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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.

DEMONSTRATION. Take of E, F any equimultiples whatever K, L, and of B, D any equimultiples whatever G, H; 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 (b), and of A and C certain equimultiples have been taken, viz. K and L; and of B and D certain equimultiples G, H; therefore if K be greater than G, L is greater than H; and if equal, equal; if less, less (c); but K, L are any equimultiples whatever of E, F, and G, H any whatever of B, D; therefore, as E is to B, so is F to D (c). And in a similar way the other case is demonstrated.

SCHOLIUM. The fourth proposition may be algebraically expressed as follows:

THEOREM. If A a :: B: b; then mAm B: na: nb.

Because A:a:: B:b

K

L

- Z

E

A

B

G M

F

C

D H

N

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multiplying both sides by a, and dividing both sides by B,

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therefore m. Am. B:: n. an. b.

If " be taken equal to unity, the above will be a demonstration of the corollary.

PROPOSITION V.

THEOREM.-If one magnitude be the same multiple of another which a part taken from the first is of a part taken from the other, the first remainder is the same multiple of the second that the first magnitude is of the second.

Let the magnitude 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.

DEMONSTRATION. 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 (a); but AE is the same multiple of CF that AB is of CD (b); therefore EG is the same multiple of CD that AB is of CD; wherefore EG is equal to AB (c); take from each of them the common magnitude AE; and the remainder AG is equal to the remainder EB. Wherefore, since AE is the same multiple of CF that AG is of FD, 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 (6); therefore EB is the same multiple of FD that AB is of CD. Therefore, if one magnitude, &c.

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SCHOLIUM. The foregoing proposition, algebraically expressed, is as follows:

THEOREM. If A is the same multiple of a that B is of b, then A - B is the same multiple of a - b. For let

then,

A m. a, and B = m.

m.b,

A-Bm.am.bm. . (a - b).

PROPOSITION VI.

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.

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 remainder GB, HD, shall be either equal to E, F, or equimultiples of them.

DEMONSTRATION. First, let GB be equal to E: HD shall be equal to F. Make CK equal to F and because AG is the same multiple of E that CH is of F (a), and that GB is equal to E, and CK 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; therefore KH is the same multiple of F that CD is of F; wherefore KH is equal to CD (b); take away the common magnitude CH, then the remainder CK is equal to the remainder HD; but CK is equal to F; therefore HD is equal to F.

G

B

K

C

.H.

D E

F

81.

(a) Hypoth.

A

G

K

C

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; and because AG is the same multiple of E that CH is of F (a); and GB the same multiple of E that CK is of F; therefore AB is the same multiple of E that KH is of F (c); but AB is the same multiple of E that CD is of F (a); therefore KH is the same multiple of F that CD is of F; wherefore KH is equal to CD (b); take away CH from both, and the remainder KC is equal to the remainder HD; and because GB is the same multiple of E that KC is of F, and that KC is equal to HD; therefore HD is the same multiple of F that GB is of E.

H

B

D E

F

(c) V. 2.

SCHOLIUM. The foregoing proposition, algebraically expressed, is as follows:

THEOREM. If A, B be equimultiples of a and b, then A are either equimultiples of a, b, or are equal to them.

- m. a, B-m.b

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which is the second case in Euclid; when n = 2, and m = 1

A

― m. a = a, and B -m.b = b,

which is the first case in Euclid.

PROPOSITION A.

THEOREM.-If the first of four magnitudes have the same ratio to the second which the third has to the fourth, then, if the first be greater than the second, the third is also greater than the fourth; and if equal, equal; if less, less.

DEMONSTRATION. Take any equimultiples of each of them, as the doubles of each; then by Def. 5 of this book, 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; but if the first be greater than the second, the double of the first is greater than the double of the second; wherefore, also, the double of the third is greater than the double of the fourth; therefore the third is greater than the fourth; in like manner, if the first be equal to the second, or less than it, the third can be proved to be equal to the fourth, or less than it.

SCHOLIUM. This proposition and the three following have been added by Simson. It may be expressed algebraically as follows:

THEOREM. If A: a :: B: b, then, according as A is >, =, or <a B is >,, or < b.

Let any equimultiples of them be taken, as

m. A, m. a, m. B, m.

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b;

>,, or <m.b.

But if A be,=, or <a, then m. A is >,=, or < m. .a;

therefore m. B is >,, or < m. .b,

and B is,=, or < b.

Therefore, according as A is >,=, or <a, B is >,, or < b.

PROPOSITION B.

THEOREM.-If four magnitudes are proportionals, they are proportionals also when taken inversely.

If the magnitude A be to B as C is to D, then also inversely B is to A as D to C.

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DEMONSTRATION. Take of B and D any equimultiples whatever E and F; and of Ă and C any equimultiples whatever G and H. First, let E be greater than G, then G is less than E; and because A is to B as C is to D (a), and of A and C, the first and third, G and Hare equimultiples; and of B and D, the second and fourth, E and F are equimultiples; and that G is less than E, therefore H is less than F (b); that is, F is greater than H; if, therefore, E be greater than G, F is greater than H; in like manner, if E be equal to G, F may be shown to be equal to H; and if less, less; but E, F, are any equimultiples whatever of B and D, and G, H, any whatever of A and C; therefore, as B is to A so is D to C (b).

SCHOLIUM. This proposition, algebraically expressed, is as follows:

THEOREM.-If A: a: B: b, then a: A :: b: B.

G

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(a) Hypoth. (b) V. Def. 5.

PROPOSITION C.

THEOREM.-If the first be the same multiple or submultiple of the second that the third is of the fourth, the first is to the second as the third is to the fourth.

Let the first A be the same multiple of the second B that the third C is of the fourth D; A is to B as C is to D.

DEMONSTRATION. Take of A and C any equimultiples whatever E and F; and of B and D any equimultiples whatever G and H; then, because A is the same multiple of B that C is of D (a); and that E is the same multiple of A that F is of C; therefore E is the same mul

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