To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.

Εὑρεῖν δύο ῥητὰς δυνάμει μόνον συμμέτρους, ὥστε τὴν μείζονα τῆς ἐλάσσονος μεῖζον δύνασθαι τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει. Ἐκκείσθω γάρ τις ῥητὴ ἡ ΑΒ καὶ δύο τετράγωνοι ἀριθμοὶ οἱ ΓΔ, ΔΕ, ὥστε τὴν ὑπεροχὴν αὐτῶν τὸν ΓΕ μὴ εἶναι τετράγωνον, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΖΒ, καὶ πεποιήσθω ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ τῆς ΒΑ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΑΖ τετράγωνον καὶ ἐπεζεύχθω ἡ ΖΒ. Ἐπεὶ [ οὖν ] ἐστιν ὡς τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, οὕτως ὁ ΔΓ πρὸς τὸν ΓΕ, τὸ ἀπὸ τῆς ΒΑ ἄρα πρὸς τὸ ἀπὸ τῆς ΑΖ λόγον ἔχει, ὃν ἀριθμὸς ὁ ΔΓ πρὸς ἀριθμὸν τὸν ΓΕ: σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΒΑ τῷ ἀπὸ τῆς ΑΖ. ῥητὸν δὲ τὸ ἀπὸ τῆς ΑΒ: ῥητὸν ἄρα καὶ τὸ ἀπὸ τῆς ΑΖ: ῥητὴ ἄρα καὶ ἡ ΑΖ. καὶ ἐπεὶ ὁ ΔΓ πρὸς τὸν ΓΕ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΒΑ ἄρα πρὸς τὸ ἀπὸ τῆς ΑΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: ἀσύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ ΑΖ μήκει: αἱ ΒΑ, ΑΖ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι. καὶ ἐπεί [ ἐστιν ] ὡς ὁ ΔΓ πρὸς τὸν ΓΕ, οὕτως τὸ ἀπὸ τῆς ΒΑ πρὸς τὸ ἀπὸ τῆς ΑΖ, ἀναστρέψαντι ἄρα ὡς ὁ ΓΔ πρὸς τὸν ΔΕ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ἀπὸ τῆς ΒΖ. ὁ δὲ ΓΔ πρὸς τὸν ΔΕ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: καὶ τὸ ἀπὸ τῆς ΑΒ ἄρα πρὸς τὸ ἀπὸ τῆς ΒΖ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν: σύμμετρος ἄρα ἐστὶν ἡ ΑΒ τῇ ΒΖ μήκει. καί ἐστι τὸ ἀπὸ τῆς ΑΒ ἴσον τοῖς ἀπὸ τῶν ΑΖ, ΖΒ: ἡ ΑΒ ἄρα τῆς ΑΖ μεῖζον δύναται τῇ ΒΖ συμμέτρῳ ἑαυτῇ. Εὕρηνται ἄρα δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ ΒΑ, ΑΖ, ὥστε τὴν μείζονα τὴν ΑΒ τῆς ἐλάσσονος τῆς ΑΖ μεῖζον δύνασθαι τῷ ἀπὸ τῆς ΒΖ συμμέτρου ἑαυτῇ μήκει: ὅπερ ἔδει δεῖξαι. | To find two rational straight lines commensurable in square only and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater. For let there be set out any rational straight line AB, and two square numbers CD, DE such that their difference CE is not square; [Lemma 1] let there be described on AB the semicircle AFB, and let it be contrived that, as DC is to CE, so is the square on BA to the square on AF. [X. 6, Por.] Let FB be joined. Since, as the square on BA is to the square on AF, so is DC to CE, therefore the square on BA has to the square on AF the ratio which the number DC has to the number CE; therefore the square on BA is commensurable with the square on AF. [X. 6] But the square on AB is rational; [X. Def. 4] therefore the square on AF is also rational; [id.] therefore AF is also rational. And, since DC has not to CE the ratio which a square number has to a square number, neither has the square on BA to the square on AF the ratio which a square number has to a square number; therefore AB is incommensurable in length with AF. [X. 9] Therefore BA, AF are rational straight lines commensurable in square only. And since, as DC is to CE, so is the square on BA to the square on AF, therefore, convertendo, as CD is to DE, so is the square on AB to the square on BF. [V. 19, Por., III. 31, I. 47] But CD has to DE the ratio which a square number has to a square number: therefore also the square on AB has to the square on BF the ratio which a square number has to a square number; therefore AB is commensurable in length with BF. [X. 9] And the square on AB is equal to the squares on AF, FB; therefore the square on AB is greater than the square on AF by the square on BF commensurable with AB. |