If as many numbers as we please beginning from a unit are in continued proportion, then by whatever prime numbers the last is measured, the next to the unit is also measured by the same.

Let *A, B, C,* and *D* be as many numbers as we please beginning from a unit in continued proportion.
I say that, by whatever prime numbers *D* is measured, *A* is also measured by the same.

Let *D* be measured by any prime number *E.*

I say that *E* measures *A.*

Suppose it does not.

Now *E* is prime, and any prime number is relatively prime to any which it does not measure, therefore *E* and *A* are relatively prime. And, since *E* measures *D,* let it measure it according to *F,* therefore *E* multiplied by *F* makes *D.*

Again, since *A* measures *D* according to the units in *C,* therefore *A* multiplied by *C* makes *D.* But, further, *E* multiplied by *F* makes *D,* therefore the product of *A* and *C* equals the product of *E* and *F.*

Therefore *A* is to *E* as *F* is to *C.* But *A* and *E* are relatively prime, primes are also least, and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent, therefore *E* measures *C.* Let it measure it according to *G.*

Therefore *E* multiplied by *G* makes *C.* But, further, by the previous theorem, *A* multiplied by *B* makes *C.* Therefore the product of *A* and *B* equals the product of *E* and *G.*

Therefore *A* is to *E* as *G* is to *B.* But *A* and *E* are relatively prime, primes are also least, and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent, therefore *E* measures *B.* Let it measure it according to *H.* Then *E* multiplied by *H* makes *B.*

But, further, *A* multiplied by itself makes *B,* therefore the product of *E* and *H* equals the square on *A.* Therefore *E* is to *A* as *A* is to *H.*

But *A* and *E* are relatively prime, primes are also least, and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent, therefore *E* measures *A* antecedent antecedent. But, again, it also does not measure it, which is impossible.

Therefore *E* and *A* are not relatively prime. Therefore they are relatively composite. But numbers relatively composite are measured by some number.

And, since *E* is by hypothesis prime, and a prime is not measured by any number other than itself, therefore *E* measures *A* and *E,* so that *E* measures *A.*

But it also measures *D,* therefore *E* measures *A* and *D.* Similarly we can prove that, by whatever prime numbers *D* is measured, *A* also is measured by the same.

Therefore, *if as many numbers as we please beginning from a unit are in continued proportion, then by whatever prime numbers the last is measured, the next to the unit is also measured by the same.*

Q.E.D.

Assume that a prime number *p* divides a power *a*^{k} of a number *a.* Suppose that *p* does not divide *a.* Then *p* is relatively prime to *a* (VII.29). From the proportion

we see that the ratio (*a*^{k}/*p*) : *a*^{k-1} reduces to *a* : *p* in lowest terms (VII.21). Therefore, *p* divides *a*^{k-1} (VII.20).

Apply this reduction step repeatedly until the conclusion *p* divides *a* is reached.