PROPOSITION VIII.

THEOREM.

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

a

F

A

a 3. 1.

Let AB, C, be unequal magnitudes, and let AB be greater than c; also let D be some other magnitude; then AB has a greater ratio to D than c has to D; also D has a greater ratio to c than it has to AB. For, because AB is greater than c, make BEa equal to c; therefore, the less of AE, EB, being multiplied, will at length be greater than D. First, let AE be less than AB, and multiply AE until it becomes greater than D; let FG be the multiple of AE, which is greater than D; then make GH the same multiple of EB, and K of C, as FG is of AE; and KHCDL MN take L, double of D, м, triple of it, and

G B

b 4 Def. 5.

so on, greater by one, until a multiple is taken greater than D, and in the first place greater than K. Let N be this magnitude, being quadruple of D, and in the first place greater than K. Therefore, because K is less than N, K will not be less than M. And since FG is the same multiple of AE as GH is of EB, FGC will also be the same multiple of AE as FH is © 1.5. of AB. But FG is the same multiple of AE as K is of c, and, consequently, FH, K, are equal multiples of AB, C. Again, because GH is the same multiple of EB as K is of c, and EB is equal to C, GH will also be equal to K. But K is not less than м; therefore GH is not less than M. But FG is greater than D; whence By con. the whole FH will be greater than D, M, together. But D, M, together are equal to N; wherefore FH exceeds N; but к does not exceed N and FH, K, are equimultiples of AB, C, and N some other multiple of D. Therefore AB has a greater ratio to D than C has to D.

Moreover, D has a greater ratio to c, than D has to AB. For, the same construction being made, in like manner we demonstrate that N exceeds K, but does not exceed FH. And N is a multiple of D, also FH, K,

e 7 Def. 5.

some other equimultiples of AB, C; therefore D has a

7 Def. 5. greater ratio to c than D has to AB.

* 4 Def. 5.

may

F

A

E

Let it

But if AB be greater than EB, therefore EB the less be multiplied until it be greater than D. be multiplied, and let GH be that multiple of EB, greater than D. And what multiple GH is of EB; the same multiples make FG of AE, and к of c. For the same reason, when we before showed FH and к are equimultiples of AB, C. And in like manner, take N the same multiple of D, but in the first place greater than FG; therefore again FG is not less than м; but GH greater than D therefore the whole FH exceeds D and м together, that is N. But K does not exceed N, because FG, which is greater than GH, that is, than к, does not exceed N. And in like manner, as before said, we finish the demonstration. Therefore of unequal magnitudes, the greater has, &c. Q. E. D.

The same by Algebra.

KH CD LMN

Let m and n be two unequal magnitudes of which m is the greater, and let p be some other magnitude; then m has a greater ratio to p than n has; and p has a greater ratio to n than it has to m. Take a and b equimultiples of m and n, so that e, a multiple of p, may be greater than b, but less than a (which will easily happen if both a and b be taken greater than p). It is

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.

For let each of the magnitudes A, B, have the same ratio to c: then is a equal to B.

For if they were not equal, A and B would not have

some other equimultiples of AB, C; therefore D has a

7 Def. 5. greater ratio to c than D has to AB.

* 4 Def. 5.

F

the less

Let it

A

E

But if AB be greater than EB, therefore EB may be multiplied until it be greater than D. be multiplied, and let GH be that multiple of EB, greater than D. And what multiple GH is of EB; the same multiples make FG of AE, and K of c. For the same reason, when we before showed FH and к are equimultiples of AB, C. And in like manner, take N the same multiple of D, but in the first place greater than FG; therefore again FG is not less than м; but GH greater than D therefore the whole FH exceeds D and м together, that is N. But K does not exceed N, because FG, which is greater than GH, that is, than K, does not exceed N. And in like manner, as before said, we finish the demonstration. Therefore of unequal magnitudes, the greater has, &c. Q. E. D.

The same by Algebra.

KH CD LMN

Let m and n be two unequal magnitudes of which m is the greater, and let p be some other magnitude; then m has a greater ratio to p than n has; and p has a greater ratio to n than it has to m. Take a and b equimultiples of m and n, so that e, a multiple of p, may be greater than b, but less than a (which will easily happen if both a and b be taken greater than p). It is

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.

For let each of the magnitudes A, B, have the same ratio to c then is A equal to B.

For if they were not equal, A and B would not have

the same ratio to c, but they have; therefore A is equal to B.

Again, let c have the same ratio to A and B, then A is equal to c.

For if it were not equal, c would not have the same ratio to A, B, but they have: therefore A is equal to B. Therefore those magnitudes which have the same ratio to the same magnitude, &c. Q. E D.

The same by Algebra.

A

Let m and n be two magnitudes, and let p be some other magnitude; then if m: p :: n : p; m = n. For ..m=n. Again, if p: m :: p : n; m = n.

m.

=

n

For P

m

==== or n p = m p, .. n = m. Q. E. D.

PROPOSITION X.
THEOREM.

Of magnitudes having a ratio to the same magnitude, that which has the greater ratio, is the greater of the two; but to that which the same magnitude has the greater ratio, is the less of the two.

For let a have to c a greater ratio, than в has to c: then is a greater than B.

For if it be not greater, it is either equal or less. But it is not equal to в, for then each of the magnitudes A, B, would have the same proportion to c.a But they have not; therefore A is not equal to в, neither is a less than B, for then A would have a less proportion to

B

a 8.5.

b

7,5.

c, than в has to c; but it has not; therefore a is not ↳ 8. 5. less than B. And it has been shown not to be equal wherefore A shall be greater than B.

Again, let c have a greater ratio to в than c has to A then в is less than A.

:

d

с

d 8. 5.

For if it be not less, it is either equal or greater. B is not equal to A; for then c would have the same ratio to both A and B. But it has not: wherefore 7. 5. A is not equal to в; neither is B greater than a; for then c would have a less ratio to в than it has to A. But it has not therefore в is not greater than A. And it has been shown not to be equal: wherefore в shall be less than A. Therefore of magnitudes having a ratio, &c. Q. E. D.