Given mA and mB any equimultiples of the two magnitudes A and B, of which A is greater than B; to prove that mAmB is the same multiple of A – B that mA is of A; that is, mA — mB m(A — B).

=

(Prep.) Let D be the excess of A above = D, and adding B to both A

D' + B.

then A B (Dem.) Therefore

take mB from both, and mA·

mD + mB;
but DAB, therefore mA ·

mB

mB =

m(A — B).

If from a multiple of a magnitude by any number a multiple of the same magnitude by a less number be taken away, the remainder will be the same multiple of that magnitude that the difference of the numbers is of unity.

Given mA and nA multiples of the magnitude A by the numbers m and n, and let m be greater than n; to prove that - nA contains A as oft as mn contains unity, — nλ = (m — n)A.

mA

(Dem.) Let m — n = q ;

(V. 2) mA = nA + qA;

qA. Therefore mA units in q; that is, in m

COR.-When the difference of the two numbers is equal to unity, or m -n= = 1, then mA

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

Given A: B:: C: D; to prove that B:A:: D: C.

(Prep.) Let mA and mC be any equimultiples of A and C'; nB and nD any equimultiples of B and D. (Dem.) Then, because A: B::C: D, if mA be less than nB, mC will be less than nD (V. Def. 10); that is, if nB be greater than mA, nD will be greater than mC. For the same reason, if nB mA, nD = mC, and if nB mA, nD <mC. But nB, nD are any equimultiples of B and D, and mA, mC any equimultiples of A and C ; therefore (V. Def. 10) B: A:: D: C.

If the first be the same multiple of the second, or the same part of it, that the third is of the fourth; the first is to the second as the third to the fourth.

Given mA, mB equimultiples of the magnitudes A and B; to prove that mA: A:: mB: B, and that A: mA::B: mB. (Prep.) Take of mA and mB equimultiples by any number n; and of A and B equimultiples by any number p; these will be nmA (V. 3), pA, nmB, pB. (Dem.) Now, if nmA be greater than pA, nm is also greater than p; and if nm is greater than p, nmB is greater than pB; therefore, when nmA is greater than pA, nmB is greater than pB. In the same manner, if nmA = pA, nmB = pB, and if nmA Now, nmA, nmB are any equimultiples of pA, pB are any equimultiples of A and B; :: mB: B (V. Def. 10).

Again, since mA: A:: mB: B, :: B: mB.

pA, nmB ≤pB. mA and mB; and therefore mA: A

by inversion (V. a), A : mA

If the first term of an analogy be a multiple or a part of the second, the third is the same multiple or the same part of the fourth.

Given A: B:: C: D, and first let A be a multiple of B; to prove that C is the same multiple of D; that is, if A = mB

C = mD.

(Prep.) Take of A and C equimultiples by any number as 2namely, 2A and 2C; and of B and D, take equimultiples by the number 2m-namely, 2mB, 2mD (V. 3); (Dem.) then, because AmB, 2A: 2mB; and since A:B:: C:D, and since 2A = 2mB, therefore 2C: = 2mD (V. Def. 10), and C = mD; that is, C contains D m times, or as often as A

contains B.

=

For, But A

and therefore, as is

Next, let A be a part of B, C is the same part of D. since AB:: C: D, inversely (V. A), B: A :: D: C. being a part of B, B is a multiple of A, shewn above, D is the same multiple of C, the same part of D that A is of B.

and therefore C is

Equal magnitudes have the same ratio to the same magnitude; and the same has the same ratio to equal magnitudes.

Given A and B any equal magnitudes, and C any other; to prove that A:C:: B: C.

(Prep.) Let mA, mB be any equimultiples of A and B ; and nC any multiple of C.

(Dem.) Because A = B, mA = mB (V. Ax. 1);

wherefore,

if mA be greater than nC, mB is greater than лC; = nC, mB = nC; or, if mAnC, mBnC.

and if mA

But mA and

mB are any equimultiples of A and B, and nC is any multiple therefore (V. Def. 10) A: C:: B: C.

of C;

Again, if A = B, C: A :: C:B; for, as has been proved, A:C::B: C, and inversely, (V. a), C: A :: C: B.

:

PROPOSITION VIII.

PROBLEM.

Of unequal magnitudes, the greater has a greater ratio to the same than the less has; and the same magnitude has a greater ratio to the less than it has to the greater.

Given A+B a magnitude greater than A, magnitude; to prove that

A has to C;

A+ B.

and C a third A+B has to C a greater ratio than and C has a greater ratio to A than it has to

(Prep.) Let m be such a of them greater than C;

number that mA and mB are each and let nC be the least multiple

of C that exceeds mA + mB; then nC C; that is (V. 1), (n-1)C will be less than mA + mB; m(A+B) is greater than (n - 1)C.

or mA+mB, that is, (Dem.) But because nC nC C is

is greater than mA + mB, and C less than mB, greater than mA, or mA is less than nCC; that is, than (n - 1)C. Therefore the multiple of A + B by m exceeds the multiple of C by n − 1, but the multiple of A by m does not exceed the multiple of C by n therefore A+B has a greater ratio to C than A has to C (V. Def. 14).

1;

Again, because the multiple of C by n 1, exceeds the multiple of A by m, but does not exceed the multiple of A+B by C has a greater ratio to A than it has to A + B.

m,

PROPOSITION IX. THEOREM.

Magnitudes which have the same ratio to the same magnitude are equal to one another; and those to which the same magnitude has the same ratio are equal to one another.

Given A: C:: B: C; to prove that A = B.

then, because

(Prep.) For, if not, let A be greater than B; A is greater than B, two numbers, m and n, may be found as in the last proposition, such that mA shall exceed nC, while mB does not exceed nC. (Dem.) But because A: C:: B: C; if mA exceed nC, mB must also exceed nC (V. Def. 10); and it is also shewn that mB does not exceed nC, which is impossible. Therefore A is not greater than B; and in the same way it is demonstrated that B is not greater than A;

equal to B.

therefore A is

Next, let C: A :: C: B, A = B. For by inversion (V. A), A:C:: B:C; and therefore by the first case, A = B.

That magnitude which has a greater ratio than another has to the same magnitude is the greater of the two; and that magnitude to which the same has a greater ratio than it has to another magnitude is the less of the two.

Given the ratio of A to C greater than that of B to C; to prove that A is greater than B.

(Dem.) Because A: CB: C,

be found, such that mAnC, and mBnC (V. Def. 14).

Therefore, also mA7 mB,

Again, let C: B>C:A;

two numbers m and n may

and AB (V. Ax. 4).

BZA. For two numbers, m

and n may be found, such that nCmB, and nCmA. Therefore, since mB is less, and mA greater than the same magnitude, nC, mB ≤ mA,

and therefore BA.

Ratios that are equal to the same ratio are equal to one another.

Given A: B:: C:D;
A:B:: E:F.

and also C:D :: E:F; to prove that

(Prep.) Take mA, mC, mE any equimultiples of A, C, and E; and nB, nD, nF any equimultiples of B, D, and F. (Dem.) Because A:B:: C:D, if mAnB, mCnD (V. Def. 10); but if mCnD, mEnF, because C:D :: E:F;

therefore if mAnB, mEnF. In the same manner, if mAnB, mE = nF; and if mAnB, mE ≤nF. Now, mA, mE are any equimultiples whatever of A and E; and nB, nF any whatever of B and F;

(V. Def. 10).

therefore A:B:: E: F

If any number of magnitudes be proportionals, as one of the antecedents is to its consequent, so is the sum of all the antecedents to that of the consequents.

Given A: B:: C:D, and C:D :: EF; to prove that A:B:: A+ C+E:B+D+ F.

(Prep.) Take mA, mC, mE any equimultiples of A, C, and E; and nB, nD, nF any equimultiples of B, D, and F. (Dem.) Then, because A: B:: CD, if mAnB, mCnD (V. Def. 8); and when mCnD, mEnF, because C: D:: E:F.

Therefore, if mAnB,

=

mA + mC+mEnB + nD +nF. In the same manner, if mΑ: nB, mA + mC + mE = nB + nD + nF ; and if mAnB, mA + mC + mE≤nB +nD+nF. Now, mA + mC + mE = m(A + C+E) (V. 1), so that mA and mA + mC+mE are any equal multiples of A and of A+ C + E. And for the same reason nB, and nB + nD + nF, are any equimultiples of B, and of B+D + F; therefore (V. Def. 10) A: B:: A + C + E:B + D+ F.

If the first has to the second the same ratio which the third has to the fourth, but the third to the fourth a greater ratio than the fifth has to the sixth; the first shall also have to the second a greater ratio than the fifth has to the sixth.

Given A: B::C:D;

A:B7E: F.

but C:DE: F; to prove that

(Dem.) Because C: DE: F,

there are some two numbers

Now, Therefore

wherefore A:BE:F.

m and n, such that mCnD, but mEnF (V. Def. 14). if mCnD, mAnB, because A: B:: C: D. mAnB, and mEnF;

If the first term of an analogy be greater than the third, the second shall be greater than the fourth; and if equal, equal; and if less, less.

Given A: B:: C: D; to prove that if AC, BD, A = C, BD, and if AC, B ≤ D.

if

(Dem.) First, let AC;

then A:BC: B (V. 8),

but

A B C D, therefore CDC: B (V. 13),

and

therefore BD (V. 10).

and

In the same manner, it is proved that if A = C, B = D; if AC, BZD.

Magnitudes have the same ratio to one another which their equimultiples have.

Given A and B two magnitudes, and m any number; to prove that A:B:: mA: mB.

(Dem.) Because A:B:: A:B (V. 7); A:B:: A+ A: B + B (V. 12), or A: B: 2A: 2B. And since it has been proved that A: B:: 2A : 2B, (V. 12), or A: B:: 3A: 3B; multiples of A and B.

A:B:: A+ 2A :: B + 2B and so on, for all the equi