1. A magnitude is a part of a(nother) magnitude, the lesser of the greater, when it measures the greater.
2. And the greater (magnitude is) a multiple of the lesser when it is measured by the lesser.
3. A ratio is a certain type of condition with respect to size of two magnitudes of the same kind.
4. (Those)magnitudes are said to have a ratio with respect to one another which, being multiplied, are capable of exceeding one another.
5. Magnitudes are said to be in the same ratio, the first to the second, and the third to the fourth, when equal multiples of the first and the third either both exceed, are both equal to, or are both less than, equal multiples of the second and the fourth, respectively, being taken in corresponding order, according to any kind of multiplication whatever.
6. And let magnitudes having the same ratio be called proportional.
7. And when for equal multiples (as in Def. 5), the multiple of the first (magnitude) exceeds the multiple of the second, and the multiple of the third (magnitude) does not exceed the multiple of the fourth, then the first (magnitude) is said to have a greater ratio to the second than the third (magnitude has) to the fourth.
8. And a proportion in three terms is the smallest (possible).
9. And when three magnitudes are proportional, the first is said to have to the third the squared" ratio of that (it has) to the
10. And when four magnitudes are (continuously) proportional, the first is said to have to the fourth the cubed ratio of that (it has) to the second. And so on, similarly, in successive order, whatever the (continuous) proportion might be.
11. These magnitudes are said to be corresponding (magnitudes): the leading to the leading (of two ratios), and the following to the following.
12. An alternate ratio is a taking of the (ratio of the) leading (magnitude) to the leading (of two equal ratios), and (setting it equal to) the (ratio of the) following (magnitude) to the following.
13. An inverse ratio is a taking of the (ratio of the) following (magnitude) as the leading and the leading (magnitude) as the following.
14. A composition of a ratio is a taking of the (ratio of the) leading plus the following (magnitudes), as one, to the following (magnitude) by itself.
15. A separation of a ratio is a taking of the (ratio of the) excess by which the leading (magnitude) exceeds the following to the following (magnitude) by itself.
16. A conversion of a ratio is a taking of the (ratio of the) leading (magnitude) to the excess by which the leading (magnitude) exceeds the following.
17. There being several magnitudes, and other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, a ratio via equality (or ex aequali) occurs when as the first is to the last in the first (set of) magnitudes, so the first (is) to the last in the second (set of) magnitudes. Or alternately, (it is) a taking of the (ratio of the) outer (magnitudes) by the removal of the inner (magnitudes).
18. There being three magnitudes, and other (magnitudes) of equal number to them, a perturbed proportion occurs when as the leading is to the following in the first (set of) magnitudes, so the leading (is) to the following in the second (set of) magnitudes, and as the following (is) to some other (i.e., the remaining magnitude) in the first (set of) magnitudes, so some other (is) to the leading in the second (set of) magnitudes.
Proposition 1
If there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second).
Let there be any number of magnitudes whatsoever, AB, CD, (which are) equal multiples, respectively, of some (other) magnitudes, E, F, of equal number (to them).
I say that as many times as AB is (divisible) by E, so many times will AB, CD also be (divisible) by E, F.
For since AB, CD are equal multiples of E, F, thus as many magnitudes as (there) are in AB equal to E, so many (are there) also in CD equal to F.
Let AB have been divided into magnitudes AG, GB, equal to E, and CD into (magnitudes) CH, HD, equal to F.
So, the number of (divisions) AG, GB will be equal to the number of (divisions) CH, HD.
And since AG is equal to E, and CH to F, AG (is) thus equal to E, and AG, CH to E, F.
So, for the same (reasons), GB is equal to E, and GB, HD to E, F.
Thus, as many (magnitudes) as (there) are in AB equal to E, so many (are there) also in AB, CD equal to E, F.
Thus, as many times as AB is (divisible) by E, so many times will AB, CD also be (divisible) by E, F.
Thus, if there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second).
(Which is) the very thing it was required to show.
Proposition 2
If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magnitude) and the fifth, being added together, and the third and the sixth, (being added together), will also be equal multiples of the second (magnitude) and the fourth (respectively).
For let a first (magnitude) AB and a third DE be equal multiples of a second C and a fourth F (respectively).
And let a fifth (magnitude) BG and a sixth EH also be (other) equal multiples of the second C and the fourth F (respectively).
I say that the first (magnitude) and the fifth, being added together, (to give) AG, and the third (magnitude) and the sixth, (being added together, to give) DH, will also be equal multiples of the second (magnitude) C and the fourth F (respectively).
For since AB and DE are equal multiples of C and F (respectively), thus as many (magnitudes) as (there) are in AB equal to C, so many (are there) also in DE equal to F.
And so, for the same (reasons), as many (magnitudes) as (there) are in BG equal to C, so many (are there) also in EH equal to F.
Thus, as many (magnitudes) as (there) are in the whole of AG equal to C, so many (are there) also in the whole of DH equal to F.
Thus, as many times as AG is (divisible) by C, so many times will DH also be divisible by F.
Thus, the first (magnitude) and the fifth, being added together, (to give) AG, and the third (magnitude) and the sixth, (being added together, to give) DH, will also be equal multiples of the second (magnitude) C and the fourth F (respectively).
Thus, if a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magnitude) and the fifth, being added together, and the third and sixth, (being added together), will also be equal multiples of the second (magnitude) and the fourth (respectively).
(Which is) the very thing it was required to show.
Proposition 3
If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and equal multiples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal multiples of the second (magnitude) and the fourth, respectively.
For let a first (magnitude) A and a third C be equal multiples of a second B and a fourth D (respectively), and let the equal multiples EF and GH have been taken of A and C (respectively).
I say that EF and GH are equal multiples of B and D (respectively).
For since EF and GH are equal multiples of A and C (respectively), thus as many (magnitudes) as (there) are in EF equal to A, so many (are there) also in GH equal to C.
Let EF have been divided into magnitudes EK, KF equal to A, and GH into (magnitudes) GL, LH equal to C.
So,the number of (magnitudes) EK, KF will be equal to the number of (magnitudes) GL, LH.
And since A and C are equal multiples of B and D (respectively), and EK (is) equal to A, and GL to C, EK and GL are thus equal multiples of B and D (respectively).
So, for the same (reasons), KF and LH are equal multiples of B and D (respectively).
Therefore, since the first (magnitude) EK and the third GL are equal multiples of the second B and the fourth D (respectively), and the fifth (magnitude) KF and the sixth LH are also equal multiples of the second B and the fourth D (respectively), then the first (magnitude) and fifth, being added together, (to give) EF, and the third (magnitude) and sixth, (being added together, to give) GH, are thus also equal multiples of the second (magnitude) B and the fourth D (respectively) [Prop. 5.2].
Thus, if a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and equal multiples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal multiples of the second (magnitude) and the fourth, respectively.
(Which is) the very thing it was required to show.
Proposition 4
If a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever.
For let a first (magnitude) A have the same ratio to a second B that a third C (has) to a fourth D.
And let equal multiples E and F have been taken of A and C (respectively), and other random equal multiples G and H of B and D (respectively).
I say that as E (is) to G, so F (is) to H.
For let equal multiples K and L have been taken of E and F (respectively), and other random equal multiples M and N of G and H (respectively).
[And] since E and F are equal multiples of A and C (respectively), and the equal multiples K and L have been taken of E and F (respectively), K and L are thus equal multiples of A and C (respectively) [Prop. 5.3].
So, for the same (reasons), M and N are equal multiples of B and D (respectively).
And since as A is to B, so C (is) to D, and the equal multiples K and L have been taken of A and C (respectively), and the other random equal multiples M and N of B and D (respectively), then if K exceeds M then L also exceeds N, and if ( K is) equal (to M then L is also) equal (to N), and if ( K is) less (than M then L is also) less (than N) [Def. 5.5].
And K and L are equal multiples of E and F (respectively), and M and N other random equal multiples of G and H (respectively).
Thus, as E (is) to G, so F (is) to H [Def. 5.5].
Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever.
(Which is) the very thing it was required to show.
Proposition 5
If a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively).
For let the magnitude AB be the same multiple of the magnitude CD that the (part) taken away AE (is) of the (part) taken away CF (respectively).
I say that the remainder EB will also be the same multiple of the remainder FD as that which the whole AB (is) of the whole CD (respectively).
For as many times as AE is (divisible) by CF, so many times let EB also have been made (divisible) by CG.
And since AE and EB are equal multiples of CF and GC (respectively), AE and AB are thus equal multiples of CF and GF (respectively) [Prop. 5.1].
And AE and AB are assumed (to be) equal multiples of CF and CD (respectively).
Thus, AB is an equal multiple of each of GF and CD.
Thus, GF (is) equal to CD.
Let CF have been subtracted from both.
Thus, the remainder GC is equal to the remainder FD.
And since AE and EB are equal multiples of CF and GC (respectively), and GC (is) equal to DF, AE and EB are thus equal multiples of CF and FD (respectively).
And AE and AB are assumed (to be) equal multiples of CF and CD (respectively).
Thus, EB and AB are equal multiples of FD and CD (respectively).
Thus, the remainder EB will also be the same multiple of the remainder FD as that which the whole AB (is) of the whole CD (respectively).
Thus, if a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively).
(Which is) the very thing it was required to show.
Proposition 6
If two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples of them (respectively).
For let two magnitudes AB and CD be equal multiples of two magnitudes E and F (respectively).
And let the (parts) taken away AG and CH be equal multiples of E and F (respectively).
I say that the remainders GB and HD are also either equal to E and F (respectively), or (are) equal multiples of them.
For let GB be, first of all, equal to E.
I say that HD is also equal to F.
For let CK be made equal to F.
Since AG and CH are equal multiples of E and F (respectively), and GB (is) equal to E, and KC to F, AB and KH are thus equal multiples of E and F (respectively) [Prop. 5.2].
And AB and CD are assumed (to be) equal multiples of E and F (respectively).
Thus, KH and CD are equal multiples of E and F (respectively).
Therefore, KH and CD are each equal multiples of F.
Thus, KH is equal to CD.
Let CH have be taken away from both.
Thus, the remainder KC is equal to the remainder HD.
But, F is equal to KC.
Thus, HD is also equal to F.
Hence, if GB is equal to E then HD will also be equal to F.
So, similarly, we can show that even if GB is a multiple of E then HD will also be the same multiple of F.
Thus, if two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples of them (respectively).
(Which is) the very thing it was required to show.
Proposition 7
Equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ratio) to the equal (magnitudes).
Let A and B be equal magnitudes, and C some other random magnitude.
I say that A and B each have the same ratio to C, and (that) C (has the same ratio) to each of A and B.
For let the equal multiples D and E have been taken of A and B (respectively), and the other random multiple F of C.
Therefore, since D and E are equal multiples of A and B (respectively), and A (is) equal to B, D (is) thus also equal to E.
And F (is) different, at random.
Thus, if D exceeds F then E also exceeds F, and if ( D is) equal (to F then E is also) equal (to F), and if ( D is) less (than F then E is also) less (than F).
And D and E are equal multiples of A and B (respectively), and F another random multiple of C.
Thus, as A (is) to C, so B (is) to C [Def. 5.5].
[So] I say that C also has the same ratio to each of A and B.
For, similarly, we can show, by the same construction, that D is equal to E.
And F (has) some other (value).
Thus, if F exceeds D then it also exceeds E, and if ( F is) equal (to D then it is also) equal (to E), and if ( F is) less (than D then it is also) less (than E).
And F is a multiple of C, and D and E other random equal multiples of A and B.
Thus, as C (is) to A, so C (is) to B [Def. 5.5].
Thus, equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ratio) to the equal (magnitudes).
Corollary
So (it is) clear, from this, that if some magnitudes are proportional then they will also be proportional inversely.
(Which is) the very thing it was required to show.
Proposition 8
For unequal magnitudes, the greater (magnitude) has a greater ratio than the lesser to the same (magnitude).
And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater.
Let AB and C be unequal magnitudes, and let AB be the greater (of the two), and D another random magnitude.
I say that AB has a greater ratio to D than C (has) to D, and (that) D has a greater ratio to C than (it has) to AB.
For since AB is greater than C, let BE be made equal to C.
So, the lesser of AE and EB, being multiplied, will sometimes be greater than D [Def. 5.4].
First of all, let AE be less than EB, and let AE have been multiplied, and let FG be a multiple of it which (is) greater than D.
And as many times as FG is (divisible) by AE, so many times let GH also have become (divisible) by EB, and K by C.
And let the double multiple L of D have been taken, and the triple multiple M, and several more, (each increasing) in order by one, until the (multiple) taken becomes the first multiple of D (which is) greater than K.
Let it have been taken, and let it also be the quadruple multiple N of D ---the first (multiple) greater than K.
Therefore, since K is less than N first, K is thus not less than M.
And since FG and GH are equal multiples of AE and EB (respectively), FG and FH are thus equal multiples of AE and AB (respectively) [Prop. 5.1].
And FG and K are equal multiples of AE and C (respectively).
Thus, FH and K are equal multiples of AB and C (respectively).
Thus, FH, K are equal multiples of AB, C.
Again, since GH and K are equal multiples of EB and C, and EB (is) equal to C, GH (is) thus also equal to K.
And K is not less than M.
Thus, GH not less than M either.
And FG (is) greater than D.
Thus, the whole of FH is greater than D and M (added) together.
But, D and M (added) together is equal to N, inasmuch as M is three times D, and M and D (added) together is four times D, and N is also four times D.
Thus, M and D (added) together is equal to N.
But, FH is greater than M and D.
Thus, FH exceeds N.
And K does not exceed N.
And FH, K are equal multiples of AB, C, and N another random multiple of D.
Thus, AB has a greater ratio to D than C (has) to D [Def. 5.7].
So, I say that D also has a greater ratio to C than D (has) to AB.
For, similarly, by the same construction, we can show that N exceeds K, and N does not exceed FH.
And N is a multiple of D, and FH, K other random equal multiples of AB, C (respectively).
Thus, D has a greater ratio to C than D (has) to AB [Def. 5.5].
And so let AE be greater than EB.
So, the lesser, EB, being multiplied, will sometimes be greater than D.
Let it have been multiplied, and let GH be a multiple of EB (which is) greater than D.
And as many times as GH is (divisible) by EB, so many times let FG also have become (divisible) by AE, and K by C.
So, similarly (to the above), we can show that FH and K are equal multiples of AB and C (respectively).
And, similarly (to the above), let the multiple N of D, (which is) the first (multiple) greater than FG, have been taken.
So, FG is again not less than M.
And GH (is) greater than D.
Thus, the whole of FH exceeds D and M, that is to say N.
And K does not exceed N, inasmuch as FG, which (is) greater than GH ---that is to say, K ---also does not exceed N.
And, following the above (arguments), we (can) complete the proof in the same manner.
Thus, for unequal magnitudes, the greater (magnitude) has a greater ratio than the lesser to the same (magnitude).
And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater.
(Which is) the very thing it was required to show.
Proposition 9
(Magnitudes) having the same ratio to the same (magnitude) are equal to one another.
And those (magnitudes) to which the same (magnitude) has the same ratio are equal.
For let A and B each have the same ratio to C.
I say that A is equal to B.
For if not, A and B would not each have the same ratio to C [Prop. 5.8].
But they do.
Thus, A is equal to B.
So, again, let C have the same ratio to each of A and B.
I say that A is equal to B.
For if not, C would not have the same ratio to each of A and B [Prop. 5.8].
But it does.
Thus, A is equal to B.
Thus, (magnitudes) having the same ratio to the same (magnitude) are equal to one another.
And those (magnitudes) to which the same (magnitude) has the same ratio are equal.
(Which is) the very thing it was required to show.
Proposition 10
For (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater.
And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser.
For let A have a greater ratio to C than B (has) to C.
I say that A is greater than B.
For if not, A is surely either equal to or less than B.
In fact, A is not equal to B.
For (then) A and B would each have the same ratio to C [Prop. 5.7].
But they do not.
Thus, A is not equal to B.
Neither, indeed, is A less than B.
For (then) A would have a lesser ratio to C than B (has) to C [Prop. 5.8].
But it does not.
Thus, A is not less than B.
And it was shown not (to be) equal either.
Thus, A is greater than B.
So, again, let C have a greater ratio to B than C (has) to A.
I say that B is less than A.
For if not, (it is) surely either equal or greater.
In fact, B is not equal to A.
For (then) C would have the same ratio to each of A and B [Prop. 5.7].
But it does not.
Thus, A is not equal to B.
Neither, indeed, is B greater than A.
For (then) C would have a lesser ratio to B than (it has) to A [Prop. 5.8].
But it does not.
Thus, B is not greater than A.
And it was shown that (it is) not equal (to A) either.
Thus, B is less than A.
Thus, for (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater.
And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser.
(Which is) the very thing it was required to show.
Proposition 11
(Ratios which are) the same with the same ratio are also the same with one another.
For let it be that as A (is) to B, so C (is) to D, and as C (is) to D, so E (is) to F.
I say that as A is to B, so E (is) to F.
For let the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively).
And since as A is to B, so C (is) to D, and the equal multiples G and H have been taken of A and C (respectively), and the other random equal multiples L and M of B and D (respectively), thus if G exceeds L then H also exceeds M, and if ( G is) equal (to L then H is also) equal (to M), and if ( G is) less (than L then H is also) less (than M) [Def. 5.5].
Again, since as C is to D, so E (is) to F, and the equal multiples H and K have been taken of C and E (respectively), and the other random equal multiples M and N of D and F (respectively), thus if H exceeds M then K also exceeds N, and if ( H is) equal (to M then K is also) equal (to N), and if ( H is) less (than M then K is also) less (than N) [Def. 5.5].
But (we saw that) if H was exceeding M then G was also exceeding L, and if ( H was) equal (to M then G was also) equal (to L), and if ( H was) less (than M then G was also) less (than L).
And, hence, if G exceeds L then K also exceeds N, and if ( G is) equal (to L then K is also) equal (to N), and if ( G is) less (than L then K is also) less (than N).
And G and K are equal multiples of A and E (respectively), and L and N other random equal multiples of B and F (respectively).
Thus, as A is to B, so E (is) to F [Def. 5.5].
Thus, (ratios which are) the same with the same ratio are also the same with one another.
(Which is) the very thing it was required to show.
Proposition 12
If there are any number of magnitudes whatsoever (which are) proportional then as one of the leading (magnitudes is) to one of the following, so will all of the leading (magnitudes) be to all of the following.
Let there be any number of magnitudes whatsoever, A, B, C, D, E, F, (which are) proportional, (so that) as A (is) to B, so C (is) to D, and E to F.
I say that as A is to B, so A, C, E (are) to B, D, F.
For let the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively).
And since as A is to B, so C (is) to D, and E to F, and the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively), thus if G exceeds L then H also exceeds M, and K (exceeds) N, and if ( G is) equal (to L then H is also) equal (to M, and K to N), and if ( G is) less (than L then H is also) less (than M, and K than N) [Def. 5.5].
And, hence, if G exceeds L then G, H, K also exceed L, M, N, and if ( G is) equal (to L then G, H, K are also) equal (to L, M, N) and if ( G is) less (than L then G, H, K are also) less (than L, M, N).
And G and G, H, K are equal multiples of A and A, C, E (respectively), inasmuch as if there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second) [Prop. 5.1].
So, for the same (reasons), L and L, M, N are also equal multiples of B and B, D, F (respectively).
Thus, as A is to B, so A, C, E (are) to B, D, F (respectively).
Thus, if there are any number of magnitudes whatsoever (which are) proportional then as one of the leading (magnitudes is) to one of the following, so will all of the leading (magnitudes) be to all of the following.
(Which is) the very thing it was required to show.
Proposition 13
If a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the third (magnitude) has a greater ratio to the fourth than a fifth (has) to a sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth.
For let a first (magnitude) A have the same ratio to a second B that a third C (has) to a fourth D, and let the third (magnitude) C have a greater ratio to the fourth D than a fifth E (has) to a sixth F.
I say that the first (magnitude) A will also have a greater ratio to the second B than the fifth E (has) to the sixth F.
For since there are some equal multiples of C and E, and other random equal multiples of D and F, (for which) the multiple of C exceeds the (multiple) of D, and the multiple of E does not exceed the multiple of F [Def. 5.7], let them have been taken.
And let G and H be equal multiples of C and E (respectively), and K and L other random equal multiples of D and F (respectively), such that G exceeds K, but H does not exceed L.
And as many times as G is (divisible) by C, so many times let M be (divisible) by A.
And as many times as K (is divisible) by D, so many times let N be (divisible) by B.
And since as A is to B, so C (is) to D, and the equal multiples M and G have been taken of A and C (respectively), and the other random equal multiples N and K of B and D (respectively), thus if M exceeds N then G exceeds K, and if ( M is) equal (to N then G is also) equal (to K), and if ( M is) less (than N then G is also) less (than K) [Def. 5.5].
And G exceeds K.
Thus, M also exceeds N.
And H does not exceeds L.
And M and H are equal multiples of A and E (respectively), and N and L other random equal multiples of B and F (respectively).
Thus, A has a greater ratio to B than E (has) to F [Def. 5.7].
Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and a third (magnitude) has a greater ratio to a fourth than a fifth (has) to a sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth.
(Which is) the very thing it was required to show.
Proposition 14
If a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the first (magnitude) is greater than the third, then the second will also be greater than the fourth.
And if (the first magnitude is) equal (to the third then the second will also be) equal (to the fourth).
And if (the first magnitude is) less (than the third then the second will also be) less (than the fourth).
For let a first (magnitude) A have the same ratio to a second B that a third C (has) to a fourth D.
And let A be greater than C.
I say that B is also greater than D.
For since A is greater than C, and B (is) another random [magnitude], A thus has a greater ratio to B than C (has) to B [Prop. 5.8].
And as A (is) to B, so C (is) to D.
Thus, C also has a greater ratio to D than C (has) to B.
And that (magnitude) to which the same (magnitude) has a greater ratio is the lesser [Prop. 5.10].
Thus, D (is) less than B.
Hence, B is greater than D.
So, similarly, we can show that even if A is equal to C then B will also be equal to D, and even if A is less than C then B will also be less than D.
Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the first (magnitude) is greater than the third, then the second will also be greater than the fourth.
And if (the first magnitude is) equal (to the third then the second will also be) equal (to the fourth).
And if (the first magnitude is) less (than the third then the second will also be) less (than the fourth).
(Which is) the very thing it was required to show.
Proposition 15
Parts have the same ratio as similar multiples, taken in corresponding order.
For let AB and DE be equal multiples of C and F (respectively).
I say that as C is to F, so AB (is) to DE.
For since AB and DE are equal multiples of C and F (respectively), thus as many magnitudes as there are in AB equal to C, so many (are there) also in DE equal to F.
Let AB have been divided into (magnitudes) AG, GH, HB, equal to C, and DE into (magnitudes) DK, KL, LE, equal to F.
So, the number of (magnitudes) AG, GH, HB will equal the number of (magnitudes) DK, KL, LE.
And since AG, GH, HB are equal to one another, and DK, KL, LE are also equal to one another, thus as AG is to DK, so GH (is) to KL, and HB to LE [Prop. 5.7].
And, thus (for proportional magnitudes), as one of the leading (magnitudes) will be to one of the following, so all of the leading (magnitudes will be) to all of the following [Prop. 5.12].
Thus, as AG is to DK, so AB (is)to DE.
And AG is equal to C, and DK to F.
Thus, as C is to F, so AB (is) to DE.
Thus, parts have the same ratio as similar multiples, taken in corresponding order.
(Which is) the very thing it was required to show.
Proposition 16
If four magnitudes are proportional then they will also be proportional alternately.
Let A, B, C and D be four proportional magnitudes, (such that) as A (is) to B, so C (is) to D.
I say that they will also be [proportional] alternately, (so that) as A (is) to C, so B (i s) to magnitude}.
For let the equal multiples E and F have been taken of A and B (respectively), and the other random equal multiples G and H of C and D (respectively).
And since E and F are equal multiples of A and B (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as A is to B, so E (is) to F.
But as A (is) to B, so C (is) to D.
And, thus, as C (is) to D, so E (is) to F [Prop. 5.11].
Again, since G and H are equal multiples of C and D (respectively), thus as C is to D, so G (is) to H [Prop. 5.15].
But as C (is) to D, [so] E (is) to F.
And, thus, as E (is) to F, so G (is) to H [Prop. 5.11].
And if four magnitudes are proportional, and the first is greater than the third then the second will also be greater than the fourth, and if (the first is) equal (to the third then the second will also be) equal (to the fourth), and if (the first is) less (than the third then the second will also be) less (than the fourth) [Prop. 5.14].
Thus, if E exceeds G then F also exceeds H, and if ( E is) equal (to G then F is also) equal (to H), and if ( E is) less (than G then F is also) less (than H).
And E and F are equal multiples of A and B (respectively), and G and H other random equal multiples of C and D (respectively).
Thus, as A is to C, so B (is) to D [Def. 5.5].
Thus, if four magnitudes are proportional then they will also be proportional alternately.
(Which is) the very thing it was required to show.
Proposition 17
If composed magnitudes are proportional then they will also be proportional (when) separarted.
Let AB, BE, CD, and DF be composed magnitudes (which are) proportional, (so that) as AB (is) to BE, so CD (is) to DF.
I say that they will also be proportional (when) separated, (so that) as AE (is) to EB, so CF (is) to DF.
For let the equal multiples GH, HK, LM, and MN have been taken of AE, EB, CF, and FD (respectively), and the other random equal multiples KO and NP of EB and FD (respectively).
And since GH and HK are equal multiples of AE and EB (respectively), GH and GK are thus equal multiples of AE and AB (respectively) [Prop. 5.1].
But GH and LM are equal multiples of AE and CF (respectively).
Thus, GK and LM are equal multiples of AB and CF (respectively).
Again, since LM and MN are equal multiples of CF and FD (respectively), LM and LN are thus equal multiples of CF and CD (respectively) [Prop. 5.1].
And LM and GK were equal multiples of CF and AB (respectively).
Thus, GK and LN are equal multiples of AB and CD (respectively).
Thus, GK, LN are equal multiples of AB, CD.
Again, since HK and MN are equal multiples of EB and FD (respectively), and KO and NP are also equal multiples of EB and FD (respectively), then, added together, HO and MP are also equal multiples of EB and FD (respectively) [Prop. 5.2].
And since as AB (is) to BE, so CD (is) to DF, and the equal multiples GK, LN have been taken of AB, CD, and the equal multiples HO, MP of EB, FD, thus if GK exceeds HO then LN also exceeds MP, and if ( GK is) equal (to HO then LN is also) equal (to MP), and if ( GK is) less (than HO then LN is also) less (than MP) [Def. 5.5].
So let GK exceed HO, and thus, HK being taken away from both, GH exceeds KO.
But (we saw that) if GK was exceeding HO then LN was also exceeding MP.
Thus, LN also exceeds MP, and, MN being taken away from both, LM also exceeds NP.
Hence, if GH exceeds KO then LM also exceeds NP.
So, similarly, we can show that even if GH is equal to KO then LM will also be equal to NP, and even if ( GH is) less (than KO then LM will also be) less (than NP).
And GH, LM are equal multiples of AE, CF, and KO, NP other random equal multiples of EB, FD.
Thus, as AE is to EB, so CF (is) to FD [Def. 5.5].
Thus, if composed magnitudes are proportional then they will also be proportional (when) separated.
(Which is) the very thing it was required to show.
Proposition 18
If separated magnitudes are proportional then they will also be proportional (when) composed.
Let AE, EB, CF, and FD be separated magnitudes (which are) proportional, (so that) as AE (is) to EB, so CF (is) to FD.
I say that they will also be proportional (when) composed, (so that) as AB (is) to BE, so CD (is) to FD.
For if (it is) not (the case that) as AB is to BE, so CD (is) to FD, then it will surely be (the case that) as AB (is) to BE, so CD is either to some (magnitude) less than DF, or (some magnitude) greater (than DF).
Let it, first of all, be to (some magnitude) less (than DF), (namely) DG.
And since composed magnitudes are proportional, (so that) as AB is to BE, so CD (is) to DG, they will thus also be proportional (when) separated [Prop. 5.17].
Thus, as AE is to EB, so CG (is) to GD.
But it was also assumed that as AE (is) to EB, so CF (is) to FD.
Thus, (it is) also (the case that) as CG (is) to GD, so CF (is) to FD [Prop. 5.11].
And the first (magnitude) CG (is) greater than the third CF.
Thus, the second (magnitude) GD (is) also greater than the fourth FD [Prop. 5.14].
But (it is) also less.
The very thing is impossible.
Thus, (it is) not (the case that) as AB is to BE, so CD (is) to less than FD.
Similarly, we can show that neither (is it the case) to greater (than FD).
Thus, (it is the case) to the same (as FD).
Thus, if separated magnitudes are proportional then they will also be proportional (when) composed.
(Which is) the very thing it was required to show.
Proposition 19
If as the whole is to the whole so the (part) taken away is to the (part) taken away then the remainder to the remainder will also be as the whole (is) to the whole.
For let the whole AB be to the whole CD as the (part) taken away AE (is) to the (part) taken away CF.
I say that the remainder EB to the remainder FD will also be as the whole AB (is) to the whole CD.
For since as AB is to CD, so AE (is) to CF, (it is) also (the case), alternately, (that) as BA (is) to AE, so DC (is) to CF [Prop. 5.16].
And since composed magnitudes are proportional then they will also be proportional (when) separated, (so that) as BE (is) to EA, so DF (is) to CF [Prop. 5.17].
Also, alternately, as BE (is) to DF, so EA (is) to FC [Prop. 5.16].
And it was assumed that as AE (is) to CF, so the whole AB (is) to the whole CD.
And, thus, as the remainder EB (is) to the remainder FD, so the whole AB will be to the whole CD.
Thus, if as the whole is to the whole so the (part) taken away is to the (part) taken away then the remainder to the remainder will also be as the whole (is) to the whole.
[And since it was shown (that) as AB (is) to CD, so EB (is) to FD, (it is) also (the case), alternately, (that) as AB (is) to BE, so CD (is) to FD.
Thus, composed magnitudes are proportional.
And it was shown (that) as BA (is) to AE, so DC (is) to CF.
And (the latter) is converted (from the former).]
Corollary
So (it is) clear, from this, that if composed magnitudes are proportional then they will also be proportional (when) converted.
(Which is) the very thing it was required to show.
Proposition 20
If there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if), via equality, the first is greater than the third then the fourth will also be greater than the sixth.
And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth).
And if (the first is) less (than the third then the fourth will also be) less (than the sixth).
Let A, B, and C be three magnitudes, and D, E, F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two, (so that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F.
And let A be greater than C, via equality.
I say that D will also be greater than F.
And if ( A is) equal (to C then D will also be) equal (to F).
And if ( A is) less (than C then D will also be) less (than F).
For since A is greater than C, and B some other (magnitude), and the greater (magnitude) has a greater ratio than the lesser to the same (magnitude) [Prop. 5.8], A thus has a greater ratio to B than C (has) to B.
But as A (is) to B, [so] D (is) to E.
And, inversely, as C (is) to B, so F (is) to E [Prop. 5.7 corr.].
Thus, D also has a greater ratio to E than F (has) to E [Prop. 5.13].
And for (magnitudes) having a ratio to the same (magnitude), that having the greater ratio is greater [Prop. 5.10].
Thus, D (is) greater than F.
Similarly, we can show that even if A is equal to C then D will also be equal to F, and even if ( A is) less (than C then D will also be) less (than F).
Thus, if there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if), via equality, the first is greater than the third, then the fourth will also be greater than the sixth.
And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth).
And (if the first is) less (than the third then the fourth will also be) less (than the sixth).
(Which is) the very thing it was required to show.
Proposition 21
If there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if) their proportion (is) perturbed, and (if), via equality, the first is greater than the third then the fourth will also be greater than the sixth.
And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth).
And if (the first is) less (than the third then the fourth will also be) less (than the sixth).
Let A, B, and C be three magnitudes, and D, E, F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two.
And let their proportion be perturbed, (so that) as A (is) to B, so E (is) to F, and as B (is) to C, so D (is) to E.
And let A be greater than C, via equality.
I say that D will also be greater than F.
And if ( A is) equal (to C then D will also be) equal (to F).
And if ( A is) less (than C then D will also be) less (than F).
For since A is greater than C, and B some other (magnitude), A thus has a greater ratio to B than C (has) to B [Prop. 5.8].
But as A (is) to B, so E (is) to F.
And, inversely, as C (is) to B, so E (is) to D [Prop. 5.7 corr.].
Thus, E also has a greater ratio to F than E (has) to D [Prop. 5.13].
And that (magnitude) to which the same (magnitude) has a greater ratio is (the) lesser (magnitude) [Prop. 5.10].
Thus, F is less than D.
Thus, D is greater than F.
Similarly, we can show that even if A is equal to C then D will also be equal to F, and even if ( A is) less (than C then D will also be) less (than F).
Thus, if there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if) their proportion (is) perturbed, and (if), via equality, the first is greater than the third then the fourth will also be greater than the sixth.
And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth).
And if (the first is) less (than the third then the fourth will also be) less (than the sixth).
(Which is) the very thing it was required to show.
Proposition 22
If there are any number of magnitudes whatsoever, and (some) other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality.
Let there be any number of magnitudes whatsoever, A, B, C, and (some) other (magnitudes), D, E, F, of equal number to them, (which are) in the same ratio taken two by two, (so that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F.
I say that they will also be in the same ratio via equality.
(That is, as A is to C, so D is to F.)
For let the equal multiples G and H have been taken of A and D (respectively), and the other random equal multiples K and L of B and E (respectively), and the yet other random equal multiples M and N of C and F (respectively).
And since as A is to B, so D (is) to E, and the equal multiples G and H have been taken of A and D (respectively), and the other random equal multiples K and L of B and E (respectively), thus as G is to K, so H (is) to L [Prop. 5.4].
And, so, for the same (reasons), as K (is) to M, so L (is) to N.
Therefore, since G, K, and M are three magnitudes, and H, L, and N other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, thus, via equality, if G exceeds M then H also exceeds N, and if ( G is) equal (to M then H is also) equal (to N), and if ( G is) less (than M then H is also) less (than N) [Prop. 5.20].
And G and H are equal multiples of A and D (respectively), and M and N other random equal multiples of C and F (respectively).
Thus, as A is to C, so D (is) to F [Def. 5.5].
Thus, if there are any number of magnitudes whatsoever, and (some) other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality.
(Which is) the very thing it was required to show.
Proposition 23
If there are three magnitudes, and others of equal number to them, (being) in the same ratio taken two by two, and (if) their proportion is perturbed, then they will also be in the same ratio via equality.
Let A, B, and C be three magnitudes, and D, E and F other (magnitudes) of equal number to them, (being) in the same ratio taken two by two.
And let their proportion be perturbed, (so that) as A (is) to B, so E (is) to F, and as B (is) to C, so D (is) to E.
I say that as A is to C, so D (is) to F.
Let the equal multiples G, H, and K have been taken of A, B, and D (respectively), and the other random equal multiples L, M, and N of C, E, and F (respectively).
And since G and H are equal multiples of A and B (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as A (is) to B, so G (is) to H.
And, so, for the same (reasons), as E (is) to F, so M (is)to N.
And as A is to B, so E (is) to F.
And, thus, as G (is) to H, so M (is) to N [Prop. 5.11].
And since as B is to C, so D (is) to E, also, alternately, as B (is) to D, so C (is) to E [Prop. 5.16].
And since H and K are equal multiples of B and D (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as B is to D, so H (is) to K.
But, as B (is) to D, so C (is) to E.
And, thus, as H (is) to K, so C (is) to E [Prop. 5.11].
Again, since L and M are equal multiples of C and E (respectively), thus as C is to E, so L (is) to M [Prop. 5.15].
But, as C (is) to E, so H (is) to K.
And, thus, as H (is) to K, so L (is) to M [Prop. 5.11].
Also, alternately, as H (is) to L, so K (is) to M [Prop. 5.16].
And it was also shown (that) as G (is) to H, so M (is) to N.
Therefore, since G, H, and L are three magnitudes, and K, M, and N other (magnitudes) of equal number to them, (being) in the same ratio taken two by two, and their proportion is perturbed, thus, via equality, if G exceeds L then K also exceeds N, and if ( G is) equal (to L then K is also) equal (to N), and if ( G is) less (than L then K is also) less (than N) [Prop. 5.21].
And G and K are equal multiples of A and D (respectively), and L and N of C and F (respectively).
Thus, as A (is) to C, so D (is) to F [Def. 5.5].
Thus, if there are three magnitudes, and others of equal number to them, (being) in the same ratio taken two by two, and (if) their proportion is perturbed, then they will also be in the same ratio via equality.
(Which is) the very thing it was required to show.
Proposition 24
If a first (magnitude) has to a second the same ratio that third (has) to a fourth, and a fifth (magnitude) also has to the second the same ratio that a sixth (has) to the fourth, then the first (magnitude) and the fifth, added together, will also have the same ratio to the second that the third (magnitude) and sixth (added together, have) to the fourth.
For let a first (magnitude) AB have the same ratio to a second C that a third DE (has) to a fourth F.
And let a fifth (magnitude) BG also have the same ratio to the second C that a sixth EH (has) to the fourth F.
I say that the first (magnitude) and the fifth, added together, AG, will also have the same ratio to the second C that the third (magnitude) and the sixth, (added together), DH, (has) to the fourth F.
For since as BG is to C, so EH (is) to F, thus, inversely, as C (is) to BG, so F (is) to EH [Prop. 5.7 corr.].
Therefore, since as AB is to C, so DE (is) to F, and as C (is) to BG, so F (is) to EH, thus, via equality, as AB is to BG, so DE (is) to EH [Prop. 5.22].
And since separated magnitudes are proportional then they will also be proportional (when) composed [Prop. 5.18].
Thus, as AG is to GB, so DH (is) to HE.
And, also, as BG is to C, so EH (is) to F.
Thus, via equality, as AG is to C, so DH (is) to F [Prop. 5.22].
Thus, if a first (magnitude) has to a second the same ratio that a third (has) to a fourth, and a fifth (magnitude) also has to the second the same ratio that a sixth (has) to the fourth, then the first (magnitude) and the fifth, added together, will also have the same ratio to the second that the third (magnitude) and the sixth (added together, have) to the fourth.
(Which is) the very thing it was required to show.
Proposition 25
If four magnitudes are proportional then the (sum of the) largest and the smallest [of them] is greater than the (sum of the) remaining two (magnitudes).
Let AB, CD, E, and F be four proportional magnitudes, (such that) as AB (is) to CD, so E (is) to F.
And let AB be the greatest of them, and F the least.
I say that AB and F is greater than CD and E.
For let AG be made equal to E, and CH equal to F.
[In fact,] since as AB is to CD, so E (is) to F, and E (is) equal to AG, and F to CH, thus as AB is to CD, so AG (is) to CH.
And since the whole AB is to the whole CD as the (part) taken away AG (is) to the (part) taken away CH, thus the remainder GB will also be to the remainder HD as the whole AB (is) to the whole CD [Prop. 5.19].
And AB (is) greater than CD.
Thus, GB (is) also greater than HD.
And since AG is equal to E, and CH to F, thus AG and F is equal to CH and E.
And [since] if [equal (magnitudes) are added to unequal (magnitudes) then the wholes are unequal, thus if] AG and F are added to GB, and CH and E to HD --- GB and HD being unequal, and GB greater---it is inferred that AB and F (is) greater than CD and E.
Thus, if four magnitudes are proportional then the (sum of the) largest and the smallest of them is greater than the (sum of the) remaining two (magnitudes).
(Which is) the very thing it was required to show.