If an odd number be prime to any number, it will also be prime to the double of it.
|Ἐὰν περισσὸς ἀριθμὸς πρός τινα ἀριθμὸν πρῶτος ᾖ, καὶ πρὸς τὸν διπλασίονα αὐτοῦ πρῶτος ἔσται. Περισσὸς γὰρ ἀριθμὸς ὁ Α πρός τινα ἀριθμὸν τὸν Β πρῶτος ἔστω, τοῦ δὲ Β διπλασίων ἔστω ὁ Γ: λέγω, ὅτι ὁ Α [ καὶ ] πρὸς τὸν Γ πρῶτός ἐστιν. Εἰ γὰρ μή εἰσιν [ οἱ Α, Γ ] πρῶτοι, μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. καί ἐστιν ὁ Α περισσός: περισσὸς ἄρα καὶ ὁ Δ. καὶ ἐπεὶ ὁ Δ περισσὸς ὢν τὸν Γ μετρεῖ, καί ἐστιν ὁ Γ ἄρτιος, καὶ τὸν ἥμισυν ἄρα τοῦ Γ μετρήσει [ ὁ Δ ]. τοῦ δὲ Γ ἥμισύ ἐστιν ὁ Β: ὁ Δ ἄρα τὸν Β μετρεῖ. μετρεῖ δὲ καὶ τὸν Α. ὁ Δ ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ὁ Α πρὸς τὸν Γ πρῶτος οὔκ ἐστιν. οἱ Α, Γ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι.||If an odd number be prime to any number, it will also be prime to the double of it. For let the odd number A be prime to any number B, and let C be double of B; I say that A is prime to C. For, if they are not prime to one another, some number will measure them. Let a number measure them, and let it be D. Now A is odd; therefore D is also odd. And since D which is odd measures C, and C is even, therefore [D] will measure the half of C also. [IX. 30] But B is half of C; therefore D measures B. But it also measures A; therefore D measures A, B which are prime to one another: which is impossible. Therefore A cannot but be prime to C.|