# Proof: By Euclid

• For let the least (number), $D$, measured by the two (numbers) $A$ and $B$ have been taken [Prop. 7.34].
• So $C$ either measures, or does not measure, $D$.
• Let it, first of all, measure ($D$).
• And $A$ and $B$ also measure $D$.
• Thus, $A$, $B$, and $C$ (all) measure $D$.
• So I say that ($D$ is) also the least (number measured by $A$, $B$, and $C$).
• For if not, $A$, $B$, and $C$ will (all) measure [some] number which is less than $D$.
• Let them measure $E$ (which is less than $D$).
• Since $A$, $B$, and $C$ (all) measure $E$ then $A$ and $B$ thus also measure $E$.
• Thus, the least (number) measured by $A$ and $B$ will also measure [$E$] [[Prop. 7.35]]bookofproofs$2365. • And$D$is the least (number) measured by$A$and$B$. • Thus,$D$will measure$E$, the greater (measuring) the lesser. • The very thing is impossible. • Thus,$A$,$B$, and$C$cannot (all) measure some number which is less than$D$. • Thus,$A$,$B$, and$C$(all) measure the least (number)$D$. • So, again, let$C$not measure$D$. • And let the least number,$E$, measured by$C$and$D$have been taken [Prop. 7.34]. • Since$A$and$B$measure$D$, and$D$measures$E$,$A$and$B$thus also measure$E$. • And$C$also measures [$E$]. • Thus,$A$,$B$, and$C$[also] measure$E$. • So I say that ($E$is) also the least (number measured by$A$,$B$, and$C$). • For if not,$A$,$B$, and$C$will (all) measure some (number) which is less than$E$. • Let them measure$F$(which is less than$E$). • Since$A$,$B$, and$C$(all) measure$F$,$A$and$B$thus also measure$F$. • Thus, the least (number) measured by$A$and$B$will also measure$F$[Prop. 7.35]. • And$D$is the least (number) measured by$A$and$B$. • Thus,$D$measures$F$. • And$C$also measures$F$. • Thus,$D$and$C$(both) measure$F$. • Hence, the least (number) measured by$D$and$C$will also measure$F$[Prop. 7.35]. • And$E$is the least (number) measured by$C$and$D$. • Thus,$E$measures$F$, the greater (measuring) the lesser. • The very thing is impossible. • Thus,$A$,$B$, and$C$cannot measure some number which is less than$E$. • Thus,$E$(is) the least (number) which is measured by$A$,$B$, and$C\$.
• (Which is) the very thing it was required to show.

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