§8.prop.11δύο τετραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καὶ ὁ τετράγωνος πρὸς τὸν τετράγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν.
Between two square numbers there is one mean proportional number, and the square has to the square a duplicate ratio of that which the side has to the side.
ἔστωσαν τετράγωνοι ἀριθμοὶ οἱ Α, Β, καὶ τοῦ μὲν Α πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ· λέγω, ὅτι τῶν Α, Β εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ.
ὁ Γ γὰρ τὸν Δ πολλαπλασιάσας τὸν Ε ποιείτω.
Let Α, Β be square numbers, and let the side of Α be Γ, and that of Β be Δ; I say that between Α, Β there is one mean proportional number, and Α has to Β a duplicate ratio of that which Γ has to Δ. For let Γ by multiplying Δ make Ε.
καὶ ἐπεὶ τετράγωνός ἐστιν ὁ Α, πλευρὰ δὲ αὐτοῦ ἐστιν ὁ Γ, ὁ Γ ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Α πεποίηκεν.
And since Α is square, and its side is Γ, therefore Γ by multiplying itself has made Α.
διὰ τὰ αὐτὰ δὴ καὶ ὁ Δ ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν.
For the same reasons also Δ by multiplying itself has made Β.
ἐπεὶ οὖν ὁ Γ ἑκάτερον τῶν Γ, Δ πολλαπλασιάσας ἑκάτερον τῶν Α, Ε πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Ε. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν Β. καὶ ὡς ἄρα ὁ Α πρὸς τὸν Ε, οὕτως ὁ Ε πρὸς τὸν Β. τῶν Α, Β ἄρα εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός.
Since then Γ by multiplying each of Γ, Δ has made each of Α, Ε, therefore, as Γ is to Δ, so is Α to Ε. For the same reasons also, as Γ is to Δ, so is Ε to Β. Therefore also, as Α is to Ε, so is Ε to Β. Therefore between Α, Β there is one mean proportional number.
λέγω δή, ὅτι καὶ ὁ Α πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. ἐπεὶ γὰρ τρεῖς ἀριθμοὶ ἀνάλογόν εἰσιν οἱ Α, Ε, Β, ὁ Α ἄρα πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ὁ Α πρὸς τὸν Ε. ὡς δὲ ὁ Α πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Δ. ὁ Α ἄρα πρὸς τὸν Β διπλασίονα λόγον ἔχει ἤπερ ἡ Γ πλευρὰ πρὸς τὴν Δ· ὅπερ ἔδει δεῖξαι.
I say next that Α also has to Β a duplicate ratio of that which Γ has to Δ. For since the three numbers Α, Ε, Β are in proportion, therefore Α has to Β a duplicate ratio of that which Α has to Ε. But, as Α is to Ε, so is Γ to Δ; therefore Α also has to Β a duplicate ratio of that which the side Γ has to Δ; which was to be proved.
§8.prop.12δύο κύβων ἀριθμῶν δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί, καὶ ὁ κύβος πρὸς τὸν κύβον τριπλασίονα λόγον ἔχει ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν.
Between two cube numbers there are two mean proportional numbers, and the cube has to the cube a triplicate ratio of that which the side has to the side.
ἔστωσαν κύβοι ἀριθμοὶ οἱ Α, Β καὶ τοῦ μὲν Α πλευρὰ ἔστω ὁ Γ, τοῦ δὲ Β ὁ Δ· λέγω, ὅτι τῶν Α, Β δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί, καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ.
ὁ γὰρ Γ ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε ποιείτω, τὸν δὲ Δ πολλαπλασιάσας τὸν Ζ ποιείτω, ὁ δὲ Δ ἑαυτὸν πολλαπλασιάσας τὸν Η ποιείτω, ἑκάτερος δὲ τῶν Γ, Δ τὸν Ζ πολλαπλασιάσας ἑκάτερον τῶν Θ, Κ ποιείτω.
Let Α, Β be cube numbers, and let the side of Α be Γ, and that of Β be Δ; I say that between Α, Β there are two mean proportional numbers, and Α has to Β a triplicate ratio of that which Γ has to Δ. For let Γ by multiplying itself make Ε, and by multiplying Δ make Ζ, and let Δ by multiplying itself make Η, and let each of Γ, Δ by multiplying Ζ make each of Θ, Κ.
καὶ ἐπεὶ κύβος ἐστὶν ὁ Α, πλευρὰ δὲ αὐτοῦ ὁ Γ, καὶ ὁ Γ ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, ὁ Γ ἄρα ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε πεποίηκεν, τὸν δὲ Ε πολλαπλασιάσας τὸν Α πεποίηκεν.
And since Α is a cube, and its side is Γ, and Γ by multiplying itself has made Ε, therefore Γ by multiplying itself has made Ε, and by multiplying Ε has made Α.
διὰ τὰ αὐτὰ δὴ καὶ ὁ Δ ἑαυτὸν μὲν πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ Η πολλαπλασιάσας τὸν Β πεποίηκεν.
For the same reasons also Δ by multiplying itself has made Η, and by multiplying Η has made Β.
καὶ ἐπεὶ ὁ Γ ἑκάτερον τῶν Γ, Δ πολλαπλασιάσας ἑκάτερον τῶν Ε, Ζ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς τὸν Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Ζ πρὸς τὸν Η. πάλιν, ἐπεὶ ὁ Γ ἑκάτερον τῶν Ε, Ζ πολλαπλασιάσας ἑκάτερον τῶν Α, Θ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Θ. ὡς δὲ ὁ Ε πρὸς τὸν Ζ, οὕτως ὁ Γ πρὸς τὸν Δ·
And since Γ by multiplying each of Γ, Δ has made each of Ε, Ζ, therefore, as Γ is to Δ, so is Ε to Ζ. For the same reasons also, as Γ is to Δ, so is Ζ to Η. Again, since Γ by multiplying each of Ε, Ζ has made each of Α, Θ, therefore, as Ε is to Ζ, so is Α to Θ. But, as Ε is to Ζ, so is Γ to Δ; therefore also, as Γ is to Δ, so is Α to Θ.
καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Θ. πάλιν, ἐπεὶ ἑκάτερος τῶν Γ, Δ τὸν Ζ πολλαπλασιάσας ἑκάτερον τῶν Θ, Κ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Θ πρὸς τὸν Κ. πάλιν, ἐπεὶ ὁ Δ ἑκάτερον τῶν Ζ, Η πολλαπλασιάσας ἑκάτερον τῶν Κ, Β πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ζ πρὸς τὸν Η, οὕτως ὁ Κ πρὸς τὸν Β. ὡς δὲ ὁ Ζ πρὸς τὸν Η, οὕτως ὁ Γ πρὸς τὸν Δ·
Again, since each of Γ, Δ by multiplying Ζ has made each of Θ, Κ, therefore, as Γ is to Δ, so is Θ to Κ. Again, since Δ by multiplying each of Ζ, Η has made each of Κ, Β, therefore, as Ζ is to Η, so is Κ to Β. But, as Ζ is to Η, so is Γ to Δ; therefore also, as Γ is to Δ, so is both Α to Θ, Θ to Κ, and Κ to Β.
καὶ ὡς ἄρα ὁ Γ πρὸς τὸν Δ, οὕτως ὅ τε Α πρὸς τὸν Θ καὶ ὁ Θ πρὸς τὸν Κ καὶ ὁ Κ πρὸς τὸν Β. τῶν Α, Β ἄρα δύο μέσοι ἀνάλογόν εἰσιν οἱ Θ, Κ.
λέγω δή, ὅτι καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. ἐπεὶ γὰρ τέσσαρες ἀριθμοὶ ἀνάλογόν εἰσιν οἱ Α, Θ, Κ, Β, ὁ Α ἄρα πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Α πρὸς τὸν Θ. ὡς δὲ ὁ Α πρὸς τὸν Θ, οὕτως ὁ Γ πρὸς τὸν Δ· καὶ ὁ Α πρὸς τὸν Β τριπλασίονα λόγον ἔχει ἤπερ ὁ Γ πρὸς τὸν Δ. ὅπερ ἔδει δεῖξαι.
Therefore between Α, Β there are two mean proportional numbers Θ, Κ. I say next that Α also has to Β a triplicate ratio of that which Γ has to Δ. For since the four numbers Α, Θ, Κ, Β are in proportion, therefore Α has to Β a triplicate ratio of that which Α has to Θ. But, as Α is to Θ, so is Γ to Δ; therefore Α also has to Β a triplicate ratio of that which Γ has to Δ; which was to be proved.