PROPOSITION 8.¶
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 1.]
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 greater than K.
Then, since K is less than N first,
And, since FG is the same multiple of AE that GH is of EB,
But FG is the same multiple of AE that K is of C ;
Again, since GH is the same multiple of EB that K is of C, and EB is equal to C,
But K is not less than M ;
And FG is greater than D ;
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 ;
while K is not in excess of N.
And FH, K are equimultiples of AB, C, while N is another, chance, multiple of D ;
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 ;
[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 ;
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.
Dependency Graph¶
Required for¶
V.9, V.10, V.14, V.16, V.18, V.19, V.20, V.21, V.22, V.23, V.24, V.25, VI.2, VI.3, VI.4, VI.5, VI.6, VI.7, VI.8, VI.9, VI.10, VI.11, VI.12, VI.13, VI.14, VI.15, VI.16, VI.17, VI.18, VI.19, VI.20, VI.22, VI.23, VI.24, VI.25, VI.26, VI.27, VI.28, VI.29, VI.30, VI.31, VI.32, VII.19, VII.24, VII.25, VII.26, VII.27, VII.30, VII.33, VII.34, VII.36, VII.39, VIII.2, VIII.3, VIII.4, VIII.5, VIII.6, VIII.7, VIII.8, VIII.9, VIII.14, VIII.15, VIII.16, VIII.17, VIII.20, VIII.21, VIII.22, VIII.23, VIII.24, VIII.25, VIII.26, VIII.27, IX.1, IX.2, IX.3, IX.4, IX.5, IX.6, IX.8, IX.9, IX.10, IX.12, IX.13, IX.14, IX.15, IX.18, IX.19, IX.32, IX.36, X.5, X.6, X.7, X.8, X.9, X.10, X.11, X.12, X.13, X.14, X.17, X.18, X.19, X.20, X.21, X.22, X.23, X.24, X.25, X.26, X.27, X.28, X.29, X.30, X.31, X.32, X.33, X.34, X.35, X.36, X.37, X.38, X.39, X.40, X.41, X.42, X.43, X.44, X.45, X.46, X.47, X.48, X.49, X.50, X.51, X.52, X.53, X.54, X.55, X.56, X.57, X.58, X.59, X.60, X.61, X.62, X.63, X.64, X.65, X.66, X.67, X.68, X.69, X.70, X.71, X.72, X.73, X.75, X.76, X.77, X.78, X.79, X.80, X.81, X.82, X.83, X.84, X.85, X.86, X.87, X.88, X.89, X.90, X.91, X.92, X.93, X.94, X.95, X.96, X.97, X.98, X.99, X.100, X.101, X.102, X.103, X.104, X.105, X.106, X.107, X.108, X.109, X.110, X.111, X.112, X.113, X.114, X.115, XI.17, XI.23, XI.27, XI.31, XI.32, XI.33, XI.34, XI.36, XI.37, XI.39, XII.1, XII.2, XII.3, XII.4, XII.5, XII.6, XII.7, XII.8, XII.9, XII.10, XII.11, XII.12, XII.13, XII.14, XII.15, XII.17, XII.18, XIII.1, XIII.4, XIII.5, XIII.6, XIII.8, XIII.9, XIII.10, XIII.11, XIII.13, XIII.16, XIII.17, XIII.18
See Commentary on Proposition V.8.