Classification of incommensurables: Book 10 Proposition 36
Translations
Ἐὰν δύο ῥηταὶ δυνάμει μόνον σύμμετροι συντεθῶσιν, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο ὀνομάτων. Συγκείσθωσαν γὰρ δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ: λέγω, ὅτι ὅλη ἡ ΑΓ ἄλογός ἐστιν. Ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει: δυνάμει γὰρ μόνον εἰσὶ σύμμετροι: ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ὑπὸ τῶν ΑΒΓ πρὸς τὸ ἀπὸ τῆς ΒΓ, ἀσύμμετρον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ τῷ ἀπὸ τῆς ΒΓ. ἀλλὰ τῷ μὲν ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, τῷ δὲ ἀπὸ τῆς ΒΓ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ: αἱ γὰρ ΑΒ, ΒΓ ῥηταί εἰσι δυνάμει μόνον σύμμετροι: ἀσύμμετρον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τοῖς ἀπὸ τῶν ΑΒ, ΒΓ. καὶ συνθέντι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μετὰ τῶν ἀπὸ τῶν ΑΒ, ΒΓ, τουτέστι τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ: ἄλογον ἄρα [ἐστὶ] τὸ ἀπὸ τῆς ΑΓ: ὥστε καὶ ἡ ΑΓ ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο ὀνομάτων: ὅπερ ἔδει δεῖξαι.
If two rational straight lines commensurable in square only be added together, the whole is irrational; and let it be called binomial. For let two rational straight lines AB, BC commensurable in square only be added together; I say that the whole AC is irrational. For, since AB is incommensurable in length with BC for they are commensurable in square only and, as AB is to BC, so is the rectangle AB, BC to the square on BC, therefore the rectangle AB, BC is incommensurable with the square on BC. [X. 11] But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [X. 6], and the squares on AB, BC are commensurable with the square on BCfor AB, BC are rational straight lines commensurable in square only [X. 15] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13] And, componendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on AC [II. 4], is incommensurable with the sum of the squares on AB, BC. [X. 16] But the sum of the squares on AB, BC is rational; therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4]