(related to Proposition: 7.33: Least Ratio of Numbers)

- Let $A$, $B$, and $C$ be any given multitude of numbers.
- So it is required to find the least of those (numbers) having the same ratio as $A$, $B$, and $C$.

- For $A$, $B$, and $C$ are either prime to one another, or not.
- In fact, if $A$, $B$, and $C$ are prime to one another then they are the least of those (numbers) having the same ratio as them [Prop. 7.22].
- And if not, let the greatest common measure, $D$, of $A$, $B$, and $C$ have be taken [Prop. 7.3].
- And as many times as $D$ measures $A$, $B$, $C$, so many units let there be in $E$, $F$, $G$, respectively.
- And thus $E$, $F$, $G$ measure $A$, $B$, $C$, respectively, according to the units in $D$ [Prop. 7.15].
- Thus, $E$, $F$, $G$ measure $A$, $B$, $C$ (respectively) an equal number of times.
- Thus, $E$, $F$, $G$ are in the same ratio as $A$, $B$, $C$ (respectively) [Def. 7.20] .
- So I say that (they are) also the least (of those numbers having the same ratio as $A$, $B$, $C$).
- For if $E$, $F$, $G$ are not the least of those (numbers) having the same ratio as $A$, $B$, $C$ (respectively), then there will be [some] [numbers]bookofproofs$2315 less than $E$, $F$, $G$ which are in the same ratio as $A$, $B$, $C$ (respectively).
- Let them be $H$, $K$, $L$.
- Thus, $H$ measures $A$ the same number of times that $K$, $L$ also measure $B$, $C$, respectively.
- And as many times as $H$ measures $A$, so many units let there be in $M$.
- Thus, $K$, $L$ measure $B$, $C$, respectively, according to the units in $M$.
- And since $H$ measures $A$ according to the units in $M$, $M$ thus also measures $A$ according to the units in $H$ [Prop. 7.15].
- So, for the same (reasons), $M$ also measures $B$, $C$ according to the units in $K$, $L$, respectively.
- Thus, $M$ measures $A$, $B$, and $C$.
- And since $H$ measures $A$ according to the units in $M$, $H$ has thus made $A$ (by) multiplying $M$.
- So, for the same (reasons), $E$ has also made $A$ (by) multiplying $D$.
- Thus, the (number created) from (multiplying) $E$ and $D$ is equal to the (number created) from (multiplying) $H$ and $M$.
- Thus, as $E$ (is) to $H$, so $M$ (is) to $D$ [Prop. 7.19].
- And $E$ (is) greater than $H$.
- Thus, $M$ (is) also greater than $D$ [Prop. 5.13].
- And ($M$) measures $A$, $B$, and $C$.
- The very thing is impossible.
- For $D$ was assumed (to be) the greatest common measure of $A$, $B$, and $C$.
- Thus, there cannot be any numbers less than $E$, $F$, $G$ which are in the same ratio as $A$, $B$, $C$ (respectively).
- Thus, $E$, $F$, $G$ are the least of (those numbers) having the same ratio as $A$, $B$, $C$ (respectively).
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"