Construction. Because AB is the same multiple of E that CD is of F, as many magnitudes as there are in AB equal to E, so many are there in CD equal to F. Divide AB into magnitudes equal to E, viz., AG, GB; and CD into CH, HD, equal each of them to F.

Demonstration. Because the number of the magnitudes CH and HD is equal to the number of the others AG and GB; and AG is equal to E, and CH to F; therefore

1. AG and CH together are equal to E and F together (I. Ax. 2). For the same reason, because GB is equal to E, and HD to F; 2. GB and HD together are equal to E and F together;

wherefore

therefore

3. As many magnitudes as there are in AB equal_to E, so many are there in AB, CD together, equal to E and F together;

4. Whatsoever multiple AB is of E, the same multiple is AB and CD together, of E and F together.

Therefore, if any magnitudes, how many soever, be equimultiples of as many, each of each; whatsoever multiple any one of them is of its part, the same multiple shall all the first magnitudes be of all the others:

"For the same demonstration holds in any number of magnitudes, which was here applied to two." Q.E.D.

PROPOSITION 2.-Theorem.

If the first magnitude be the same multiple of the second that the third is of the fourth, and the fifth the same multiple of the second that the sixth is of the fourth; then shall the first together with the fifth be the same multiple of the second, that the third together with the sixth is of the fourth.

Let AB the first be the same multiple of C the second, that DE the third is of F the fourth; and BG the fifth the same multiple of C the second, that EH the sixth is of F the fourth.

Then shall AG, the first together with the fifth, be the same multiple of C the second, that DH, the third together with the sixth, is of F the fourth.

Demonstration. Because AB is the same multiple of C that DE

is of F;

1. There are as many magnitudes in AB equal to C, as there are in DE equal to F.

In like manner,

therefore

2. There are as many magnitudes in BG equal to C, as there are in EH equal to ř1;

3. As many as there are in the whole AG equal to C, so many are there in the whole DH equal to F;

therefore AG is the same multiple of C that DH is of F; that is, 4. AG, the first and fifth together, is the same multiple of the second C, that DH, the third and sixth together, is of the fourth F

If therefore, the first be the same multiple, &c. Q.E.D. Cor. From this it is plain, that if any number of magnitudes AB, BG, GH be multiples of another C; and as many DE, EK, KL be the same multiples of F, each of each; then the whole of the first, viz., AH, is the same multiple of C, that the whole of the last, viz., DL, is of F.

If the first be the same multiple of the second, which the d is of the fourth; and if of the first and third there be taken equimultiples, these shall be equimultiples, the one of the second, and the other of the fourth.

Let A the first be the same multiple of B the second, that C the third is of D the fourth; and of A, U let equimultiples EF, GH be

taken.

Then EF shall be the same multiple of B, that GH is of D.

Construction. Divide EF into the magnitudes EK, KF, each equal to A, and GH into GL, LH, each equal to C.

Demonstration. Because the number of the magnitudes EK and KF, is equal to the number of the others GL and LH; and A is the same multiple of B, that C is of D, and EK is equal to A, and GL to C (constr.); therefore

1. EK is the same multiple of B, that GL is of D;

for the same reason,

and

2. KF is the same multiple of B, that LH is of D;

So, if there be more parts in EF, GH, equal to A, C; therefore, because the first EK is the same multiple of the second B, which the third GL is of the fourth D, and that the fifth KF is the same multiple of the second B, which the sixth LH is of the fourth D;

3.

EF the first, together with the fifth, is the same multiple of the second B(V. 2), which GH the third, together with the sixth, is of the fourth D.

If, therefore, the first, &c. Q.E.D.

PROPOSITION 4.-Theorem.

If the first of four magnitudes has the same ratio to the second which the third has to the fourth; then any equimultiples whatever of the first and third shall have the same ratio to any equimultiples of the second and fourth, viz., "the equimultiple of the first shall have the same ratio to that of the second, which the equimultiple of the third has to that of the fourth."

Let A the first have to B the second, the same ratio which the third C has to the fourth D; and of A and C let there be taken any equimultiples whatever E, F; and of B and D any equimultiples whatever G, H.

Then E shall have the same ratio to G, which F has to H.

Construction. Take of E and Fany equimultiples whatever K, L, and of G, H any equimultiples whatever M, N.

Demonstration. Then because E is the same multiple of A, that F is of C; and of E and F have been taken equimultiples K, L;

1. K is the same multiple of A, that L is of C (V. 3)

for the same reason,

2. M is the same multiple of B, that N is of D.

And because, as A is to B, so is C to D (hyp.), and of A and C have been taken certain equimultiples K, L, and of B and D have been taken certain equimultiples M, N; therefore

3. If K be greater than M, L is greater than N; and if equal, equal; if less, less (V. Def. 5);

but K, L are any equimultiples whatever of E, F (constr.), and M, N any whatever of G, H; therefore

4. As E is to G, so is F to H (V. Def. 5).

Therefore, if the first, &c. Q.E.D.

Cor. Likewise, if the first has the same ratio to the second, which the third has to the fourth, then also any equimultiples whatever of the firstand third shall have the same ratio to the second and fourth; and in like manner, the first and the third shall have the same ratio to any equimultiples whatever of the second and fourth.

Let A the first have to B the second the same ratio which the third C has to the fourth D; and of A and C let E and F be any equimultiples whatever.

Then E shall be to B as F to D.

Construction. Take of E, F any equimultiples whatever K, L, and of B, D any equimultiples whatever G, H.

Demonstration. Then it may be demonstrated, as before, that K is the same multiple of A, that L is of C; and because A is to B, as C is to D (hyp.), and of A and C certain equimultiples have been taken, viz., K and L; and of B and D certain equimultiples G, H; therefore

1. If K be greater than G, L is greater than H; and if equal, equal; if less, less (V. Def. 5);

but K, L are any equimultiples whatever of E, F (constr.), and G, H any whatever of B, D; therefore

2. As E is to B, so is F to D (V. Def. 5).

And in the same way the other case is demonstrated.

PROPOSITION 5.—Theorem.

If one magnitude be the same multiple of another, which a magnitude taken from the first is of a magnitude taken from the other; the remainder shall be the same multiple of the remainder, that the whole is of the whole.

Let the magnitude AB be the same multiple of CD, that AE taken from the first, is of CF taken from the other.

Then the remainder EB shall be the same multiple of the remainder FD, that the whole AB is of the whole CD.

Construction. Take AG the same multiple of FD, that AE is of

CF.

Demonstration. Then AE is the same multiple of CF, that EG is of CD (V. 1); but AE, by the hypothesis, is the same multiple of CF, that AB is of CD; therefore EG is the same multiple of CD that AB is of CD; wherefore

1. EG is equal to AB (V. Ax. 1) ;

take from each of them the common magnitude AE; and

2. The remainder AG is equal to the remainder EB. Wherefore, since AE is the same multiple of CF, that AG is of FD, (constr.), and that AG has been proved equal to EB, therefore

3. AE is the same multiple of CF, that EB is of FD; but AE is the same multiple of CF that AB is of CD (hyp.); therefore

4. EB is the same multiple of FD, that AB is of CD.

Therefore, if one magnitude, &c. Q.E.D.

PROPOSITION 6.-Theorem.

If two magnitudes be equimultiples of two others, and if equimultiples of these be taken from the first two; the remainders are either equal to these others, or equimultiples of them.