Classification of incommensurables: Book 10 Proposition 16
Translations
Ἐὰν δύο μεγέθη ἀσύμμετρα συντεθῇ, καὶ τὸ ὅλον ἑκατέρῳ αὐτῶν ἀσύμμετρον ἔσται: κἂν τὸ ὅλον ἑνὶ αὐτῶν ἀσύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη ἀσύμμετρα ἔσται. Συγκείσθω γὰρ δύο μεγέθη ἀσύμμετρα τὰ ΑΒ, ΒΓ: λέγω, ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἀσύμμετρόν ἐστιν. Εἰ γὰρ μή ἐστιν ἀσύμμετρα τὰ ΓΑ, ΑΒ, μετρήσει τι [αὐτὰ] μέγεθος. μετρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει. μετρεῖ δὲ καὶ τὸ ΑΒ: τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρεῖ. σύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΒΓ: ὑπέκειντο δὲ καὶ ἀσύμμετρα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΓΑ, ΑΒ μετρήσει τι μέγεθος: ἀσύμμετρα ἄρα ἐστὶ τὰ ΓΑ, ΑΒ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τὰ ΑΓ, ΓΒ ἀσύμμετρά ἐστιν. τὸ ΑΓ ἄρα ἑκατέρῳ τῶν ΑΒ, ΒΓ ἀσύμμετρόν ἐστιν. Ἀλλὰ δὴ τὸ ΑΓ ἑνὶ τῶν ΑΒ, ΒΓ ἀσύμμετρον ἔστω. ἔστω δὴ πρότερον τῷ ΑΒ: λέγω, ὅτι καὶ τὰ ΑΒ, ΒΓ ἀσύμμετρά ἐστιν. εἰ γὰρ ἔσται σύμμετρα, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸ ΑΓ μετρήσει. μετρεῖ δὲ καὶ τὸ ΑΒ: τὸ Δ ἄρα τὰ ΓΑ, ΑΒ μετρεῖ. σύμμετρα ἄρα ἐστὶ τὰ ΓΑ, ΑΒ: ὑπέκειτο δὲ καὶ ἀσύμμετρα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΑΒ, ΒΓ μετρήσει τι μέγεθος: ἀσύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΒΓ. Ἐὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς.Λῆμμα. Ἐὰν παρά τινα εὐθεῖαν παραβληθῇ παραλληλόγραμμον ἐλλεῖπον εἴδει τετραγώνῳ, τὸ παραβληθὲν ἴσον ἐστὶ τῷ ὑπὸ τῶν ἐκ τῆς παραβολῆς γενομένων τμημάτων τῆς εὐθείας. Παρὰ γὰρ εὐθεῖαν τὴν ΑΒ παραβεβλήσθω παραλληλόγραμμον τὸ ΑΔ ἐλλεῖπον εἴδει τετραγώνῳ τῷ ΔΒ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΔ τῷ ὑπὸ τῶν ΑΓ, ΓΒ. Καί ἐστιν αὐτόθεν φανερόν: ἐπεὶ γὰρ τετράγωνόν ἐστι τὸ ΔΒ, ἴση ἐστὶν ἡ ΔΓ τῇ ΓΒ, καί ἐστι τὸ ΑΔ τὸ ὑπὸ τῶν ΑΓ, ΓΔ, τουτέστι τὸ ὑπὸ τῶν ΑΓ, ΓΒ. Ἐὰν ἄρα παρά τινα εὐθεῖαν, καὶ τὰ ἑξῆς.
If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable. For let the two incommensurable magnitudes AB, BC be added together; I say that the whole AC is also incommensurable with each of the magnitudes AB, BC. For, if CA, AB are not incommensurable, some magnitude will measure them. Let it measure them, if possible, and let it be D. Since then D measures CA, AB, therefore it will also measure the remainder BC. But it measures AB also; therefore D measures AB, BC. Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure CA, AB; therefore CA, AB are incommensurable. [X. Def. 1] Similarly we can prove that AC, CB are also incommensurable. Therefore AC is incommensurable with each of the magnitudes AB, BC. Next, let AC be incommensurable with one of the magnitudes AB, BC. First, let it be incommensurable with AB; I say that AB, BC are also incommensurable. For, if they are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC. therefore it will also measure the whole AC. But it measures AB also; therefore D measures CA, AB. Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [X. Def. 1] Therefore etc.Lemma. If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application. For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB; I say that AD is equal to the rectangle contained by AC, CB. This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle AC, CD, that is, the rectangle AC, CB.