Thesauros Literature mathematics Στοιχεῖα

Στοιχεῖα

Στοιχεῖα Euclid

Book 5

7 Then, since D is the same multiple of A that E is of B, while A is equal to B, therefore D is equal to E.
7 But F is another, chance, magnitude.
7 If therefore D is in excess of F, E is also in excess of F, if equal to it, equal; and, if less, less.
7 And D, E are equimultiples of A, B, while F is another, chance, multiple of C; therefore, as A is to C, so is B to C. [V. Def. 5]
7 I say next that C also has the same ratio to each of the magnitudes A, B.
7 For, with the same construction, we can prove similarly that D is equal to E; and F is some other magnitude.
7 If therefore F is in excess of D, it is also in excess of E, if equal, equal; and, if less, less.
7 And F is a multiple of C, while D, E are other, chance, equimultiples of A, B; therefore, as C is to A, so is C to B. [V. Def. 5]
7 Therefore etc.
7 Porism. From this it is manifest that, if any magnitudes are proportional, they will also be proportional inversely. Q. E. D.

PROPOSITION 8.

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.
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.
8 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]
8 [Case I.]
8 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.
8 Then, since K is less than N first, therefore K is not less than M.

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