§10.prop1.1δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους.
Two unequal magnitudes being set out, if from the greater there be subtracted more than its half, and from that which is left more than its half, and this be done continually, there will be left some magnitude which will be less than the lesser magnitude set out.
ἔστω δύο μεγέθη ἄνισα τὰ ΑΒ, Γ, ὧν μεῖζον τὸ ΑΒ· λέγω, ὅτι, ἐὰν ἀπὸ τοῦ ΑΒ ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ Γ μεγέθους.
Let ΑΒ, Γ be two unequal magnitudes, of which ΑΒ is the greater; I say that, if from ΑΒ there be subtracted more than its half, and from that which is left more than its half, and this be done continually, there will be left some magnitude which will be less than the magnitude Γ.
τὸ Γ γὰρ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ ΑΒ μεῖζον.
For Γ, being multiplied, will sometime be greater than ΑΒ.
πεπολλαπλασιάσθω, καὶ ἔστω τὸ ΔΕ τοῦ μὲν Γ πολλαπλάσιον, τοῦ δὲ ΑΒ μεῖζον, καὶ διῃρήσθω τὸ ΔΕ εἰς τὰ τῷ Γ ἴσα τὰ ΔΖ, ΖΗ, ΗΕ, καὶ ἀφῃρήσθω ἀπὸ μὲν τοῦ ΑΒ μεῖζον ἢ τὸ ἥμισυ τὸ ΒΘ, ἀπὸ δὲ τοῦ ΑΘ μεῖζον ἢ τὸ ἥμισυ τὸ ΘΚ, καὶ τοῦτο ἀεὶ γιγνέσθω, ἕως ἂν αἱ ἐν τῷ ΑΒ διαιρέσεις ἰσοπληθεῖς γένωνται ταῖς ἐν τῷ ΔΕ διαιρέσεσιν.
Let it be multiplied, and let ΔΕ be a multiple of Γ, but greater than ΑΒ, and let ΔΕ be divided into parts equal to Γ, namely ΔΖ, ΖΗ, ΗΕ, and let there be subtracted from ΑΒ more than its half, ΒΘ, and from ΑΘ more than its half, ΘΚ, and let this continually happen until the divisions in ΑΒ become equal in multitude to the divisions in ΔΕ.
ἔστωσαν οὖν αἱ ΑΚ, ΚΘ, ΘΒ διαιρέσεις ἰσοπληθεῖς οὖσαι ταῖς ΔΖ, ΖΗ, ΗΕ·
Let then the divisions ΑΚ, ΚΘ, ΘΒ be equal in multitude to ΔΖ, ΖΗ, ΗΕ.
καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΔΕ τοῦ ΑΒ, καὶ ἀφῄρηται ἀπὸ μὲν τοῦ ΔΕ ἔλασσον τοῦ ἡμίσεος τὸ ΕΗ, ἀπὸ δὲ τοῦ ΑΒ μεῖζον ἢ τὸ ἥμισυ τὸ ΒΘ, λοιπὸν ἄρα τὸ ΗΔ λοιποῦ τοῦ ΘΑ μεῖζόν ἐστιν.
And since ΔΕ is greater than ΑΒ, and from ΔΕ there has been subtracted ΕΗ, which is less than its half, and from ΑΒ there has been subtracted ΒΘ, more than its half, therefore the remainder ΗΔ is greater than the remainder ΘΑ.
καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΗΔ τοῦ ΘΑ, καὶ ἀφῄρηται τοῦ μὲν ΗΔ ἥμισυ τὸ ΗΖ, τοῦ δὲ ΘΑ μεῖζον ἢ τὸ ἥμισυ τὸ ΘΚ, λοιπὸν ἄρα τὸ ΔΖ λοιποῦ τοῦ ΑΚ μεῖζόν ἐστιν.
And since ΗΔ is greater than ΘΑ, and from ΗΔ there has been subtracted ΗΖ, the half, and from ΘΑ ΘΚ, more than its half, therefore the remainder ΔΖ is greater than the remainder ΑΚ.
ἴσον δὲ τὸ ΔΖ τῷ Γ·
But ΔΖ is equal to Γ; therefore Γ also is greater than ΑΚ.
καὶ τὸ Γ ἄρα τοῦ ΑΚ μεῖζόν ἐστιν.
Therefore ΑΚ is less than Γ.
ἔλασσον ἄρα τὸ ΑΚ τοῦ Γ.
καταλείπεται ἄρα ἀπὸ τοῦ ΑΒ μεγέθους τὸ ΑΚ μέγεθος ἔλασσον ὂν τοῦ ἐκκειμένου ἐλάσσονος μεγέθους τοῦ Γ·
Therefore there is left from the magnitude ΑΒ the magnitude ΑΚ, which is less than the lesser magnitude set out, Γ.
ὅπερ ἔδει δεῖξαι.
Which it was required to prove.
¯ὁμοίως δὲ δειχθήσεται, κἂν ἡμίση ᾖ τὰ ἀφαιρούμενα.
And it will be proved similarly even if the parts subtracted be halves.
§10.prop1.2ἐὰν δύο μεγεθῶν ἀνίσων ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος τὸ καταλειπόμενον μηδέποτε καταμετρῇ τὸ πρὸ ἑαυτοῦ, ἀσύμμετρα ἔσται τὰ μεγέθη.
If, two unequal magnitudes being set out, and the lesser being continually subtracted from the greater, that which is left never measures that which is before it, the magnitudes will be incommensurable.
δύο γὰρ μεγεθῶν ὄντων ἀνίσων τῶν ΑΒ, ΓΔ καὶ ἐλάσσονος τοῦ ΑΒ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος τὸ περιλειπόμενον μηδέποτε καταμετρείτω τὸ πρὸ ἑαυτοῦ· λέγω, ὅτι ἀσύμμετρά ἐστι τὰ ΑΒ, ΓΔ μεγέθη.
For, two unequal magnitudes being ΑΒ, ΓΔ, and ΑΒ being the lesser, and the lesser being continually subtracted from the greater, let that which is left never measure that which is before it; I say that the magnitudes ΑΒ, ΓΔ are incommensurable.
εἰ γάρ ἐστι σύμμετρα, μετρήσει τι αὐτὰ μέγεθος.
For, if they are commensurable, some magnitude will measure them.
μετρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Ε· καὶ τὸ μὲν ΑΒ τὸ ΖΔ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΓΖ, τὸ δὲ ΓΖ τὸ ΒΗ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΑΗ, καὶ τοῦτο ἀεὶ γινέσθω, ἕως οὗ λειφθῇ τι μέγεθος, ὅ ἐστιν ἔλασσον τοῦ Ε. γεγονέτω, καὶ λελείφθω τὸ ΑΗ ἔλασσον τοῦ Ε. ἐπεὶ οὖν τὸ Ε τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΔΖ μετρεῖ, καὶ τὸ Ε ἄρα τὸ ΖΔ μετρήσει.
Let it measure them, if possible, and let it be Ε; and let ΑΒ, measuring ΖΔ, leave ΓΖ less than itself, and let ΓΖ, measuring ΒΗ, leave ΑΗ less than itself, and let this continually happen until some magnitude is left which is less than Ε. Let it have happened, and let ΑΗ be left, less than Ε. Since then Ε measures ΑΒ, but ΑΒ measures ΔΖ, therefore Ε will also measure ΖΔ.
μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ· καὶ λοιπὸν ἄρα τὸ ΓΖ μετρήσει.
But it also measures the whole ΓΔ; therefore it will also measure the remainder ΓΖ.
ἀλλὰ τὸ ΓΖ τὸ ΒΗ μετρεῖ· καὶ τὸ Ε ἄρα τὸ ΒΗ μετρεῖ.
But ΓΖ measures ΒΗ; therefore Ε also measures ΒΗ.
μετρεῖ δὲ καὶ ὅλον τὸ ΑΒ· καὶ λοιπὸν ἄρα τὸ ΑΗ μετρήσει, τὸ μεῖζον τὸ ἔλασσον.
But it also measures the whole ΑΒ; therefore it will also measure the remainder ΑΗ, the greater measuring the less.
ὅπερ ἐστὶν ἀδύνατον.
Which is impossible.
οὐκ ἄρα τὰ ΑΒ, ΓΔ μεγέθη μετρήσει τι μέγεθος· ἀσύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΓΔ μεγέθη.
Therefore no magnitude will measure the magnitudes ΑΒ, ΓΔ; therefore the magnitudes ΑΒ, ΓΔ are incommensurable.
ἐὰν ἄρα δύο μεγεθῶν ἀνίσων, καὶ τὰ ἑξῆς.
Therefore, if two unequal magnitudes, and the rest.