Proposition 9

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.