G.186 unequal magnitudes greater — Heath

V.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 / A B /, C be unequal magnitudes, and let / A B / be greater ; let D be another, chance, magnitude ; I say that / A B / has to D a greater ratio than C has to D, and D has to C a greater ratio than it has to / A B /.

A E B C F G H K D L M N

For, since / A B / is greater than C, let / E B / be made equal to C ; then the less of the magnitudes / A E /, / E B /, if multiplied, will sometime be greater than D. [G.184]

[Case 1.]

First, let / A E / be less than / E B / ; let / A E / be multiplied, and let / F G / be a multiple of it which is greater than D ; then, whatever multiple / F G / is of / A E /, let / G H / be made the same multiple of / E B / 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 greater than K.

Then, since K is less than N first,

therefore K is not less than M.

And, since / F G / is the same multiple of / A E / that / G H / is of / E B /,

[G.171]therefore / F G / is the same multiple of / A E / that / F H / is of / A B /.

But / F G / is the same multiple of / A E / that K is of C ;

therefore / F H / is the same multiple of / A B / that K is of C ; therefore / F H /, K are equimultiples of / A B /, C.

Again, since / G H / is the same multiple of / E B / that K is of C, and / E B / is equal to C,

therefore / G H / is equal to K.

But K is not less than M ;

therefore neither is / G H / less than M.

And / F G / is greater than D ;

therefore the whole / F H / 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 / F H / is greater than M, D ;

therefore / F H / is in excess of N,

while K is not in excess of N.

And / F H /, K are equimultiples of / A B /, C, while N is another, chance, multiple of D ;

[G.185]therefore / A B / has to D a greater ratio than C has to D.

I say next, that D also has to C a greater ratio than D has to / A B /.

For, with the same construction, we can prove similarly that N is in excess of K, while N is not in excess of / F H /.

And N is a multiple of D, while / F H /, K are other, chance, equimultiples of / A B /, C ;

[G.185]therefore D has to C a greater ratio than D has to / A B /.

[Case 2.]

Again, let / A E / be greater than / E B /.

Then the less, / E B /, if multiplied, will sometime be greater than D. [G.184]

Let it be multiplied, and let / G H / be a multiple of / E B / and greater than D ; and, whatever multiple / G H / is of / E B /, let / F G / be made the same multiple of / A E /, and K of C.

A E B C F G H K D L M N

Then we can prove similarly that / F H /, K are equimultiples of / A B /, C ; and, similarly, let N be taken a multiple of D but the first that is greater than / F G /, so that / F G / is again not less than M.

But / G H / is greater than D ;

therefore the whole / F H / is in excess of D, M, that is, of N.

Now K is not in excess of N, inasmuch as / F G / also, which is greater than / G H /, 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.

Dependency Graph

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