Continued Proportion

Theorem 80 (Euclid VIII.1). If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the numbers are the least of those which have the same ratio with them.

Proof: Let A, B, C, D be in continued proportion, with A, D prime to one another. If they are not the least, let E, F, G, H be less, in the same ratio. Ex aequali, A : D = E : H (Theorem 54). A, D are prime, so least (Theorem 60), so A measures E (Theorem 59), the greater the less: which is impossible.

Theorem 81 (Euclid VIII.2). To find numbers in continued proportion, as many as may be prescribed, and the least that are in a given ratio.

Proof: Let the given ratio in least numbers be that of A to B. Let A by multiplying itself make C, by multiplying B make D, and let B by multiplying itself make E. Let A by multiplying C, D, E make F, G, H, and let B by multiplying E make K.

A by multiplying A, B makes C, D, so A : B = C : D (Theorem 56). A, B by multiplying B make D, E, so A : B = D : E (Theorem 57). So C, D, E are in the ratio of A to B. Again, A by multiplying C, D, E makes F, G, H, so C : D = F : G and D : E = G : H (Theorem 56); and A, B by multiplying E make H, K, so A : B = H : K (Theorem 57). So F, G, H, K are in the ratio of A to B.

They are the least. A, B are least, so prime to one another (Theorem 61). So C, E are prime to one another, and F, K likewise (Theorem 66). The extremes being prime, the numbers are the least (Theorem 80).

Porism. If three numbers in continued proportion are the least of those which have the same ratio with them, the extremes are squares; and if four, cubes.

Theorem 82 (Euclid VIII.3). If as many numbers as we please in continued proportion be the least of those which have the same ratio with them, the extremes of them are prime to one another.

Proof: Let A, B, C, D be such. Take E, F the least in their ratio (Theorem 72), then three, G, H, K, and four, L, M, N, O, as in Theorem 81. L, M, N, O are the least in that ratio, and so are A, B, C, D, with equal multitude; so each is equal to each, and A is equal to L, D to O. E, F are prime to one another, so L, O, which are their cubes, are prime to one another (Theorem 66).

Theorem 83 (Euclid VIII.4). Given as many ratios as we please in least numbers, to find numbers in continued proportion which are the least in the given ratios.

Proof: Let the ratios be A : B, C : D, E : F. Let G be the least number measured by B, C (Theorem 73). As many times as B measures G, so many times let A measure H; and as many times as C measures G, so many times let D measure K.

First let E measure K, and as many times as E measures K, so many times let F measure L. Then A : B = H : G, C : D = G : K, E : F = K : L (Definition 18). So H, G, K, L are in the given ratios. They are the least: if N, O, M, P were less, in the same ratios, then, A, B being least, B would measure O (Theorem 59), and likewise C would measure O; so G would measure O (Theorem 74), the greater the less: which is impossible.

Next let E not measure K. Let M be the least number measured by E, K (Theorem 73). As many times as K measures M, so many times let H, G measure N, O; and as many times as E measures M, so many times let F measure P. Then N, O, M, P are in the given ratios. They are the least: if Q, R, S, T were less, in the same ratios, then B, C would measure R, so G would measure R (Theorem 74). And G : R = K : S, so K measures S. E also measures S (Theorem 59). So M measures S (Theorem 74), the greater the less: which is impossible.

Theorem 84 (Euclid VIII.5). Plane numbers have to one another the ratio compounded of the ratios of their sides.

Proof: Let A be plane with sides C, D, and B with sides E, F. Take G, H, K the least with G : H = C : E and H : K = D : F (Theorem 83). Let D by multiplying E make L. D by multiplying C, E makes A, L, so C : E = A : L (Theorem 56), and so G : H = A : L. E by multiplying D, F makes L, B, so D : F = L : B, and so H : K = L : B. Ex aequali, G : K = A : B (Theorem 54). And G has to K the ratio compounded of the ratios of the sides.

Theorem 85 (Euclid VIII.6). If there be as many numbers as we please in continued proportion, and the first do not measure the second, neither will any other measure any other.

Proof: Let A, B, C, D, E be in continued proportion, and A not measure B. That none measures the next is manifest. Say A does not measure C. Take F, G, H the least in the ratio of A, B, C (Theorem 72). Ex aequali, A : C = F : H (Theorem 54). Since A : B = F : G and A does not measure B, F does not measure G; so F is not a unit. F, H are prime to one another (Theorem 82). So F does not measure H, and A does not measure C. Likewise for any other.

Theorem 86 (Euclid VIII.7). If there be as many numbers as we please in continued proportion, and the first measure the last, it will measure the second also.

Proof: If A does not measure B, neither will any other measure any other (Theorem 85). But A measures D.

Theorem 87 (Euclid VIII.8). If between two numbers there fall numbers in continued proportion with them, then, however many numbers fall between them in continued proportion, so many will also fall in continued proportion between the numbers which have the same ratio with the original numbers.

Proof: Let C, D fall between A, B, and let E : F = A : B. Take G, H, K, L the least in the ratio of A, C, D, B (Theorem 72). G, L are prime to one another (Theorem 82). Ex aequali, G : L = A : B = E : F. So G measures E the same number of times that L measures F (Theorem 60, Theorem 59). As many times as G measures E, so many times let H, K measure M, N. Then G, H, K, L measure E, M, N, F the same number of times, so E, M, N, F are in the same ratio with G, H, K, L (Definition 18), and are in continued proportion.

Theorem 88 (Euclid VIII.9). If two numbers be prime to one another, and numbers fall between them in continued proportion, then, however many numbers fall between them in continued proportion, so many will also fall between each of them and an unit in continued proportion.

Proof: Let A, B be prime to one another, let C, D fall between them, and let E be the unit. Take F, G the least in the ratio of A, C, D, B, then three, H, K, L, and four, M, N, O, P (Theorem 81), so that F by multiplying itself makes H and by multiplying H makes M, and G by multiplying itself makes L and by multiplying L makes P. M, N, O, P are the least in the ratio, and so are A, C, D, B (Theorem 80); so M is equal to A and P to B. F measures H according to the units in F, and the unit E measures F according to the units in it, so E : F = F : H. Again H measures M according to the units in F, so E : F = H : M. So E, F, H, A are in continued proportion. Likewise E, G, L, B.

Theorem 89 (Euclid VIII.10). If numbers fall between each of two numbers and an unit in continued proportion, however many numbers fall between each of them and an unit in continued proportion, so many also will fall between the numbers themselves in continued proportion.

Proof: Let D, E fall between the unit C and A, and F, G between C and B. Then D by multiplying itself makes E and by multiplying E makes A; and F by multiplying itself makes G and by multiplying G makes B. Let D by multiplying F make H, and let D, F by multiplying H make K, L. D by multiplying D, F makes E, H, so D : F = E : H (Theorem 56). D, F by multiplying F make H, G, so D : F = H : G (Theorem 57). D by multiplying E, H makes A, K, so E : H = A : K. D, F by multiplying H make K, L, so D : F = K : L. F by multiplying H, G makes L, B, so H : G = L : B. So A : K = K : L = L : B.

Theorem 90 (Euclid IX.16). If two numbers be prime to one another, the second will not be to any other number as the first is to the second.

Proof: Let A, B be prime to one another. If A : B = B : C, then, A, B being least, A measures B (Theorem 60, Theorem 59). It also measures itself. So A measures A, B, which are prime to one another: which is impossible.

Theorem 91 (Euclid IX.17). If there be as many numbers as we please in continued proportion, and the extremes of them be prime to one another, the last will not be to any other number as the first to the second.

Proof: Let A, B, C, D be in continued proportion, with A, D prime to one another. If A : B = D : E, then alternately A : D = B : E, and A measures B (Theorem 60, Theorem 59). Since A : B = B : C, B measures C, so A measures C. Since B : C = C : D, C measures D, so A measures D. It also measures itself. So A measures A, D, which are prime to one another: which is impossible.

Theorem 92 (Euclid IX.18). Given two numbers, to investigate whether it is possible to find a third proportional to them.

Proof: Let them be A, B. If they are prime to one another, it is impossible (Theorem 90). If not, let B by multiplying itself make C. If A measures C, according to D, then the product of A, D is equal to the square on B, so A : B = B : D (Theorem 58): D is found. If A does not measure C, it is impossible: for if D were a third proportional, the product of A, D would be equal to the square on B, and A would measure C.

Theorem 93 (Euclid IX.19). Given three numbers, to investigate when it is possible to find a fourth proportional to them.

Proof: Let them be A, B, C, and let B by multiplying C make D. If A measures D, according to E, then the product of A, E is equal to the product of B, C, so A : B = C : E (Theorem 58): E is found. If A does not measure D, it is impossible, as in Theorem 92.

The text of this proposition is corrupt in the manuscripts: Euclid's division into cases, as it has come down, does not hold. What is given here is the sound part, which is all that later propositions need.

Theorem 94 (Euclid IX.15). If three numbers in continued proportion be the least of those which have the same ratio with them, any two whatever added together will be prime to the remaining number.

Proof: Let them be A, B, C. Take DE, EF the least in their ratio (Theorem 81), so that DE by multiplying itself makes A, by multiplying EF makes B, and EF by multiplying itself makes C. DE, EF are prime to one another (Theorem 61), so the sum DF is prime to each (Theorem 67). DF, DE are each prime to EF, so their product is prime to EF (Theorem 63), and so to the square on EF (Theorem 64). The product of FD, DE is the square on DE together with the product of DE, EF (Euclid II.3): that is, A, B together. So A, B together are prime to C. Likewise B, C to A. Again DF is prime to each of DE, EF, so the square on DF is prime to the product of DE, EF (Theorem 63, Theorem 64). The square on DF is the squares on DE, EF with twice the product of DE, EF (Euclid II.4): that is, A, C with twice B. So A, C with twice B are prime to B; so, separando, A, C, B together are prime to B, and again A, C together are prime to B (Theorem 67).

The proof supposes: Euclid II.3 and II.4 read of numbers. Both are Theorem 32.