Mathematical Collections (Commandino 1588)

OTHERWISE

Let A be equal to E, but B greater than F: and let it be done as A is to B, so is B to C, and C to D. And as E is to F, so is F to G, and G to H. I say that D is greater than H.

8 of the fifth

11 of the fifth

A geometric diagram shows two horizontal rows of points (A, B, C, D and E, F, G, H) with dashed lines connecting them vertically, demonstrating comparative ratios.
10 of the fifth

For since B is greater than F, and A is equal to E; B will have a greater ratio to A than F has to E: and conversely, A will have a smaller ratio to B than E has to F. But as E is to F, so is F to G, and as A is to B, so is B to C; therefore B has a smaller ratio to C than F has to G. And as B is to C, so is C to D. Wherefore C has a smaller ratio to D than F has to G. But as F is to G, so is G to H. Therefore C has a smaller ratio to D than G has to H. Since, therefore, A has a smaller ratio to B than E has to F; and B indeed to C a smaller one than F to G; and C to D a smaller one than G to H; by equality, A will have a smaller ratio to D than E has to H, as will be demonstrated subsequently. And A and E are equal to each other. Therefore D is greater than H, which was to be demonstrated.

THEOREM III. PROPOSITION III.

Let A have a smaller ratio to B than C has to D. I say also that by permuting, A has a smaller ratio to C than B has to D.

A simple diagram shows four vertical lines of varying heights labeled A, B, C, and D.
10 of the fifth

8 of the fifth

Let it be done as A is to B, so is C to E. Therefore E is greater than D. And since as A is to B, so is C to E; it will be by permuting, as A is to C, so is B to E. But B has a smaller ratio to E than B has to D. Therefore A also will have a smaller ratio to C than B has to D.

THEOREM IIII. PROPOSITION IIII.

With this demonstrated, let A have a smaller ratio to B than D has to E: and B a smaller ratio to C...