# Proof: By Euclid

• For let $D$, $E$, and $F$ be numbers having the same names as the parts $A$, $B$, and $C$ (respectively).
• And let the least number, $G$, measured by $D$, $E$, and $F$, have been taken [Prop. 7.36].
• Thus, $G$ has parts called the same as $D$, $E$, and $F$ [Prop. 7.37].
• And $A$, $B$, and $C$ are parts called the same as $D$, $E$, and $F$ (respectively).
• Thus, $G$ has the parts $A$, $B$, and $C$.
• So I say that ($G$) is also the least (number having the parts $A$, $B$, and $C$).
• For if not, there will be some number less than $G$ which will have the parts $A$, $B$, and $C$.
• Let it be $H$.
• Since $H$ has the parts $A$, $B$, and $C$, $H$ will thus be measured by numbers called the same as the parts $A$, $B$, and $C$ [Prop. 7.38].
• And $D$, $E$, and $F$ are numbers called the same as the parts $A$, $B$, and $C$ (respectively).
• Thus, $H$ is measured by $D$, $E$, and $F$.
• And ($H$) is less than $G$.
• The very thing is impossible.
• Thus, there cannot be some number less than $G$ which will have the parts $A$, $B$, and $C$.
• (Which is) the very thing it was required to show.

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