G.186 unequal magnitudes greater

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.
unequal_magnitudes_greater(/ A B /, C)
given / A B /, C be unequal magnitudes, let / A B / be greater
prove / 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
A E B C F G H K D L M N
given / A B /, C be unequal magnitudes, let / A B / be greater
assert Let D be another, chance, magnitude
construct Since / A B / is greater than C, let / E B / be made equal to C
∵ magnitudes_said_ratio_def_v_def_4assert Then the less of the magnitudes / A E /, / E B /, if multiplied, will sometime be greater than D
assert [Case 1.]
assert First, let / A E / be less than / E B /
assert Let / A E / be multiplied, and let / F G / be a multiple of it which > D
construct Then, whatever multiple / F G / is of / A E /, let / G H / be made the same multiple of / E B / and K of C
construct 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
construct Let it be taken
assert Let it be N which is quadruple of D and the first multiple of it that > K
assert K < N first → therefore K is not less than M
assert since / F G / is the same multiple of / A E / that / G H / is of / E B /,
∵ there_number_magnitudesassert / F G / is the same multiple of / A E / that / F H / is of / A B /
assert / F G / is the same multiple of / A E / that K is of C
assert / F H / is the same multiple of / A B / that K is of C ; therefore / F H /, K are equimultiples of / A B /, C
assert / G H / is the same multiple of / E B / that K is of C, and / E B / = C → therefore / G H / = K
assert K is not less than M
assert neither is / G H / less than M
assert / F G / > D
assert / F H / > D, M together
assert 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
assert Whence M, D together are equal to N
assert / F H / > M, D
assert / F H / is in excess of N,
assert While K is not in excess of N
assert / F H /, K are equimultiples of / A B /, C, while N is another, chance, multiple of D
∵ equimultiples_multiple_magnitude_defassert / A B / has to D a greater ratio than C has to D
assert I say next, that D also has to C a greater ratio than D has to / A B /
assert With the same construction, we can prove similarly that N is in excess of K, while N is not in excess of / F H /
assert N is a multiple of D, while / F H /, K are other, chance, equimultiples of / A B /, C
∵ equimultiples_multiple_magnitude_defassert D has to C a greater ratio than D has to / A B /
assert [Case 2.]
assert Let / A E / be greater than / E B /
∵ magnitudes_said_ratio_def_v_def_4assert Then the less, / E B /, if multiplied, will sometime be greater than D
assert Let it be multiplied, and let / G H / be a multiple of / E B / and greater than D
construct And, whatever multiple / G H / is of / E B /, let / F G / be made the same multiple of / A E /, and K of C.
assert Then we can prove similarly that / F H /, K are equimultiples of / A B /, C
construct 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.
assert / G H / > D
assert / F H / is in excess of D, M, that is, of N
assert Now K is not in excess of N, inasmuch as / F G / also, which > / G H /, that is, than K, is not in excess of N
∴ In the same manner, by following the above argument , we complete the demonstration

Dependency Graph

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