# Proof: By Euclid

(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.

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