by) multiplying $B$.
• Thus, $B$ has also made $C$ (by) multiplying $A$ [Prop. 7.16].
• Thus, $A$ and $B$ (both) measure $C$.
• So I say that ($C$) is also the least (number which they both measure).
• For if not, $A$ and $B$ will (both) measure some (other) number which is less than $C$.
• Let them (both) measure $D$ (which is less than $C$).
• And as many times as $A$ measures $D$, so many units let there be in $E$.
• And as many times as $B$ measures $D$, so many units let there be in $F$.
• Thus, $A$ has made $D$ (by) multiplying $E$, and $B$ has made $D$ (by) multiplying $F$.
• Thus, the (number created) from (multiplying) $A$ and $E$ is equal to the (number created) from (multiplying) $B$ and $F$.
• Thus, as $A$ (is) to $B$, so $F$ (is) to $E$ [Prop. 7.19].
• And $A$ and $B$ are prime (to one another), and prime (numbers) are the least (of those numbers having the same ratio) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser [Prop. 7.20].
• Thus, $B$ measures $E$, as the following (number measuring) the following.
• And since $A$ has made $C$ and $D$ (by) multiplying $B$ and $E$ (respectively), thus as $B$ is to $E$, so $C$ (is) to $D$ [Prop. 7.17].
• And $B$ measures $E$.
• Thus, $C$ also measures $D$, the greater (measuring) the lesser.
• The very thing is impossible.
• Thus, $A$ and $B$ do not (both) measure some number which is less than $C$.
• Thus, $C$ is the least (number) which is measured by (both) $A$ and $B$.

• So let $A$ and $B$ be not prime to one another.
• And let the least numbers, $F$ and $E$, have been taken having the same ratio as $A$ and $B$ (respectively) [Prop. 7.33].
• Thus, the (number created) from (multiplying) $A$ and $E$ is equal to the (number created) from (multiplying) $B$ and $F$ [Prop. 7.19].
• And let $A$ make $C$ (by) multiplying $E$.
• Thus, $B$ has also made $C$ (by) multiplying $F$.
• Thus, $A$ and $B$ (both) measure $C$.
• So I say that ($C$) is also the least (number which they both measure).
• For if not, $A$ and $B$ will (both) measure some number which is less than $C$.
• Let them (both) measure $D$ (which is less than $C$).
• And as many times as $A$ measures $D$, so many units let there be in $G$.
• And as many times as $B$ measures $D$, so many units let there be in $H$.
• Thus, $A$ has made $D$ (by) multiplying $G$, and $B$ has made $D$ (by) multiplying $H$.
• Thus, the (number created) from (multiplying) $A$ and $G$ is equal to the (number created) from (multiplying) $B$ and $H$.
• Thus, as $A$ is to $B$, so $H$ (is) to $G$ [Prop. 7.19].
• And as $A$ (is) to $B$, so $F$ (is) to $E$.
• Thus, also, as $F$ (is) to $E$, so $H$ (is) to $G$.
• And $F$ and $E$ are the least (numbers having the same ratio as $A$ and $B$), and the least (numbers) measure those (numbers) having the same ratio an equal number of times, the greater (measuring) the greater, and the lesser the lesser [Prop. 7.20].
• Thus, $E$ measures $G$.
• And since $A$ has made $C$ and $D$ (by) multiplying $E$ and $G$ (respectively), thus as $E$ is to $G$, so $C$ (is) to $D$ [Prop. 7.17].
• And $E$ measures $G$.
• Thus, $C$ also measures $D$, the greater (measuring) the lesser.
• The very thing is impossible.
• Thus, $A$ and $B$ do not (both) measure some (number) which is less than $C$.
• Thus, $C$ (is) the least (number) which is measured by (both) $A$ and $B$.
• (Which is) the very thing it was required to show.

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### References

#### Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"
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