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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