If numbers fall between two numbers and a unit in continued proportion, then however many numbers fall between each of them and a unit in continued proportion, so many also fall between the numbers themselves in continued proportion.

Let the numbers *D* and *E* and the numbers *F* and *G* respectively fall between the two numbers *A* and *B* and the unit *C* in continued proportion.

I say that, as many numbers have fallen between each of the numbers *A* and *B* and the unit *C* in continued proportion as fall between *A* and *B* in continued proportion.

Multiply *D* by *F* to make *H,* and multiply the numbers *D* and *F* by *H* to make *K* and *L* respectively.

Now, since the unit *C* is to the number *D* as *D* is to *E,* therefore the unit *C* measures the number *D* the same number of times as *D* measures *E.* But the unit *C* measures the number *D* according to the units in *D,* therefore the number *D* also measures *E* according to the units in *D.* Therefore *D* multiplied by itself makes *E.*

Again, since *C* is to the number *D* as *E* is to *A,* therefore the unit *C* measures the number *D* the same number of times as *E* measures *A.* But the unit *C* measures the number *D* according to the units in *D,* therefore *E* also measures *A* according to the units in *D.* Therefore *D* multiplied by *E* makes *A.*

For the same reason also *F* multiplied by itself makes *G,* and multiplied by *G* makes *B.*

And, since *D* multiplied by itself makes *E* and multiplied by *F* makes *H,* therefore *D* is to *F* as *E* is to *H.*

For the same reason also *D* is to *F* as *H* is to *G.* Therefore *E* is to *H* as *H* is to *G.*

Again, since *D* multiplied by the numbers *E* and *H* makes *A* and *K* respectively, therefore *E* is to *H* as *A* is to *K.* But *E* is to *H* as *D* is to *F,* therefore *D* is to *F* as *A* is to *K.*

Again, since the numbers *D* and *F* multiplied by *H* make *K* and *L* respectively, therefore *D* is to *F* as *K* is to *L.* But *D* is to *F* as *A* is to *K,* therefore *A* is to *K* as *K* is to *L.* Further, since *F* multiplied by the numbers *H* and *G* makes *L* and *B* respectively, therefore *H* is to *G* as *L* is to *B.*

But *H* is to *G* as *D* is to *F,* therefore *D* is to *F* as *L* is to *B.* But it was also proved that *D* is to *F* as *A* is to *K* and as *K* is to *L,* therefore *A* is to *K* as *K* is to *L* and as *L* is to *B.* Therefore *A, K, L,* and *B* are in continued proportion.

Therefore, as many numbers as fall between each of the numbers *A* and *B* and the unit *C* in continued proportion, so many also fall between *A* and *B* in continued proportion.

Therefore, *if numbers fall between two numbers and a unit in continued proportion, then however many numbers fall between each of them and a unit in continued proportion, so many also fall between the numbers themselves in continued proportion.*

Q.E.D.

Essentially, the proposition says that if *a* and *b* are (*n*–1)^{st} powers of *d* and *f*, respectively, then the sequence

is in continued proportion.

This is a partial converse to the previous proposition; it doesn’t require that the generating numbers *d* and *f* be relatively prime.

The next two propositions are special cases of this proposition when the powers are squares and cubes.