Hoc in volumine haec opera continentur

[continues from previous page] It will be, therefore, that b of c is a multiple, of which it was not impossible; therefore d of c is a multiple; and neither is it superparticular, for no proportional mean falls into a superparticular interval, but b falls into d c, which is impossible; therefore [it is impossible for] d of c to be either a multiple or a superparticular. If an interval twice composed does not make a multiple, not even it itself will be a multiple; for let the interval be b c, and let it be such that c is to b, [and] b is to d; and let d not be a multiple of c, and I claim b of c will not be a multiple; for if b of c were a multiple, d of c would also be a multiple. But it is not; therefore b of c will not be a multiple. A double interval consists of two maximum superparticulars, from the sesqualter 3:2 ratio and the sesquitertius 4:3 ratio; let b c of d f be sesqualter, but d f of h be sesqtertius. I claim b c of h is double. By taking away, for let the part equal to h be f g, and to d f be c l. Therefore, since b c of d f is sesqualter, b l therefore is the third part of b c, but [it is] the half of d f. Again, d f of h is sesqtertius, but d K is indeed the fourth part of d f, and the third part of h; therefore, since d K is the fourth part of d f, and b is the half of d f, therefore b l will be half of d K; but b l was the third part of b c; therefore d K is the sixth part of b c; but it will be that d K is the third part of h; therefore b c of h is double.

A decorative initial 'S' is depicted. Let a of b be sesqualter, but b of c be sesquitertius; I claim a of c is double; for since a of b is sesqualter, therefore a has b and its half; therefore, two a are equal to three b. Again, because b of d is sesquitertius, therefore b has c and its third; therefore, three b are equal to four c; but three b are equal to two a; therefore, two a are equal to four c; therefore, a is equal to two c; therefore, a of c is double. From a double interval and a sesqualter interval, a triple interval is made; for let a of b be double, but b of c be sesqualter; I claim a of c is triple; for because a of b is double, therefore a is equal to two b. Again, because b of c is sesqualter, therefore b has c and its half; therefore, two b are equal to three c; but two b are a; therefore, a is equal to three c; therefore, a of c is triple.

A decorative initial 'S' is depicted. If a sesquitertius interval is taken away from a sesqualter interval, the remainder is sesquioctavum 9:8 ratio; for let a of b be sesqualter, but c of b be sesquitertius; I claim a of c is sesquioctavus; for because a of b is sesqualter, therefore a has b and its half; therefore, eight a are equal to twelve b. Again, because c of b is sesquitertius, therefore c has b and its third; therefore, nine c are equal to twelve b; but twelve b are equal to a; therefore, eight a are equal to nine c; therefore, a is equal to c and its eighth part; therefore, a of c is sesquioctavus. Six sesquioctava intervals are greater than one double interval; for let the number be a, and the sesqoctauus of a be b, and the sesqoctauus of b be c; of c the sesqoctauus be d, of d the sesqoctauus be e, of e the sesqoctauus be f, of f the sesqoctauus be g; I claim g of a is greater than double; for I have learned to find seven numbers consecutive to each other [as] sesqoctauos. Let a, b, c, d, f, g be [numbers], and let a be