(related to Proposition: 9.36: Theorem of Even Perfect Numbers (First Part))

- 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.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"