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D. 1 Hyp.
28, V.
3 13, V.

4 10, V.

5 Conc.

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But the M to which the same has the gr. ratio is the less,

.. DB, i. e., B > D.

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D. 1 | Hyp. & 9,V. | . A : B = C, i. e. A: D, .. B = D.

CASE III. If A < C, B < D.

D.1 Hyp.

2 Case 1. 3. Rec.

:: C > A, and C : D = A : B;

D > B, i. e. B < D.

Therefore, if the first has the same ratio,&c. Q.E.D.

Briefly. 1°. Let A > C; then, by 8, V., A: B> C: B; but A: B= C: D, .. by 13, V., C: D > C: B, and .. by 10, V., B > D.

Similarly, 2°. if A = C, by 9, V., B = D, and 3°. if A < C, B < D.

COR. Hence also, if A: B = or < the 4th D, then the 1st A >

C: D, and if the 2nd B >= or the 3rd C.

=

PROP. XV.-THEOR.

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

CON. Pst. 2, V. Any given m of a M may be divided into parts, each of which is equal to that M.

DEM. 7, V. Equal Ms, with the same ratio. 12, V. Antecedents to consequents.

E. 1 Hyp. 2 Conc.

Let A B be the same m of C, that D E is of F; then C: FAB: DE.

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D. 1 C.

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Divide A B into Ms each

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2 C. 1 & 2.

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i. e., AG GH = HB

DE
i. e., DK

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each

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C,

=

F,

: C.

: F.

Now AG, GH, HB in number = DK, KL,

& LE;

and A G GH= HB, and D K=KL=LE, .. AG: DK =GH: KL

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.. C: FAB: DE.

Wherefore, Magnitudes have the same ratio, &c.

Q. E. D.

Otherwise, Take many number, A & B two magnitudes;

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D. 17, V. & 12, V. A: BA: B,.'.A:B=A+A: B+B, or 2A=2B;

2 D. 1,

3 12, V.

4 Sim.

5 Conc.

and . A: B = 2A : 2 B,

.. A: B = 2 A + A: 2B + B, or 3 A = 3 B.
And so on for all equims. of A and B;

.. A: B

mA: m B.

Magnitudes have the same ratio to one another which

their equal submultiples or like parts have.

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COR. 2. In Proportionals the equims. of the 1st and 2nd have the same ratio as the equims. of the 3rd and 4th. If A: BC: D. then m Am B = n C: n D ;

Arith. If 2:3 =

4:6, then 2 x 2 : 2 × 3 = 3 × 4:3 × 6; i. e., 4: 6 = 12 : 18.

Con. 3. In Proportionals also, any like parts of the 1st and 2nd, and also of the 3rd and 4th are proportional;

as

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Sch. The Proportion is sometimes explained, "one magnitude shall have the same ratio to another magnitude of the same kind which any multiple of the first has to the same multiple of the second."-HOSE.

PROP. XVI.-THEOR.

Alternando, or Permutando. If four Magnitudes of the same kind be proportionals, they shall also be proportionals when taken alternately.

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C. Pst. 1, V.

D.1 C.

2 15, V.

Take E, F, equims. of A, B, and G, H, equims.

of C, D.

E same m of A, that F is of B,

.. A : B = E: F;

3 H. & 11, V. but

4 C.

5 15, V.

7 14, V.

= .. C: D E: F.

A: BC: D,
Again G & H are equims. of C & D,
C:D=G: H.

6 D.3,11,V but C: D = E: F, .. E: F = G: H.

In 4 proportionals, if 1st, > = or < 3rd,
the 2nd is > = or < 4th;

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if E > = or < G, F > = or < H; Now E, F, are equims. of A, B ; & G, H, of C,D; .. A: C B: D.

If then four magnitudes of the same kind, &c.

Q. E. D.

Take m A, m B, n C, n D, equims. of A, B, C, D.
. m A: mBA: B, .. m A: mB: = C: D.
also, . n C:n DC: D, .. mA : m B = nC : nD;
now m B> or < n D, as m A> or < nC;
A:CB: D.

Alg. & Arith. Hyp. Let

Alg.

=

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(by cd), and

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USE & APP I. From principles established, especially from Def. 7, V,and from Props. 7, 11, 14, & 15, bk. V, the following Theorem, Chambers' Ex. p. 56, may be demonstrated;

If to the terms of a ratio, A: B, the same magnitude, C, be added, the ratio will be unchanged, increased, or diminished, accordiny as it is a ratio of equality, of less inequality, or of greater inequality.

CASE I. If A = B, then A+ C:B+C=A: B.

D. 1 Ax. 2, I.

27, V. & 16, V.

3 Sim. & 11, V.

A=B,... A+C=B+C;

.. A+ C:C=B+C: C, and A+ C: B÷C= C: C. So A: B = C: C. and .. A+ C:B + C = A : B.

CASE II. If A < B, then A + C:B + C>A: B.

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A< B, if B-AD; then B = A + D.

Let p C > A, and n D the least m of D exceeding p A, so that p AnD;

hence p A+An D, or = n D ;

&..pA+p C > n D, or p (A + C) > n D.
Next, let m=n+p, or p = m—n;

then . p A < n Dˆ; p = m−n, and D :

B-A:

.. (m-n) A < n (B-A); or m A-n A<a B−n A:

8 Add.

9 D. 4. 10 D. 6.

11 Add.

12 Ax. 4, I.

13 D. 8. Def. 7, V.] CASE III.

D. 1. H.
2 Sup.

3 D. 2.

4 D. 2 & 3,

5 Sup.

To both sides udd n A, ... mA < n B.
Again. ':'p (A + C) < n D,

..PA+pC > n (B-A); or m A-n A+ mC―n C
>n B-n A.

Add n A+ n C to each side of the equation;

.. m A + m C > n B+ n C; or m (A + C) >n (B+ C);

but m A < n B, .•. A + C:B + C>A: B.
If A > B, then A + C: B + C < A : B.

. A > B, if A-B = D, then AB+ D.
Let p C > B; and m D the least m of D,
so that p Bm D ;

hence Р B+B > or = m D;

&..pB+p C > m D, or p (B + C) > m D.
Next, let n = m + p; or p = n―m ;

6 D. 2, 5, & 1. then . p B <m D, p = n—m, and D = A—B,

8 Add.

9 D. 4.

10 D. 6.

11 Add.

12 Ax. 4, I.

.. (n-m) Bm(A-B); or n B-m B <mA—m B ; Add m B, n B <mA; mA > n B.

Again . p (B+ C) > m D,

..pB+p C > m (A-B); or n B−m B+ n C — m € >m A m B ;

Add m Bm C to each side of the equation;

n (B+C) > m (A+C); orm (A +C) <n (B+ C). 13 D.8, Def. 7, V. But m A > n B, .. A +C:B+C< A : B. 14 Rec.

Wherefore, If to the terms of a ratio &c.

II. Props. 11, 15 & 16, bk. V., also furnish the principles by which to establish the useful Theorem,-Chambers' Ex. p. 54;-that,

"If all the terms, or any two homologous terms, or the terms of either of the ratios of a proportion, be multiplied or divided by the same number, the resulting magnitudes will remain proportional.

E. 1| Hyp.

CASE I.

D. 115, V.

Let A B C : D, and m, n be any two numbers; then,

mAm Bm C : mD;

A : Bm Am B; and C: D= mC : mD; .. mAm Bm C : m D.

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211, V.

1

CASE II.

A:

B = C:

ID.

m

D. 1 H.

2 15, V.

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