# Proof: By Euclid

• For let the least number measured by $A$, $B$, $C$ have been taken, and let it be $DE$ [Prop. 7.36].
• And let the unit $DF$ have been added to $DE$.
• So $EF$ is either prime, or not.
• Let it, first of all, be prime.
• Thus, the (set of) prime numbers $A$, $B$, $C$, $EF$, (which is) more numerous than $A$, $B$, $C$, has been found.
• And so let $EF$ not be prime.
• Thus, it is measured by some prime number [Prop. 7.31].
• Let it be measured by the prime (number) $G$.
• I say that $G$ is not the same as any of $A$, $B$, $C$.
• For, if possible, let it be (the same).
• And $A$, $B$, $C$ (all) measure $DE$.
• Thus, $G$ will also measure $DE$.
• And it also measures $EF$.
• (So) $G$ will also measure the remainder, unit $DF$, (despite) being a number [Prop. 7.28].
• The very thing (is) absurd.
• Thus, $G$ is not the same as one of $A$, $B$, $C$.
• And it was assumed (to be) prime.
• Thus, the (set of) prime numbers $A$, $B$, $C$, $G$, (which is) more numerous than the assigned multitude (of prime numbers), $A$, $B$, $C$, has been found.
• (Which is) the very thing it was required to show.

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