If as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is square, then all the rest are also square; and if the number after the unit is cubic, then all the rest are also cubic.

Let there be as many numbers as we please, *A, B, C, D, E,* and *F,* beginning from a unit and in continued proportion, and let *A,* the number after the unit, be square.

I say that all the rest are also square.

Now it was proved that *B,* the third from the unit, is square as are all those which leave out one.

I say that all the rest are also square.

Since *A, B,* and *C* are in continued proportion, and *A* is square, therefore *C* is also square. Again, since *B, C,* and *D* are in continued proportion, and *B* is square, therefore *D* is also square. Similarly we can prove that all the rest are also square.

Next, let *A* be a cube.

I say that all the rest are also cubes.

Now it was proved that *C,* the fourth from the unit, is a cube as are all those which leave out two.

I say that all the rest are also cubic.

Since the unit is to *A* as *A* is to *B,* therefore the unit measures *A* the same number of times as *A* measures *B.* But the unit measures *A* according to the units in it, therefore *A* also measures *B* according to the units in itself, therefore *A* multiplied by itself makes *B.*

And *A* is cubic. But, if cubic number multiplied by itself makes some number, then the product is also a cube, therefore *B* is also a cube.

And, since the four numbers *A, B, C,* and *D* are in continued proportion, and *A* is a cube, therefore *D* also is a cube.

For the same reason *E* is also a cube, and similarly all the rest are cubes.

Therefore, *if as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is square, then all the rest are also square; and if the number after the unit is cubic, then all the rest are also cubic.*

Q.E.D.

The following theorem is a converse of this one.