Let A, B, C, D be any multitude whatsoever of continuously proportional numbers. And let the outermost of them, A and D, be prime to one another. I say that A, B, C, D are the least of those (numbers) having the same ratio as them.
For if not, let E, F, G, H be less than A, B, C, D (respectively), being in the same ratio as them. And since A, B, C, D are in the same ratio as E, F, G, H, and the multitude [of A, B, C, D ] is equal to the multitude [of E, F, G, H ], thus, via equality, as A is to D, (so) E (is) to H [Prop. 7.14]. And A and D (are) prime (to one another). And prime (numbers are) also the least of those (numbers having the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser---that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, A measures E, the greater (measuring) the lesser. The very thing is impossible. Thus, E, F, G, H, being less than A, B, C, D, are not in the same ratio as them. Thus, A, B, C, D are the least of those (numbers) having the same ratio as them. (Which is) the very thing it was required to show.
Let the given ratio, (expressed) in the least numbers, be that of A to B. So it is required to find the least numbers, as many as may be prescribed, (which are) in the ratio of A to B.
Let four (numbers) have been prescribed. And let A make C (by) multiplying itself, and let it make D (by) multiplying B. And, further, let B make E (by) multiplying itself. And, further, let A make F, G, H (by) multiplying C, D, E. And let B make K (by) multiplying E.
And since A has made C (by) multiplying itself, and has made D (by) multiplying B, thus as A is to B, [so] C (is) to D [Prop. 7.17]. Again, since A has made D (by) multiplying B, and B has made E (by) multiplying itself, A, B have thus made D, E, respectively, (by) multiplying B. Thus, as A is to B, so D (is) to E [Prop. 7.18]. But, as A (is) to B, (so) C (is) to D. And thus as C (is) to D, (so) D (is) to E. And since A has made F, G (by) multiplying C, D, thus as C is to D, [so] F (is) to G [Prop. 7.17]. And as C (is) to D, so A was to B. And thus as A (is) to B, (so) F (is) to G. Again, since A has made G, H (by) multiplying D, E, thus as D is to E, (so) G (is) to H [Prop. 7.17]. But, as D (is) to E, (so) A (is) to B. And thus as A (is) to B, so G (is) to H. And since A, B have made H, K (by) multiplying E, thus as A is to B, so H (is) to K. But, as A (is) to B, so F (is) to G, and G to H. And thus as F (is) to G, so G (is) to H, and H to K. Thus, C, D, E and F, G, H, K are (both continuously) proportional in the ratio of A to B. So I say that (they are) also the least (sets of numbers continuously proportional in that ratio). For since A and B are the least of those (numbers) having the same ratio as them, and the least of those (numbers) having the same ratio are prime to one another [Prop. 7.22], A and B are thus prime to one another. And A, B have made C, E, respectively, (by) multiplying themselves, and have made F, K by multiplying C, E, respectively. Thus, C, E and F, K are prime to one another [Prop. 7.27]. And if there are any multitude whatsoever of continuously proportional numbers, and the outermost of them are prime to one another, then the (numbers) are the least of those (numbers) having the same ratio as them [Prop. 8.1]. Thus, C, D, E and F, G, H, K are the least of those (continuously proportional sets of numbers) having the same ratio as A and B. (Which is) the very thing it was required to show.
Corollary
So it is clear, from this, that if three continuously proportional numbers are the least of those (numbers) having the same ratio as them then the outermost of them are square, and, if four (numbers), cube.
Let A, B, C, D be any multitude whatsoever of continuously proportional numbers (which are) the least of those (numbers) having the same ratio as them. I say that the outermost of them, A and D, are prime to one another.
For let the two least (numbers) E, F (which are) in the same ratio as A, B, C, D have been taken [Prop. 7.33]. And the three (least numbers) G, H, K [Prop. 8.2]. And (so on), successively increasing by one, until the multitude of (numbers) taken is made equal to the multitude of A, B, C, D. Let them have been taken, and let them be L, M, N, O.
And since E and F are the least of those (numbers) having the same ratio as them they are prime to one another [Prop. 7.22]. And since E, F have made G, K, respectively, (by) multiplying themselves [Prop. 8.2 corr.], and have made L, O (by) multiplying G, K, respectively, G, K and L, O are thus also prime to one another [Prop. 7.27]. And since A, B, C, D are the least of those (numbers) having the same ratio as them, and L, M, N, O are also the least (of those numbers having the same ratio as them), being in the same ratio as A, B, C, D, and the multitude of A, B, C, D is equal to the multitude of L, M, N, O, thus A, B, C, D are equal to L, M, N, O, respectively. Thus, A is equal to L, and D to O. And L and O are prime to one another. Thus, A and D are also prime to one another. (Which is) the very thing it was required to show.
Let the given ratios, (expressed) in the least numbers, be the (ratios) of A to B, and of C to D, and, further, of E to F. So it is required to find the least numbers continuously proportional in the ratio of A to B, and of C to D, and, further, of E to F.
For let the least number, G, measured by (both) B and C have be taken [Prop. 7.34]. And as many times as B measures G, so many times let A also measure H. And as many times as C measures G, so many times let D also measure K. And E either measures, or does not measure, K. Let it, first of all, measure ( K). And as many times as E measures K, so many times let F also measure L. And since A measures H the same number of times that B also (measures) G, thus as A is to B, so H (is) to G [Prop. 7.13] [Def. 7.20]. And so, for the same (reasons), as C (is) to D, so G (is) to K, and, further, as E (is) to F, so K (is) to L. Thus, H, G, K, L are continuously proportional in the ratio of A to B, and of C to D, and, further, of E to F. So I say that (they are) also the least (numbers continuously proportional in these ratios). For if H, G, K, L are not the least numbers continuously proportional in the ratios of A to B, and of C to D, and of E to F, let N, O, M, P be (the least such numbers). And since as A is to B, so N (is) to O, and A and B are the least (numbers which have the same ratio as them), and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser---that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20], B thus measures O. So, for the same (reasons), C also measures O. Thus, B and C (both) measure O. Thus, the least number measured by (both) B and C will also measure O [Prop. 7.35]. And G (is) the least number measured by (both) B and C. Thus, G measures O, the greater (measuring) the lesser. The very thing is impossible. Thus, there cannot be any numbers less than H, G, K, L (which are) continuously (proportional) in the ratio of A to B, and of C to D, and, further, of E to F.
So let E not measure K. And let the least number, M, measured by (both) E and K have been taken [Prop. 7.34]. And as many times as K measures M, so many times let H, G also measure N, O, respectively. And as many times as E measures M, so many times let F also measure P. Since H measures N the same number of times as G (measures) O, thus as H is to G, so N (is) to O [Prop. 7.13] [Def. 7.20]. And as H (is) to G, so A (is) to B. And thus as A (is) to B, so N (is) to O. And so, for the same (reasons), as C (is) to D, so O (is) to M. Again, since E measures M the same number of times as F (measures) P, thus as E is to F, so M (is) to P [Prop. 7.13] [Def. 7.20]. Thus, N, O, M, P are continuously proportional in the ratios of A to B, and of C to D, and, further, of E to F. So I say that (they are) also the least (numbers) in the ratios of A B, C D, E F. For if not, then there will be some numbers less than N, O, M, P (which are) continuously proportional in the ratios of A B, C D, E F. Let them be Q, R, S, T. And since as Q is to R, so A (is) to B, and A and B (are) the least (numbers having the same ratio as them), and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20], B thus measures R. So, for the same (reasons), C also measures R. Thus, B and C (both) measure R. Thus, the least (number) measured by (both) B and C will also measure R [Prop. 7.35]. And G is the least number measured by (both) B and C. Thus, G measures R. And as G is to R, so K (is) to S. Thus, K also measures S [Def. 7.20]. And E also measures S [Prop. 7.20]. Thus, E and K (both) measure S. Thus, the least (number) measured by (both) E and K will also measure S [Prop. 7.35]. And M is the least (number) measured by (both) E and K. Thus, M measures S, the greater (measuring) the lesser. The very thing is impossible. Thus there cannot be any numbers less than N, O, M, P (which are) continuously proportional in the ratios of A to B, and of C to D, and, further, of E to F. Thus, N, O, M, P are the least (numbers) continuously proportional in the ratios of A B, C D, E F. (Which is) the very thing it was required to show.
Let A and B be plane numbers, and let the numbers C, D be the sides of A, and (the numbers) E, F (the sides) of B. I say that A has to B the ratio compounded out of (the ratios of) their sides.
For given the ratios which C has to E, and D (has) to F, let the least numbers, G, H, K, continuously proportional in the ratios C E, D F have been taken [Prop. 8.4], so that as C is to E, so G (is) to H, and as D (is) to F, so H (is) to K. And let D make L (by) multiplying E.
And since D has made A (by) multiplying C, and has made L (by) multiplying E, thus as C is to E, so A (is) to L [Prop. 7.17]. And as C (is) to E, so G (is) to H. And thus as G (is) to H, so A (is) to L. Again, since E has made L (by) multiplying D [Prop. 7.16], but, in fact, has also made B (by) multiplying F, thus as D is to F, so L (is) to B [Prop. 7.17]. But, as D (is) to F, so H (is) to K. And thus as H (is) to K, so L (is) to B. And it was also shown that as G (is) to H, so A (is) to L. Thus, via equality, as G is to K, [so] A (is) to B [Prop. 7.14]. And G has to K the ratio compounded out of (the ratios of) the sides (of A and B). Thus, A also has to B the ratio compounded out of (the ratios of) the sides (of A and B). (Which is) the very thing it was required to show.
Let A, B, C, D, E be any multitude whatsoever of continuously proportional numbers, and let A not measure B. I say that no other (number) will measure any other (number) either.
Now, (it is) clear that A, B, C, D, E do not successively measure one another. For A does not even measure B. So I say that no other (number) will measure any other (number) either. For, if possible, let A measure C. And as many (numbers) as are A, B, C, let so many of the least numbers, F, G, H, have been taken of those (numbers) having the same ratio as A, B, C [Prop. 7.33]. And since F, G, H are in the same ratio as A, B, C, and the multitude of A, B, C is equal to the multitude of F, G, H, thus, via equality, as A is to C, so F (is) to H [Prop. 7.14]. And since as A is to B, so F (is) to G, and A does not measure B, F does not measure G either [Def. 7.20]. Thus, F is not a unit. For a unit measures all numbers. And F and H are prime to one another [Prop. 8.3] [and thus F does not measure H ]. And as F is to H,so A (is) to C. And thus A does not measure C either [Def. 7.20]. So, similarly, we can show that no other (number) can measure any other (number) either. (Which is) the very thing it was required to show.
Let A, B, C, D be any number whatsoever of continuously proportional numbers. And let A measure D. I say that A also measures B.
For if A does not measure B then no other (number) will measure any other (number) either [Prop. 8.6]. But A measures D. Thus, A also measures B. (Which is) the very thing it was required to show.
For let the numbers, C and D, fall between two numbers, A and B, in continued proportion, and let it have been contrived (that) as A (is) to B, so E (is) to F. I say that, as many numbers as have fallen in between A and B in continued proportion, so many (numbers) will also fall in between E and F in continued proportion.
For as many as A, B, C, D are in multitude, let so many of the least numbers, G, H, K, L, having the same ratio as A, B, C, D, have been taken [Prop. 7.33]. Thus, the outermost of them, G and L, are prime to one another [Prop. 8.3]. And since A, B, C, D are in the same ratio as G, H, K, L, and the multitude of A, B, C, D is equal to the multitude of G, H, K, L, thus, via equality, as A is to B, so G (is) to L [Prop. 7.14]. And as A (is) to B, so E (is) to F. And thus as G (is) to L, so E (is) to F. And G and L (are) prime (to one another). And (numbers) prime (to one another are) also the least (numbers having the same ratio as them) [Prop. 7.21]. And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser---that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, G measures E the same number of times as L (measures) F. So as many times as G measures E, so many times let H, K also measure M, N, respectively. Thus, G, H, K, L measure E, M, N, F (respectively) an equal number of times. Thus, G, H, K, L are in the same ratio as E, M, N, F [Def. 7.20]. But, G, H, K, L are in the same ratio as A, C, D, B. Thus, A, C, D, B are also in the same ratio as E, M, N, F. And A, C, D, B are continuously proportional. Thus, E, M, N, F are also continuously proportional. Thus, as many numbers as have fallen in between A and B in continued proportion, so many numbers have also fallen in between E and F in continued proportion. (Which is) the very thing it was required to show.
Let A and B be two numbers (which are) prime to one another, and let the (numbers) C and D fall in between them in continued proportion. And let the unit E be set out. I say that, as many numbers as have fallen in between A and B in continued proportion, so many (numbers) will also fall between each of A and B and the unit in continued proportion.
For let the least two numbers, F and G, which are in the ratio of A, C, D, B, have been taken [Prop. 7.33]. And the (least) three (numbers), H, K, L. And so on, successively increasing by one, until the multitude of the (least numbers taken) is made equal to the multitude of A, C, D, B [Prop. 8.2]. Let them have been taken, and let them be M, N, O, P. So (it is) clear that F has made H (by) multiplying itself, and has made M (by) multiplying H. And G has made L (by) multiplying itself, and has made P (by) multiplying L [Prop. 8.2 corr.]. And since M, N, O, P are the least of those (numbers) having the same ratio as F, G, and A, C, D, B are also the least of those (numbers) having the same ratio as F, G [Prop. 8.2], and the multitude of M, N, O, P is equal to the multitude of A, C, D, B, thus M, N, O, P are equal to A, C, D, B, respectively. Thus, M is equal to A, and P to B. And since F has made H (by) multiplying itself, F thus measures H according to the units in F [Def. 7.15]. And the unit E also measures F according to the units in it. Thus, the unit E measures the number F as many times as F (measures) H. Thus, as the unit E is to the number F, so F (is) to H [Def. 7.20]. Again, since F has made M (by) multiplying H, H thus measures M according to the units in F [Def. 7.15]. And the unit E also measures the number F according to the units in it. Thus, the unit E measures the number F as many times as H (measures) M. Thus, as the unit E is to the number F, so H (is) to M [Prop. 7.20]. And it was shown that as the unit E (is) to the number F, so F (is) to H. And thus as the unit E (is) to the number F, so F (is) to H, and H (is) to M. And M (is) equal to A. Thus, as the unit E is to the number F, so F (is) to H, and H to A. And so, for the same (reasons), as the unit E (is) to the number G, so G (is) to L, and L to B. Thus, as many (numbers) as have fallen in between A and B in continued proportion, so many numbers have also fallen between each of A and B and the unit E in continued proportion. (Which is) the very thing it was required to show.
For let the numbers D, E and F, G fall between the numbers A and B (respectively) and the unit C in continued proportion. I say that, as many numbers as have fallen between each of A and B and the unit C in continued proportion, so many will also fall in between A and B in continued proportion.
For let D make H (by) multiplying F. And let D, F make K, L, respectively, by multiplying H.
As since as the unit C is to the number D, so D (is) to E, the unit C thus measures the number D as many times as D (measures) E [Def. 7.20]. And the unit C measures the number D according to the units in D. Thus, the number D also measures E according to the units in D. Thus, D has made E (by) multiplying itself. Again, since as the [unit] C is to the number D, so E (is) to A, the unit C thus measures the number D as many times as E (measures) A [Def. 7.20]. And the unit C measures the number D according to the units in D. Thus, E also measures A according to the units in D. Thus, D has made A (by) multiplying E. And so, for the same (reasons), F has made G (by) multiplying itself, and has made B (by) multiplying G. And since D has made E (by) multiplying itself, and has made H (by) multiplying F, thus as D is to F, so E (is) to H [Prop. 7.17]. And so, for the same reasons, as D (is) to F, so H (is) to G [Prop. 7.18]. And thus as E (is) to H, so H (is) to G. Again, since D has made A, K (by) multiplying E, H, respectively, thus as E is to H, so A (is) to K [Prop. 7.17]. But, as E (is) to H, so D (is) to F. And thus as D (is) to F, so A (is) to K. Again, since D, F have made K, L, respectively, (by) multiplying H, thus as D is to F, so K (is) to L [Prop. 7.18]. But, as D (is) to F, so A (is) to K. And thus as A (is) to K, so K (is) to L. Further, since F has made L, B (by) multiplying H, G, respectively, thus as H is to G, so L (is) to B [Prop. 7.17]. And as H (is) to G, so D (is) to F. And thus as D (is) to F, so L (is) to B. And it was also shown that as D (is) to F, so A (is) to K, and K to L. And thus as A (is) to K, so K (is) to L, and L to B. Thus, A, K, L, B are successively in continued proportion. Thus, as many numbers as fall between each of A and B and the unit C in continued proportion, so many will also fall in between A and B in continued proportion. (Which is) the very thing it was required to show.
Let A and B be square numbers, and let C be the side of A, and D (the side) of B. I say that there exists one number in mean proportion to A and B, and that A has to B a squared ratio with respect to (that) C (has) to D.
For let C make E (by) multiplying D. And since A is square, and C is its side, C has thus made A (by) multiplying itself. And so, for the same (reasons), D has made B (by) multiplying itself. Therefore, since C has made A, E (by) multiplying C, D, respectively, thus as C is to D, so A (is) to E [Prop. 7.17]. And so, for the same (reasons), as C (is) to D, so E (is) to B [Prop. 7.18]. And thus as A (is) to E, so E (is) to B. Thus, one number (namely, E) is in mean proportion to A and B.
So I say that A also has to B a squared ratio with respect to (that) C (has) to D. For since A, E, B are three (continuously) proportional numbers, A thus has to B a squared ratio with respect to (that) A (has) to E [Def. 5.9]. And as A (is) to E, so C (is) to D. Thus, A has to B a squared ratio with respect to (that) side C (has) to (side) D. (Which is) the very thing it was required to show.
Let A and B be cube numbers, and let C be the side of A, and D (the side) of B. I say that there exist two numbers in mean proportion to A and B, and that A has to B a cubed ratio with respect to (that) C (has) to D.
For let C make E (by) multiplying itself, and let it make F (by) multiplying D. And let D make G (by) multiplying itself, and let C, D make H, K, respectively, (by) multiplying F.
And since A is cube, and C (is) its side, and C has made E (by) multiplying itself, C has thus made E (by) multiplying itself, and has made A (by) multiplying E. And so, for the same (reasons), D has made G (by) multiplying itself, and has made B (by) multiplying G. And since C has made E, F (by) multiplying C, D, respectively, thus as C is to D, so E (is) to F [Prop. 7.17]. And so, for the same (reasons), as C (is) to D, so F (is) to G [Prop. 7.18]. Again, since C has made A, H (by) multiplying E, F, respectively, thus as E is to F, so A (is) to H [Prop. 7.17]. And as E (is) to F, so C (is) to D. And thus as C (is) to D, so A (is) to H. Again, since C, D have made H, K, respectively, (by) multiplying F, thus as C is to D, so H (is) to K [Prop. 7.18]. Again, since D has made K, B (by) multiplying F, G, respectively, thus as F is to G, so K (is) to B [Prop. 7.17]. And as F (is) to G, so C (is) to D. And thus as C (is) to D, so A (is) to H, and H to K, and K to B. Thus, H and K are two (numbers) in mean proportion to A and B.
So I say that A also has to B a cubed ratio with respect to (that) C (has) to D. For since A, H, K, B are four (continuously) proportional numbers, A thus has to B a cubed ratio with respect to (that) A (has) to H [Def. 5.10]. And as A (is) to H, so C (is) to D. And [thus] A has to B a cubed ratio with respect to (that) C (has) to D. (Which is) the very thing it was required to show.
Let A, B, C be any multitude whatsoever of continuously proportional numbers, (such that) as A (is) to B, so B (is) to C. And let A, B, C make D, E, F (by) multiplying themselves, and let them make G, H, K (by) multiplying D, E, F. I say that D, E, F and G, H, K are continuously proportional.
For let A make L (by) multiplying B. And let A, B make M, N, respectively, (by) multiplying L. And, again, let B make O (by) multiplying C. And let B, C make P, Q, respectively, (by) multplying O.
So, similarly to the above, we can show that D, L, E and G, M, N, H are continuously proportional in the ratio of A to B, and, further, (that) E, O, F and H, P, Q, K are continuously proportional in the ratio of B to C. And as A is to B, so B (is) to C. And thus D, L, E are in the same ratio as E, O, F, and, further, G, M, N, H (are in the same ratio) as H, P, Q, K. And the multitude of D, L, E is equal to the multitude of E, O, F, and that of G, M, N, H to that of H, P, Q, K. Thus, via equality, as D is to E, so E (is) to F, and as G (is) to H, so H (is) to K [Prop. 7.14]. (Which is) the very thing it was required to show.
Let A and B be square numbers, and let C and D be their sides (respectively). And let A measure B. I say that C also measures D.
For let C make E (by) multiplying D. Thus, A, E, B are continuously proportional in the ratio of C to D [Prop. 8.11]. And since A, E, B are continuously proportional, and A measures B, A thus also measures E [Prop. 8.7]. And as A is to E, so C (is) to D. Thus, C also measures D [Def. 7.20].
So, again, let C measure D. I say that A also measures B.
For similarly, with the same construction, we can show that A, E, B are continuously proportional in the ratio of C to D. And since as C is to D, so A (is) to E, and C measures D, A thus also measures E [Def. 7.20]. And A, E, B are continuously proportional. Thus, A also measures B.
Thus, if a square (number) measures a(nother) square (number) then the side (of the former) will also measure the side (of the latter). And if the side (of a square number) measures the side (of another square number) then the (former) square (number) will also measure the (latter) square (number). (Which is) the very thing it was required to show.
For let the cube number A measure the cube (number) B, and let C be the side of A, and D (the side) of B. I say that C measures D.
For let C make E (by) multiplying itself. And let D make G (by) multiplying itself. And, further, [let] C [make] F (by) multiplying D, and let C, D make H, K, respectively, (by) multiplying F. So it is clear that E, F, G and A, H, K, B are continuously proportional in the ratio of C to D [Prop. 8.12]. And since A, H, K, B are continuously proportional, and A measures B, ( A) thus also measures H [Prop. 8.7]. And as A is to H, so C (is) to D. Thus, C also measures D [Def. 7.20].
And so let C measure D. I say that A will also measure B.
For similarly, with the same construction, we can show that A, H, K, B are continuously proportional in the ratio of C to D. And since C measures D, and as C is to D, so A (is) to H, A thus also measures H [Def. 7.20]. Hence, A also measures B. (Which is) the very thing it was required to show.
Let A and B be square numbers, and let C and D be their sides (respectively). And let A not measure B. I say that C does not measure D either.
For if C measures D then A will also measure B [Prop. 8.14]. And A does not measure B. Thus, C will not measure D either.
[So], again, let C not measure D. I say that A will not measure B either.
For if A measures B then C will also measure D [Prop. 8.14]. And C does not measure D. Thus, A will not measure B either. (Which is) the very thing it was required to show.
For let the cube number A not measure the cube number B. And let C be the side of A, and D (the side) of B. I say that C will not measure D.
For if C measures D then A will also measure B [Prop. 8.15]. And A does not measure B. Thus, C does not measure D either.
And so let C not measure D. I say that A will not measure B either. For if A measures B then C will also measure D [Prop. 8.15]. And C does not measure D. Thus, A will not measure B either. (Which is) the very thing it was required to show.
Let A and B be two similar plane numbers. And let the numbers C, D be the sides of A, and E, F (the sides) of B. And since similar numbers are those having proportional sides [Def. 7.21], thus as C is to D, so E (is) to F. Therefore, I say that there exists one number in mean proportion to A and B, and that A has to B a squared ratio with respect to that C (has) to E, or D to F ---that is to say, with respect to (that) a corresponding side (has) to a corresponding [side].
For since as C is to D, so E (is) to F, thus, alternately, as C is to E, so D (is) to F [Prop. 7.13]. And since A is plane, and C, D its sides, D has thus made A (by) multiplying C. And so, for the same (reasons), E has made B (by) multiplying F. So let D make G (by) multiplying E. And since D has made A (by) multiplying C, and has made G (by) multiplying E, thus as C is to E, so A (is) to G [Prop. 7.17]. But as C (is) to E, [so] D (is) to F. And thus as D (is) to F, so A (is) to G. Again, since E has made G (by) multiplying D, and has made B (by) multiplying F, thus as D is to F, so G (is) to B [Prop. 7.17]. And it was also shown that as D (is) to F, so A (is) to G. And thus as A (is) to G, so G (is) to B. Thus, A, G, B are continously proportional. Thus, there exists one number (namely, G) in mean proportion to A and B.
So I say that A also has to B a squared ratio with respect to (that) a corresponding side (has) to a corresponding side---that is to say, with respect to (that) C (has) to E, or D to F. For since A, G, B are continuously proportional, A has to B a squared ratio with respect to (that A has) to G [Prop. 5.9]. And as A is to G, so C (is) to E, and D to F. And thus A has to B a squared ratio with respect to (that) C (has) to E, or D to F. (Which is) the very thing it was required to show.
Let A and B be two similar solid numbers, and let C, D, E be the sides of A, and F, G, H (the sides) of B. And since similar solid (numbers) are those having proportional sides [Def. 7.21], thus as C is to D, so F (is) to G, and as D (is) to E, so G (is) to H. I say that two numbers fall (between) A and B in mean proportion, and (that) A has to B a cubed ratio with respect to (that) C (has) to F, and D to G, and, further, E to H.
For let C make K (by) multiplying D, and let F make L (by) multiplying G. And since C, D are in the same ratio as F, G, and K is the (number created) from (multiplying) C, D, and L the (number created) from (multiplying) F, G, [thus] K and L are similar plane numbers [Def. 7.21]. Thus, there exits one number in mean proportion to K and L [Prop. 8.18]. Let it be M. Thus, M is the (number created) from (multiplying) D, F, as shown in the theorem before this (one). And since D has made K (by) multiplying C, and has made M (by) multiplying F, thus as C is to F, so K (is) to M [Prop. 7.17]. But, as K (is) to M, (so) M (is) to L. Thus, K, M, L are continuously proportional in the ratio of C to F. And since as C is to D, so F (is) to G, thus, alternately, as C is to F, so D (is) to G [Prop. 7.13]. And so, for the same (reasons), as D (is) to G, so E (is) to H. Thus, K, M, L are continuously proportional in the ratio of C to F, and of D to G, and, further, of E to H. So let E, H make N, O, respectively, (by) multiplying M. And since A is solid, and C, D, E are its sides, E has thus made A (by) multiplying the (number created) from (multiplying) C, D. And K is the (number created) from (multiplying) C, D. Thus, E has made A (by) multiplying K. And so, for the same (reasons), H has made B (by) multiplying L. And since E has made A (by) multiplying K, but has, in fact, also made N (by) multiplying M, thus as K is to M, so A (is) to N [Prop. 7.17]. And as K (is) to M, so C (is) to F, and D to G, and, further, E to H. And thus as C (is) to F, and D to G, and E to H, so A (is) to N. Again, since E, H have made N, O, respectively, (by) multiplying M, thus as E is to H, so N (is) to O [Prop. 7.18]. But, as E (is) to H, so C (is) to F, and D to G. And thus as C (is) to F, and D to G, and E to H, so (is) A to N, and N to O. Again, since H has made O (by) multiplying M, but has, in fact, also made B (by) multiplying L, thus as M (is) to L, so O (is) to B [Prop. 7.17]. But, as M (is) to L, so C (is) to F, and D to G, and E to H. And thus as C (is) to F, and D to G, and E to H, so not only (is) O to B, but also A to N, and N to O. Thus, A, N, O, B are continuously proportional in the aforementioned ratios of the sides.
So I say that A also has to B a cubed ratio with respect to (that) a corresponding side (has) to a corresponding side---that is to say, with respect to (that) the number C (has) to F, or D to G, and, further, E to H. For since A, N, O, B are four continuously proportional numbers, A thus has to B a cubed ratio with respect to (that) A (has) to N [Def. 5.10]. But, as A (is) to N, so it was shown (is) C to F, and D to G, and, further, E to H. And thus A has to B a cubed ratio with respect to (that) a corresponding side (has) to a corresponding side---that is to say, with respect to (that) the number C (has) to F, and D to G, and, further, E to H. (Which is) the very thing it was required to show.
For let one number C fall between the two numbers A and B in mean proportion. I say that A and B are similar plane numbers.
[For] let the least numbers, D and E, having the same ratio as A and C have been taken [Prop. 7.33]. Thus, D measures A as many times as E (measures) C [Prop. 7.20]. So as many times as D measures A, so many units let there be in F. Thus, F has made A (by) multiplying D [Def. 7.15]. Hence, A is plane, and D, F (are) its sides. Again, since D and E are the least of those (numbers) having the same ratio as C and B, D thus measures C as many times as E (measures) B [Prop. 7.20]. So as many times as E measures B, so many units let there be in G. Thus, E measures B according to the units in G. Thus, G has made B (by) multiplying E [Def. 7.15]. Thus, B is plane, and E, G are its sides. Thus, A and B are (both) plane numbers. So I say that (they are) also similar. For since F has made A (by) multiplying D, and has made C (by) multiplying E, thus as D is to E, so A (is) to C ---that is to say, C to B [Prop. 7.17]. Again, since E has made C, B (by) multiplying F, G, respectively, thus as F is to G, so C (is) to B [Prop. 7.17]. And as C (is) to B, so D (is) to E. And thus as D (is) to E, so F (is) to G. And, alternately, as D (is) to F, so E (is) to G [Prop. 7.13]. Thus, A and B are similar plane numbers. For their sides are proportional [Def. 7.21]. (Which is) the very thing it was required to show.
For let the two numbers C and D fall between the two numbers A and B in mean proportion. I say that A and B are similar solid (numbers).
For let the three least numbers E, F, G having the same ratio as A, C, D have been taken [Prop. 8.2]. Thus, the outermost of them, E and G, are prime to one another [Prop. 8.3]. And since one number, F, has fallen (between) E and G in mean proportion, E and G are thus similar plane numbers [Prop. 8.20]. Therefore, let H, K be the sides of E, and L, M (the sides) of G. Thus, it is clear from the (proposition) before this (one) that E, F, G are continuously proportional in the ratio of H to L, and of K to M. And since E, F, G are the least (numbers) having the same ratio as A, C, D, and the multitude of E, F, G is equal to the multitude of A, C, D, thus, via equality, as E is to G, so A (is) to D [Prop. 7.14]. And E and G (are) prime (to one another), and prime (numbers) are also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser---that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures A the same number of times as G (measures) D. So as many times as E measures A, so many units let there be in N. Thus, N has made A (by) multiplying E [Def. 7.15]. And E is the (number created) from (multiplying) H and K. Thus, N has made A (by) multiplying the (number created) from (multiplying) H and K. Thus, A is solid, and its sides are H, K, N. Again, since E, F, G are the least (numbers) having the same ratio as C, D, B, thus E measures C the same number of times as G (measures) B [Prop. 7.20]. So as many times as E measures C, so many units let there be in O. Thus, G measures B according to the units in O. Thus, O has made B (by) multiplying G. And G is the (number created) from (multiplying) L and M. Thus, O has made B (by) multiplying the (number created) from (multiplying) L and M. Thus, B is solid, and its sides are L, M, O. Thus, A and B are (both) solid.
[So] I say that (they are) also similar. For since N, O have made A, C (by) multiplying E, thus as N is to O, so A (is) to C ---that is to say, E to F [Prop. 7.18]. But, as E (is) to F, so H (is) to L, and K to M. And thus as H (is) to L, so K (is) to M, and N to O. And H, K, N are the sides of A, and L, M, O the sides of B. Thus, A and B are similar solid numbers [Def. 7.21]. (Which is) the very thing it was required to show.
Let A, B, C be three continuously proportional numbers, and let the first A be square. I say that the third C is also square.
For since one number, B, is in mean proportion to A and C, A and C are thus similar plane (numbers) [Prop. 8.20]. And A is square. Thus, C is also square [Def. 7.21]. (Which is) the very thing it was required to show.
Let A, B, C, D be four continuously proportional numbers, and let A be cube. I say that D is also cube.
For since two numbers, B and C, are in mean proportion to A and D, A and D are thus similar solid numbers [Prop. 8.21]. And A (is) cube. Thus, D (is) also cube [Def. 7.21]. (Which is) the very thing it was required to show.
For let two numbers, A and B, have to one another the ratio which the square number C (has) to the square number D. And let A be square. I say that B is also square.
For since C and D are square, C and D are thus similar plane (numbers). Thus, one number falls (between) C and D in mean proportion [Prop. 8.18]. And as C is to D, (so) A (is) to B. Thus, one number also falls (between) A and B in mean proportion [Prop. 8.8]. And A is square. Thus, B is also square [Prop. 8.22]. (Which is) the very thing it was required to show.
For let two numbers, A and B, have to one another the ratio which the cube number C (has) to the cube number D. And let A be cube. [So] I say that B is also cube.
For since C and D are cube (numbers), C and D are (thus) similar solid (numbers). Thus, two numbers fall (between) C and D in mean proportion [Prop. 8.19]. And as many (numbers) as fall in between C and D in continued proportion, so many also (fall) in (between) those (numbers) having the same ratio as them (in continued proportion) [Prop. 8.8]. And hence two numbers fall (between) A and B in mean proportion. Let E and F (so) fall. Therefore, since the four numbers A, E, F, B are continuously proportional, and A is cube, B (is) thus also cube [Prop. 8.23]. (Which is) the very thing it was required to show.
Let A and B be similar plane numbers. I say that A has to B the ratio which (some) square number (has) to a(nother) square number.
For since A and B are similar plane numbers, one number thus falls (between) A and B in mean proportion [Prop. 8.18]. Let it (so) fall, and let it be C. And let the least numbers, D, E, F, having the same ratio as A, C, B have been taken [Prop. 8.2]. The outermost of them, D and F, are thus square [Prop. 8.2 corr.]. And since as D is to F, so A (is) to B, and D and F are square, A thus has to B the ratio which (some) square number (has) to a(nother) square number. (Which is) the very thing it was required to show.
Let A and B be similar solid numbers. I say that A has to B the ratio which (some) cube number (has) to a(nother) cube number.
For since A and B are similar solid (numbers), two numbers thus fall (between) A and B in mean proportion [Prop. 8.19]. Let C and D have (so) fallen. And let the least numbers, E, F, G, H, having the same ratio as A, C, D, B, (and) equal in multitude to them, have been taken [Prop. 8.2]. Thus, the outermost of them, E and H, are cube [Prop. 8.2 corr.]. And as E is to H, so A (is) to B. And thus A has to B the ratio which (some) cube number (has) to a(nother) cube number. (Which is) the very thing it was required to show.