To find the first apotome.

Εὑρεῖν τὴν πρώτην ἀποτομήν. Ἐκκείσθω ῥητὴ ἡ Α, καὶ τῇ Α μήκει σύμμετρος ἔστω ἡ ΒΗ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΒΗ. καὶ ἐκκείσθωσαν δύο τετράγωνοι ἀριθμοὶ οἱ ΔΕ, ΕΖ, ὧν ἡ ὑπεροχὴ ὁ ΖΔ μὴ ἔστω τετράγωνος: οὐδ' ἄρα ὁ ΕΔ πρὸς τὸν ΔΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ πεποιήσθω ὡς ὁ ΕΔ πρὸς τὸν ΔΖ, οὕτως τὸ ἀπὸ τῆς ΒΗ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΗΓ τετράγωνον: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΗ τῷ ἀπὸ τῆς ΗΓ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΒΗ: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΗΓ: ῥητὴ ἄρα ἐστὶ καὶ ἡ ΗΓ. καὶ ἐπεὶ ὁ ΕΔ πρὸς τὸν ΔΖ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδ' ἄρα τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ ΗΓ μήκει. καί εἰσιν ἀμφότεραι ῥηταί: αἱ ΒΗ, ΗΓ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἡ ἄρα ΒΓ ἀποτομή ἐστιν. Λέγω δή, ὅτι καὶ πρώτη. Ὧι γὰρ μεῖζόν ἐστι τὸ ἀπὸ τῆς ΒΗ τοῦ ἀπὸ τῆς ΗΓ, ἔστω τὸ ἀπὸ τῆς Θ. καὶ ἐπεί ἐστιν ὡς ὁ ΕΔ πρὸς τὸν ΖΔ, οὕτως τὸ ἀπὸ τῆς ΒΗ πρὸς τὸ ἀπὸ τῆς ΗΓ, καὶ ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΔΕ πρὸς τὸν ΕΖ, οὕτως τὸ ἀπὸ τῆς ΗΒ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΔΕ πρὸς τὸν ΕΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἑκάτερος γὰρ τετράγωνός ἐστιν: καὶ τὸ ἀπὸ τῆς ΗΒ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: σύμμετρος ἄρα ἐστὶν ἡ ΒΗ τῇ Θ μήκει. καὶ δύναται ἡ ΒΗ τῆς ΗΓ μεῖζον τῷ ἀπὸ τῆς Θ: ἡ ΒΗ ἄρα τῆς ΗΓ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. καί ἐστιν ἡ ὅλη ἡ ΒΗ σύμμετρος τῇ ἐκκειμένῃ ῥητῇ μήκει τῇ Α. ἡ ΒΓ ἄρα ἀποτομή ἐστι πρώτη. Εὕρηται ἄρα ἡ πρώτη ἀποτομὴ ἡ ΒΓ: ὅπερ ἔδει εὑρεῖν. | To find the first apotome. Let a rational straight line A be set out, and let BG be commensurable in length with A; therefore BG is also rational. Let two square numbers DE, EF be set out, and let their difference FD not be square; therefore neither has ED to DF the ratio which a square number has to a square number. Let it be contrived that, as ED is to DF, so is the square on BG to the square on GC; [X. 6, Por.] therefore the square on BG is commensurable with the square on GC. [X. 6] But the square on BG is rational; therefore the square on GC is also rational; therefore GC is also rational. And, since ED has not to DF the ratio which a square number has to a square number, therefore neither has the square on BG to the square on GC the ratio which a square number has to a square number; therefore BG is incommensurable in length with GC. [X. 9] And both are rational; therefore BG, GC are rational straight lines commensurable in square only; therefore BC is an apotome. [X. 73] I say next that it is also a first apotome. For let the square on H be that by which the square on BG is greater than the square on GC. Now since. as ED is to FD, so is the square on BG to the square on GC, therefore also, convertendo, [v. 19, Por.] as DE is to EF, so is the square on GB to the square on H. But DE has to EF the ratio which a square number has to a square number, for each is square; therefore the square on GB also has to the square on H the ratio which a square number has to a square number; therefore BG is commensurable in length with H. [X. 9] And the square on BG is greater than the square on GC by the square on a straight line commensurable in length with BG. And the whole BG is commensurable in length with the rational straight line A set out. Therefore BC is a first apotome. [X. Deff. III. 1] |