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Each of the numbers which are continually doubled beginning from a dyad is even-times even only.
| Τῶν ἀπὸ δυάδος διπλασιαζομένων ἀριθμῶν ἕκαστος ἀρτιάκις ἄρτιός ἐστι μόνον. Ἀπὸ γὰρ δυάδος τῆς Α δεδιπλασιάσθωσαν ὁσοιδηποτοῦν ἀριθμοὶ οἱ Β, Γ, Δ: λέγω, ὅτι οἱ Β, Γ, Δ ἀρτιάκις ἄρτιοί εἰσι μόνον. Ὅτι μὲν οὖν ἕκαστος [ τῶν Β, Γ, Δ ] ἀρτιάκις ἄρτιός ἐστιν, φανερόν: ἀπὸ γὰρ δυάδος ἐστὶ διπλασιασθείς. λέγω, ὅτι καὶ μόνον. ἐκκείσθω γὰρ μονάς. ἐπεὶ οὖν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ μετὰ τὴν μονάδα ὁ Α πρῶτός ἐστιν, ὁ μέγιστος τῶν Α, Β, Γ, Δ ὁ Δ ὑπ' οὐδενὸς ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ. καί ἐστιν ἕκαστος τῶν Α, Β, Γ ἄρτιος: ὁ Δ ἄρα ἀρτιάκις ἄρτιός ἐστι μόνον. ὁμοίως δὴ δείξομεν, ὅτι [ καὶ ] ἑκάτερος τῶν Β, Γ ἀρτιάκις ἄρτιός ἐστι μόνον: ὅπερ ἔδει δεῖξαι. | Each of the numbers which are continually doubled beginning from a dyad is even-times even only. For let as many numbers as we please, B, C, D, have been continually doubled beginning from the dyad A; I say that B, C, D are eventimes even only. Now that each of the numbers B, C, D is even-times even is manifest; for it is doubled from a dyad. I say that it is also even-times even only. For let an unit be set out. Since then as many numbers as we please beginning from an unit are in continued proportion, and the number A after the unit is prime, therefore D, the greatest of the numbers A, B, C, D, will not be measured by any other number except A, B, C. [IX. 13] And each of the numbers A, B, C is even; therefore D is even-times even only. [VII. Def. 8] |