Mathematical Collections (Commandino 1588) (1588) - Page 16

half of CB; and as AD is to DB, so is BD to DC, and the third of its CB: and as AE is to EB, so is BE to EC, and the fourth of its CB: and as AF is to FB, so is BF to FC, and the fifth of CB: Finally as AG is to GB, so is BG to GC, and the sixth of its CB.

A diagram shows a straight horizontal line divided into segments marked by points A, B, C, D, E, F, G, indicating equal intervals.

It is manifest that with numbers always assumed in this manner, as the given number of equal straight lines from point A to a number less by one, so is the number less by one to another yet smaller by one, and the small part of its CB, which corresponds to the given multitude of equal straight lines.

COMMENTARY

I say that as AC is to CB, so is BC to half of its CB.] For as AC is to its half CB, so is BC to its half.

And as AD is to DB, so is BD to DC, and the third of its CB.] For let it be done as AB is to BD, so is BH to HD. By composing, it will be as AD is to DB, so is BD to DH; and of the parts of which AD is 9, DB is 6: and DH 4. Therefore BH is 2, and HC 1. Therefore as AD is to DB, so is BD to DC and CH, that is, the third of its CB.

A geometric diagram illustrates the proportions described, showing line segments with points A, B, H, K, C, D, E, F, G marked.

And as AE is to EB, so is BE to EC, and the fourth of its EB.] Again, let it be done as AB is to BE, so is BK to KE: wherefore also by composing, as AE is to EB, so will BE be to EK. Of the parts of which AE is 16, EB is 12, and EK 9. Therefore BK will be 3, and KC 1. Therefore as AE is to EB, so is BE to EC and CK, namely the fourth of its CB. In the same way the rest will be demonstrated.

THEOREM II. PROPOSITION II.

Let A, B be equal straight lines; and CD smaller than each of them, A and B; but greater than N: and let it be done as A is to CD, so is CD to EF, and EF to GH: but as B is to N, so let it be done as N is to P, and P to R. I say that R is smaller than GH.

9 of the fifth

For since CD is greater than N, let CK be set equal to N. Therefore as A is to CK, so is B to N. Again, since as A is to CD, so is CD to EF; let it be done as A is to CK, so is CK to EL. But it is also as B is to N, so N is to P: and A is indeed equal to B; but CK is equal to N. Therefore, from equality, as A is to EL, so is B to P. And therefore EL will be equal to P. By the same reasoning, since it is as A is to CD, so is CD to EF, and EF to GH; it will also be as A is to CK, so is CK to EL, and EL to a value smaller than GH is to GM. Therefore, since as CK is to EL, so is EL to GM; and as N is to P, so is P to R; and CK is equal to N, and EL is equal to P; GM will be equal to R; and therefore R is smaller than GH.

A diagram depicts vertical and horizontal segments labeled A, B, C, D, E, F, G, H, K, L, M, N, P, R, illustrating the proportional relationship in Theorem II.