PROPOSITION 8.
Of unequal magnitudes,
the greater has to the same a greater ratio than the less has;
and the same has to the less a greater ratio than it has to the greater.
Let
AB,
C be unequal magnitudes, and let
AB be greater; let
D be another, chance, magnitude; I say that
AB has to
D a greater ratio than
C has to
D, and
D has to
C a greater ratio than it has to
AB.
For, since
AB is greater than
C, let
BE be made equal to
C; then the less of the magnitudes
AE,
EB, if multiplied, will sometime be greater than
D. [
V. Def. 4]
[
Case I.]
First, let
AE be less than
EB; let
AE be multiplied, and let
FG be a multiple of it which is greater than
D; then, whatever multiple
FG is of
AE, let
GH be made the same multiple of
EB and
K of
C; and let
L be taken double of
D,
M triple of it, and successive multiples increasing by one, until what is taken is a multiple of
D and the first that is greater than
K. Let it be taken, and let it be
N which is quadruple of
D and the first multiple of it that is greather than
K.
Then, since
K is less than
N first, therefore
K is not less than
M.
And, since
FG is the same multiple of
AE that
GH is of
EB, therefore
FG is the same multiple of
AE that
FH is of
AB. [
V. 1]
But
FG is the same multiple of
AE that
K is of
C;
therefore FH is the same multiple of AB that K is of C; therefore FH, K are equimultiples of AB, C.
Again, since
GH is the same multiple of
EB that
K is of
C, and
EB is equal to
C,
therefore GH is equal to K.
But
K is not less than
M;
therefore neither is GH less than M.
And
FG is greater than
D; therefore the whole
FH is greater than
D,
M together.
But
D,
M together are equal to
N, inasmuch as
M is triple of
D, and
M,
D together are quadruple of
D, while
N is also quadruple of
D; whence
M,
D together are equal to
N.
But
FH is greater than
M,
D;
therefore FH is in excess of N, while
K is not in excess of
N.
And
FH,
K are equimultiples of
AB,
C, while
N is another, chance, multiple of
D;
therefore AB has to D a greater ratio than C has to D. [V. Def. 7]
I say next, that
D also has to
C a greater ratio than
D has to
AB.
For, with the same construction, we can prove similarly that
N is in excess of
K, while
N is not in excess of
FH.
And
N is a multiple of
D, while
FH,
K are other, chance, equimultiples of
AB,
C;
therefore D has to C a greater ratio than D has to AB. [V. Def. 7]
[
Case 2.]
Again, let
AE be greater than
EB.
Then the less,
EB, if multiplied, will sometime be greater than
D. [
V. Def. 4]
Let it be multiplied, and let
GH be a multiple of
EB and greater than
D; and, whatever multiple
GH is of
EB, let
FG be made the same multiple of
AE, and
K
of
C.
Then we can prove similarly that
FH,
K are equimultiples of
AB,
C; and, similarly, let
N be taken a multiple of
D but the first that is greater than
FG, so that
FG is again not less than
M.
But
GH is greater than
D; therefore the whole
FH is in excess of
D,
M, that is, of
N.
Now
K is not in excess of
N, inasmuch as
FG also, which is greater than
GH, that is, than
K, is not in excess of
N.
And in the same manner, by following the above argument, we complete the demonstration.
Therefore etc. Q. E. D.
