De institutione musica IV.2: Diversae de intervallis speculationes.
If a multiple interval is multiplied by two, what results from this multiplication will be a multiple interval.
Let BC be a multiple interval, and B a multiple of C, and let it be that as C is to B, so B is to D. Since B is a multiple of C, the term C measures B, either twice or three times or so on. But as C is to B, so is B to D. The term B therefore measures D. Therefore the term C also will measure D; D is therefore a multiple of C, and DC is the interval produced from the interval BC compounded, joined to itself and multiplied by two. The same is proved in numbers also. Let B to C be double, as two to one, and let it be that as C is to B, so B is to D. D will therefore be four. B to C, that is, two to one, is multiple; D, four, is therefore multiple to C, one. For four is the quadruple of unity, and the half of it, multiplied by two, is the interval BC. If an interval multiplied by two produces a multiple interval, it too will be multiple. Let the interval be BC, and let it be that as C is to B, so B is to D, and let D be multiple to C. I say that B is a multiple of C. For since D is a multiple of C, C measures D. But it has been shown that if numbers are in proportion, and the first is compared naturally with the last, then if the first has measured the last, it will measure the middle also. C therefore measures B. B is therefore a multiple of C. This again from numbers. Let C be a unit, and let D, from the duplicated proportion BC, be four, and it is a multiple of C. For it is a quadruple. Since, then, this quadruple is generated from the duplicated proportion BC, the proportion BC will be its half. The proportion BC is therefore double. But the double is multiple. The proportion BC will therefore be multiple.
No middle number, neither one nor several, will fall proportionally within a superparticular interval. Let the proportion BC be superparticular, and let DF and G be the least in the same proportion. Since DF and G are the least in the same proportion, they are prime in the same proportion. Therefore only unity will measure them. Let G be taken away from DF, and let D be left. This is therefore the common measure of both. This will therefore be unity. Therefore no number will fall between FD and G which would be less than FD but greater than G. For only unity lies between.
As many as fall proportionally in superparticular proportions between the least terms of the same proportion, so many will fall also between the rest of the same proportion. But none can intervene between FD and G, the least of the same proportion; none, therefore, will fall proportionally between B and C. And in numbers. Let there be any superparticular proportion, as the sesquialter. Let these be 10 and 15. In the same proportion the least are 2 and 3. I take two from three; one remains, and it measures both. There will be no number, then, between two and three which is greater than two and less than three. Otherwise unity would be divided, which is unfitting. Therefore not even between 10 and 15 will any number be found which holds to 10 such a proportion as 15 holds to it. If an interval that is not multiple is multiplied by two, it is neither multiple nor superparticular. Let the interval BC be not multiple, and let it be that as C is to B, so B is to D. I say that D to C is neither multiple nor superparticular. For let D be, if it can be, first a multiple of C. And since it is known that if an interval multiplied by two is a multiple interval, the interval that was multiplied twice is multiple, BC will be multiple. But this was not supposed; D will not therefore be a multiple of C. Nor superparticular. For no term falls proportionally in the middle of a superparticular proportion. But between D and C a term is constituted proportionally, that is B. For as C is to B, so is B to D. It will be impossible, therefore, for D to be either a multiple or a superparticular of C, which was to be shown. And in numbers. Let the interval 6 to 4 be not multiple, and let it be that as 4 is to 6, so 6 is to some number. This will be nine, which is neither multiple nor superparticular of four.
If an interval is multiplied by two, and what is produced from that multiplication is not multiple, the interval itself will not be multiple. Let the interval BC be, and let it be that as C is to B, so B is to D, and let D not be a multiple of C. I say that B also will not be a multiple of C. For if it is, D also is a multiple of C. But it is not. B will not therefore be a multiple of C.
The double interval is joined from the two greatest superparticulars, the sesquialter and the sesquitertian. Let A be the sesquialter of B, and B the sesquitertian of C. I say that A is the double of C. Since A is the sesquialter of B, A has in itself the whole of B and its half. Two A are therefore equal to three B. Again, since B is the sesquitertian of C, B has C and its third part. Three B are therefore equal to four C. But three B were equal to two A. Two A are therefore equal to four C. One A is therefore equal to two C. A will therefore be the double of C. And in numbers. Let the sesquialter be 12 to 8, and the sesquitertian 8 to 6; 12 to 6 are therefore double.
From a double interval and a sesquialter a triple interval arises. Let A be the double of B, and B the sesquialter of C. I say that A is the triple of C. For since A is the double of B, A is equal to two B. Again, since B is the sesquialter of C, B has in itself the whole of C and its half part. Two B are therefore equal to three C. But two B were equal to one A. One A is therefore equal to three C. A is therefore the triple of C. And in numbers. Let six be the double of three, and three the sesquialter of two; six is therefore the triple of two.
If a sesquitertian interval is taken from a sesquialter interval, what remains will be sesquioctave. Let A be the sesquialter of B, and C the sesquitertian of B. I say that A is the sesquioctave of C. For since A is the sesquialter of B, A has in itself B and its half part. Eight A are therefore equal to twelve B. Again, since C is the sesquitertian of B, C has in itself B and its third part. Nine C are therefore equal to twelve B. But twelve B were equal to eight A. Eight A are therefore equal to nine C. A is therefore equal to C and its eighth part. A is therefore the sesquioctave of C. And in numbers. Let the sesquialter interval be nine to six, and the sesquitertian eight to six. Nine to eight is therefore the sesquioctave proportion. Six sesquioctave proportions are greater than one double interval. For let there be some number A, and let B be its sesquioctave, C the sesquioctave of this, D the sesquioctave of this, F the sesquioctave of this, G the sesquioctave of this, and K the sesquioctave of this. Let this be done according to the manner described in arithmetic. And let the numbers be A, B, C, D, F, G, K. And let A be 262,144, its sesquioctave B 294,912, its sesquioctave C 331,776, its sesquioctave D 373,248, its sesquioctave F 419,904, its sesquioctave G 472,392, its sesquioctave K 531,441. And 531,441, which is K, is more than the double of 262,144, which is A. Six sesquioctave proportions are therefore greater than one double interval.