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

For, because EF is the same multiple of A that GA is of C, [Hypothesis. as many magnitudes as there are in EF equal to A, so many are there in GH equal to C. Divide EF into the magnitudes EK, KF, each equal to A; and GH into the magnitudes GL, LH, each equal to C.

the magnitudes EK, KF, will be equal to the number of the magnitudes GL, LH.

And because A is the same multiple of B that is of D, .

(Hypothesis. and that EK is equal to A, and GL is equal to C'; [Constr. therefore EK is the same multiple of B that GL is of D.

For the same reason KF is the same multiple of B that LH is of D.

Therefore because EK the first is the same multiple of B the second, that GL the third is of D the fourth, and that KF the fifth is the same multiple of B the second, that LH the sixth is of D the fourth; EF the first together with the fifth, is the same multiple of B the second, that GH the third together with the sixth, is of D the fourth.

[V. 2. In the same manner, if there be more parts in EF equal to A and in GH equal to C, it may be shewn that is the same multiple of B that GH is of D. (V. 2, Cor. Wherefore, if the first &c. Q.E.D.

PROPOSITION 4. THEOREM. If the first have the same ratio to the second that the third has to the fourth, and if there be taken any equi

multiples whatever of the first and the third, and also any equimultiples whatever of the second and the fourth, then the multiple of the first shall have the same ratio to the multiple of the second, that the multiple of the third has to the multiple of the fourth.

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

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

Then, because E is the same multiple of A that F is of C, and of E and F have been taken equimultiples K and L; therefore K is the same mul- ġ É À B Ġ Ý tiple of A that L is of C. [V. 3. Į F Ç D H N

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

And because A is to B as c is to D,

[Hypothesis. and of A and C have been taken certain equimultiples K and L, and of B and D have been taken certain equimultiples M and N; therefore if K be greater than M, L is greater than N; and if equal, equal; and if less, less.

[V. Definition 5. But K and L are any equimultiples whatever of E and F, and M and N are any equimultiples whatever of G and H; therefore E is to G as F is to H.

[V. Definition 5. Wherefore, if the first &c. Q.E.D.

COROLLARY. Also if the first have the same ratio to the second that the third has to the fourth, then any equimultiples whatever of the first and third shall have the same ratio to the second and fourth; and the first and

third shall have the same ratio to any equimultiples whatever of the second and fourth.

Let A the first have the same ratio to B the second, that C the third has to D the fourth ; and of A and C let there be taken any equimultiples whatever E and F: E shall be to B as F is to D.

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

Then it may be shewn, 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, [Hypothesis. and of A and C have been taken certain equimultiples K and L, and of B and D have been taken certain equimultiples G and H; therefore if K be greater than G, L is greater than H; and if equal, equal; and if less, less.

[V. Definition 5. But K and I are any equimultiples whatever of E and F, and G and H are any equimultiples whatever of B and D; therefore E is to B as Fis to D.

[V. Definition 5. In the same way the other case may be demonstrated.

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PROPOSITION 5. THEOREM. If one magnitude be the same multiple of another that 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 AB be the same multiple of CD, that AE taken from the first, is of CF taken from the other: the remainder EB shall be the same multiple of the remainder FD, that the whole AB is of the whole CD.

Take AG the same multiple of FD, that AE is of CF; therefore AE is the same multiple of CF that EG is of CD.

[V. 1. But AE is the same multiple of CF that AB is of CD; therefore EG is the same multiple of CD that AB is of CD; therefore EG is equal to AB.

[V. Axiom 1.

From each of these take the common magnitude AE; then the remainder AG is equal to the remainder EB.

Then, because AE is the same multiple of CF that AG is of FD, [Construction. and that AG is equal to EB; therefore 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;

[Hypothesis. therefore EB is the same multiple of FD that AB is of CD.

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

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PROPOSITION 6. THEOREM. If two magnitudes be equimultiples of two others, and if equimultiples of these be taken from the first two, the remainders shall be either equal to these others, or equimultiples of them.

Let the two magnitudes AB, CD be equimultiples of the two E, F; and let AG, CH, taken from the first two, be equimultiples of the same E, F: the remainders GB, HD shall be either equal to E, F, or equimultiples of them.

First, let GB be equal to E: HD shall be equal to F. Make CK equal to F. Then, because AG is the same multiple of E that CH is of F, [Hyp. and that GB is equal to E, and CK is equal to F; therefore AB is the same multiple of E that KH is of F.

But AB is the same multiple of E that CD is of F; [Hypothesis. B D E F therefore KH is the same multiple of F that CD is of F; therefore KH is equal to CD.

[V. Axiom 1. From each of these take the common magnitude CH; then the remainder CK is equal to the remainder HD. But CK is equal to F;

[Construction. therefore HD is equal to F.

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Next let GB be a multiple of E: HD shall be the same multiple of F. Make CK the same multiple of F that GB is of E. Then, because AG is the same multiple of E that CH is of

[Hypothesis. and GB is the same multiple of E that CK is of F; [Constr. therefore AB is the same multiple of Ethat KH is of F. [V.2. But AB is the same multi

в D E F ple of E that CD is of F; [Hyp. therefore KH is the same multiple of F that CD is of F; therefore KH is equal to CD.

[V. Axiom 1. From each of these take the common magnitude CH; then the remainder CK is equal to the remainder HD.

And because CK is the same multiple of F that GB is of E,

[Construction. and that CK is equal to HD; therefore HD is the same multiple of F that GB is of E.

Wherefore, if two magnitudes &c. Q.E.D.

PROPOSITION A. THEOREM. If the first of four magnitudes have the same ratio to the second that the third has to the fourth, then, if the first be greater than the second, the third shall also be greater than the fourth, and if equal equal, and if less less.

Take any equimultiples of each of them, as the doubles of each. Then if the double of the first be greater than the double of the second, the double of the third is greater than the double of the fourth.

[V. Definition 5. But if the first be greater than the second, the double of the first is greater than the double of the second;

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