# Proof: By Euclid

• For as many as is the multitude of $A$, $B$, $C$, $D$, let so many (numbers), $E$, $HK$, $L$, $M$, have been taken in a double proportion, (starting) from $E$.
• Thus, via equality, as $A$ is to $D$, so $E$ (is) to $M$ [Prop. 7.14].
• Thus, the (number created) from (multiplying) $E, D$ is equal to the (number created) from (multiplying) $A$, $M$.
• And $FG$ is the (number created) from (multiplying) $E$, $D$.
• Thus, $FG$ is also the (number created) from (multiplying) $A$, $M$ [Prop. 7.19].
• Thus, $A$ has made $FG$ (by) multiplying $M$.
• Thus, $M$ measures $FG$ according to the units in $A$.
• And $A$ is a dyad.
• Thus, $FG$ is double $M$.
• And $M$, $L$, $HK$, $E$ are also continuously double one another.
• Thus, $E$, $HK$, $L$, $M$, $FG$ are in continued proportion in a double proportion.
• So let $HN$ and $FO$, each equal to the first (number) $E$, have been subtracted from the second (number) $HK$ and the last $FG$ (respectively).

• Thus, as the excess of the second number is to the first, so the excess of the last (is) to (the sum of) all those (numbers) before it [Prop. 9.35].
• Thus, as $NK$ is to $E$, so $OG$ (is) to $M$, $L$, $KH$, $E$.
• And $NK$ is equal to $E$.
• And thus $OG$ is equal to $M$, $L$, $HK$, $E$.
• And $FO$ is also equal to $E$, and $E$ to $A$, $B$, $C$, $D$, and a unit.
• Thus, the whole of $FG$ is equal to $E$, $HK$, $L$, $M$, and $A$, $B$, $C$, $D$, and a unit.
• And it is measured by them.
• I also say that $FG$ will be measured by no other (numbers) except $A$, $B$, $C$, $D$, $E$, $HK$, $L$, $M$, and a unit.
• For, if possible, let some (number) $P$ measure $FG$, and let $P$ not be the same as any of $A$, $B$, $C$, $D$, $E$, $HK$, $L$, $M$.
• And as many times as $P$ measures $FG$, so many units let there be in $Q$.
• Thus, $Q$ has made $FG$ (by) multiplying $P$.
• But, in fact, $E$ has also made $FG$ (by) multiplying $D$.
• Thus, as $E$ is to $Q$, so $P$ (is) to $D$ [Prop. 7.19].
• And since $A$, $B$, $C$, $D$ are continually proportional, (starting) from a unit, $D$ will thus not be measured by any other numbers except $A$, $B$, $C$ [Prop. 9.13].
• And $P$ was assumed not (to be) the same as any of $A$, $B$, $C$.
• Thus, $P$ does not measure $D$.
• But, as $P$ (is) to $D$, so $E$ (is) to $Q$.
• Thus, $E$ does not measure $Q$ either [Def. 7.20] .
• And $E$ is a prime (number) .
• And every prime number [is] [prime to]bookofproofs$1288 every (number) which it does not measure [Prop. 7.29]. • Thus,$E$and$Q$are prime to one another. • And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20]. • And as$E$is to$Q$, (so)$P$(is) to$D$. • Thus,$E$measures$P$the same number of times as$Q$(measures)$D$. • And$D$is not measured by any other (numbers) except$A$,$B$,$C$. • Thus,$Q$is the same as one of$A$,$B$,$C$. • Let it be the same as$B$. • And as many as is the multitude of$B$,$C$,$D$, let so many (of the set out numbers) have been taken, (starting) from$E$, (namely)$E$,$HK$,$L$. • And$E$,$HK$,$L$are in the same ratio as$B$,$C$,$D$. • Thus, via equality, as$B$(is) to$D$, (so)$E$(is) to$L$[Prop. 7.14]. • Thus, the (number created) from (multiplying)$B$,$L$is equal to the (number created) from multiplying$D$,$E$[Prop. 7.19]. • But, the (number created) from (multiplying)$D$,$E$is equal to the (number created) from (multiplying)$Q$,$P$. • Thus, the (number created) from (multiplying)$Q$,$P$is equal to the (number created) from (multiplying)$B$,$L$. • Thus, as$Q$is to$B$, (so)$L$(is) to$P$[Prop. 7.19]. • And$Q$is the same as$B$. • Thus,$L$is also the same as$P$. • The very thing (is) impossible. • For$P$was assumed not (to be) the same as any of the (numbers) set out. • Thus,$FG$cannot be measured by any number except$A$,$B$,$C$,$D$,$E$,$HK$,$L$,$M$, and a unit. • And$FG$was shown (to be) equal to (the sum of)$A$,$B$,$C$,$D$,$E$,$HK$,$L$,$M$, and a unit. • And a perfect number is one which is equal to (the sum of) its own parts [Def. 7.22] . • Thus,$FG\$ is a perfect (number) .
• (Which is) the very thing it was required to show.

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### References

#### Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"