§7.prop.27ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ πολλαπλασιάσας ἑκάτερος ἑαυτὸν ποιῇ τινα, οἱ γενόμενοι ἐξ αὐτῶν πρῶτοι πρὸς ἀλλήλους ἔσονται, κἂν οἱ ἐξ ἀρχῆς τοὺς γενομένους πολλαπλασιάσαντες ποιῶσί τινας, κἀκεῖνοι πρῶτοι πρὸς ἀλλήλους ἔσονται.
If two numbers be prime to one another, and each by multiplying itself make some number, the numbers produced from them will be prime to one another; and even if the original numbers by multiplying the products make some numbers, those also will be prime to one another.
ἔστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, καὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Γ ποιείτω, τὸν δὲ Γ πολλαπλασιάσας τὸν Δ ποιείτω, ὁ δὲ Β ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε ποιείτω, τὸν δὲ Ε πολλαπλασιάσας τὸν Ζ ποιείτω· λέγω, ὅτι οἵ τε Γ, Ε καὶ οἱ Δ, Ζ πρῶτοι πρὸς ἀλλήλους εἰσίν.
Let two numbers A, B be prime to one another, and let A by multiplying itself make Γ, and by multiplying Γ make Δ, and let B by multiplying itself make E, and by multiplying E make Z; I say that both Γ, E and Δ, Z are prime to one another.
ἐπεὶ γὰρ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν, οἱ Γ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν.
For since A, B are prime to one another, and A by multiplying itself has made Γ, therefore Γ, B are prime to one another.
ἐπεὶ οὖν οἱ Γ, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, οἱ Γ, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν.
Since then Γ, B are prime to one another, and B by multiplying itself has made E, therefore Γ, E are prime to one another.
πάλιν, ἐπεὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, οἱ Α, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν.
Again, since A, B are prime to one another, and B by multiplying itself has made E, therefore A, E are prime to one another.
ἐπεὶ οὖν δύο ἀριθμοὶ οἱ Α, Γ πρὸς δύο ἀριθμοὺς τοὺς Β, Ε ἀμφότεροι πρὸς ἑκάτερον πρῶτοί εἰσιν, καὶ ὁ ἐκ τῶν Α, Γ ἄρα γενόμενος πρὸς τὸν ἐκ τῶν Β, Ε πρῶτός ἐστιν.
Since then two numbers A, Γ are prime to two numbers B, E, both to each, therefore the product of A, Γ is also prime to the product of B, E.
καί ἐστιν ὁ μὲν ἐκ τῶν Α, Γ ὁ Δ, ὁ δὲ ἐκ τῶν Β, Ε ὁ Ζ. οἱ Δ, Ζ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι.
And the product of A, Γ is Δ, and the product of B, E is Z. Therefore Δ, Z are prime to one another; which was to be proved.
§7.prop.28ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ συναμφότερος πρὸς ἑκάτερον αὐτῶν πρῶτος ἔσται· καὶ ἐὰν συναμφότερος πρὸς ἕνα τινὰ αὐτῶν πρῶτος ᾖ, καὶ οἱ ἐξ ἀρχῆς ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ἔσονται.
If two numbers be prime to one another, their sum also will be prime to each of them; and if their sum be prime to any one of them, the original numbers also will be prime to one another.
Συγκείσθωσαν γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ ΑΒ, ΒΓ· λέγω, ὅτι καὶ συναμφότερος ὁ ΑΓ πρὸς ἑκάτερον τῶν ΑΒ, ΒΓ πρῶτός ἐστιν.
For let two numbers AB, BΓ prime to one another be added; I say that their sum AΓ is also prime to each of AB, BΓ.
εἰ γὰρ μή εἰσιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς ΓΑ, ΑΒ ἀριθμός.
For if ΓA, AB be not prime to one another, some number will measure ΓA, AB.
μετρείτω, καὶ ἔστω ὁ Δ. ἐπεὶ οὖν ὁ Δ τοὺς ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸν ΒΓ μετρήσει.
Let it measure them, and let it be Δ. Since then Δ measures ΓA, AB, therefore it will also measure the remaining part BΓ.
μετρεῖ δὲ καὶ τὸν ΒΑ· ὁ Δ ἄρα τοὺς ΑΒ, ΒΓ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον.
But it also measures BA; therefore Δ measures AB, BΓ, which are prime to one another; which is impossible.
οὐκ ἄρα τοὺς ΓΑ, ΑΒ ἀριθμοὺς ἀριθμός τις μετρήσει· οἱ ΓΑ, ΑΒ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν.
Therefore no number will measure the numbers ΓA, AB; therefore ΓA, AB are prime to one another.
διὰ τὰ αὐτὰ δὴ καὶ οἱ ΑΓ, ΓΒ πρῶτοι πρὸς ἀλλήλους εἰσίν.
For the same reason indeed AΓ, ΓB are prime to one another.
ὁ ΓΑ ἄρα πρὸς ἑκάτερον τῶν ΑΒ, ΒΓ πρῶτός ἐστιν.
Therefore ΓA is prime to each of AB, BΓ.
ἔστωσαν δὴ πάλιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους· λέγω, ὅτι καὶ οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους εἰσίν.
Let then again ΓA, AB be prime to one another; I say that AB, BΓ are also prime to one another.
εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς ΑΒ, ΒΓ ἀριθμός.
For if AB, BΓ be not prime to one another, some number will measure AB, BΓ.
μετρείτω, καὶ ἔστω ὁ Δ. καὶ ἐπεὶ ὁ Δ ἑκάτερον τῶν ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸν ΓΑ μετρήσει.
Let it measure them, and let it be Δ. And since Δ measures each of AB, BΓ, therefore it will also measure the whole ΓA.
μετρεῖ δὲ καὶ τὸν ΑΒ· ὁ Δ ἄρα τοὺς ΓΑ, ΑΒ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον.
But it also measures AB; therefore Δ measures ΓA, AB, which are prime to one another; which is impossible.
οὐκ ἄρα τοὺς ΑΒ, ΒΓ ἀριθμοὺς ἀριθμός τις μετρήσει.
Therefore no number will measure the numbers AB, BΓ.
οἱ ΑΒ, ΒΓ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι.
Therefore AB, BΓ are prime to one another; which was to be proved.