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For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfils the given conditions; [ Ἐὰν ἀπὸ εὐθείας εὐθεῖα ἀφαιρεθῇ δυνάμει ἀσύμμετρος οὖσα τῇ ὅλῃ, μετὰ δὲ τῆς ὅλης ποιοῦσα τὸ μὲν συγκείμενον ἐκ τῶν ἀπ' αὐτῶν τετραγώνων μέσον, τὸ δὲ δὶς ὑπ' αὐτῶν ῥητόν, ἡ λοιπὴ ἄλογός ἐστιν: καλείσθω δὲ ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα. Ἀπὸ γὰρ εὐθείας τῆς ΑΒ εὐθεῖα ἀφῃρήσθω ἡ ΒΓ δυνάμει ἀσύμμετρος οὖσα τῇ ΑΒ ποιοῦσα τὰ προκείμενα: λέγω, ὅτι ἡ λοιπὴ ἡ ΑΓ ἄλογός ἐστιν ἡ προειρημένη. Ἐπεὶ γὰρ τὸ μὲν συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τετραγώνων μέσον ἐστίν, τὸ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ῥητόν, ἀσύμμετρα ἄρα ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: καὶ λοιπὸν ἄρα τὸ ἀπὸ τῆς ΑΓ ἀσύμμετρόν ἐστι τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. καί ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ῥητόν: τὸ ἄρα ἀπὸ τῆς ΑΓ ἄλογόν ἐστιν: ἄλογος ἄρα ἐστὶν ἡ ΑΓ: καλείσθω δὲ ἡ μετὰ ῥητοῦ μέσον τὸ ὅλον ποιοῦσα. ὅπερ ἔδει δεῖξαι. For from the straight line AB let there be subtracted the straight line BC which is incommensurable in square with AB and fulfils the given conditions; [X. 34] I say that the remainder AC is the irrational straight line aforesaid. For, since the sum of the squares on AB, BC is medial, while twice the rectangle AB, BC is rational, therefore the squares on AB, BC are incommensurable with twice the rectangle AB, BC; therefore the remainder also, the square on AC, is incommensurable with twice the rectangle AB, BC. [II. 7, X. 16] And twice the rectangle AB, BC is rational; therefore the square on AC is irrational; therefore AC is irrational.