#### 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.