---
book: 9
number: 36
id: "IX.36"
kind: "theorem"
uses: ["[[book-7/proposition-14]]", "[[book-7/proposition-19]]", "[[book-9/proposition-35]]", "[[book-9/proposition-13]]", "[[book-7/proposition-29]]", "[[book-7/proposition-21]]", "[[book-7/proposition-20]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:9.prop.36"
license: "CC-BY-SA-4.0"
---

# IX.36

*If as many numbers as we please beginning from an unit be set out continuously in double proportion, until the sum of all becomes prime, and if the sum multiplied into the last make some number, the product will be perfect.*

## Proof

For let as many numbers as we please, *A*, *B*, *C*, *D*, beginning from an unit be set out in double proportion, until the sum of all becomes prime, let *E* be equal to the sum, and let *E* by multiplying *D* make *FG*; I say that *FG* is perfect.

For, however many *A*, *B*, *C*, *D* are in multitude, let so many *E*, *HK*, *L*, *M* be taken in double proportion beginning from *E*; therefore, ex aequali, as *A* is to *D*, so is *E* to *M*. [[book-7/proposition-14|VII. 14]]

Therefore the product of *E*, *D* is equal to the product of *A*, *M*. [[book-7/proposition-19|VII. 19]]

And the product of *E*, *D* is *FG*; therefore the product of *A*, *M* is also *FG*.

Therefore *A* by multiplying *M* has made *FG*; therefore *M* measures *FG* according to the units in *A*.

And *A* is a dyad; therefore *FG* is double of *M*.

But *M*, *L*, *HK*, *E* are continuously double of each other; therefore *E*, *HK*, *L*, *M*, *FG* are continuously proportional in double proportion.

Now let there be subtracted from the second *HK* and the last *FG* the numbers *HN*, *FO*, each equal to the first *E*; therefore, as the excess of the second is to the first, so is the excess of the last to all those before it. [[book-9/proposition-35|IX. 35]]

Therefore, as *NK* is to *E*, so is *OG* to *M*, *L*, *KH*, *E*.

And *NK* is equal to *E*; therefore *OG* is also equal to *M*, *L*, *HK*, *E*.

But *FO* is also equal to *E*, and *E* is equal to *A*, *B*, *C*, *D* and the unit.

Therefore the whole *FG* is equal to *E*, *HK*, *L*, *M* and *A*, *B*, *C*, *D* and the unit; and it is measured by them.

I say also that *FG* will not be measured by any other number except *A*, *B*, *C*, *D*, *E*, *HK*, *L*, *M* and the unit.

For, if possible, let some number *P* measure *FG*, and let *P* not be the same with any of the numbers *A*, *B*, *C*, *D*, *E*, *HK*, *L*, *M*.

And, as many times as *P* measures *FG*, so many units let there be in *Q*; therefore *Q* by multiplying *P* has made *FG*.

But, further, *E* has also by multiplying *D* made *FG*; therefore, as *E* is to *Q*, so is *P* to *D*. [[book-7/proposition-19|VII. 19]]

And, since *A*, *B*, *C*, *D* are continuously proportional beginning from an unit, therefore *D* will not be measured by any other number except *A*, *B*, *C*. [[book-9/proposition-13|IX. 13]]

And, by hypothesis, *P* is not the same with any of the numbers *A*, *B*, *C*; therefore *P* will not measure *D*.

But, as *P* is to *D*, so is *E* to *Q*; therefore neither does *E* measure *Q*. [[book-7/definitions#Definition 20|VII. Def. 20]]

And *E* is prime; and any prime number is prime to any number which it does not measure. [[book-7/proposition-29|VII. 29]]

Therefore *E*, *Q* are prime to one another.

But primes are also least, [[book-7/proposition-21|VII. 21]] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [[book-7/proposition-20|VII. 20]] and, as *E* is to *Q*, so is *P* to *D*; therefore *E* measures *P* the same number of times that *Q* measures *D*.

But *D* is not measured by any other number except *A*, *B*, *C*; therefore *Q* is the same with one of the numbers *A*, *B*, *C*.

Let it be the same with *B*.

And, however many *B*, *C*, *D* are in multitude, let so many *E*, *HK*, *L* be taken beginning from *E*.

Now *E*, *HK*, *L* are in the same ratio with *B*, *C*, *D*; therefore, ex aequali, as *B* is to *D*, so is *E* to *L*. [[book-7/proposition-14|VII. 14]]

Therefore the product of *B*, *L* is equal to the product of *D*, *E*. [[book-7/proposition-19|VII. 19]]

But the product of *D*, *E* is equal to the product of *Q*, *P*; therefore the product of *Q*, *P* is also equal to the product of *B*, *L*.

Therefore, as *Q* is to *B*, so is *L* to *P*. [[book-7/proposition-19|VII. 19]]

And *Q* is the same with *B*; therefore *L* is also the same with *P*; which is impossible, for by hypothesis *P* is not the same with any of the numbers set out.

Therefore no number will measure *FG* except *A*, *B*, *C*, *D*, *E*, *HK*, *L*, *M* and the unit.

And *FG* was proved equal to *A*, *B*, *C*, *D*, *E*, *HK*, *L*, *M* and the unit; and a perfect number is that which is equal to its own parts; [[book-7/definitions#Definition 22|VII. Def. 22]] therefore *FG* is perfect. Q. E. D.