§9.prop.4ἐὰν κύβος ἀριθμὸς κύβον ἀριθμὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἔσται.
If a cube number by multiplying a cube number make some number, the product will be a cube.
κύβος γὰρ ἀριθμὸς ὁ Α κύβον ἀριθμὸν τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ κύβος ἐστίν.
For let a cube number A by multiplying a cube number B make Γ; I say that Γ is a cube.
ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· ὁ Δ ἄρα κύβος ἐστίν.
For let A by multiplying itself make Δ; therefore Δ is a cube.
καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Α, Β κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν οἱ Α, Β. τῶν Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί· ὥστε καὶ τῶν Δ, Γ δύο μέσοι ἀνάλογον ἐμπεσοῦνται ἀριθμοί.
And since A by multiplying itself has made Δ, and by multiplying B has made Γ, therefore, as A is to B, so is Δ to Γ. And since A, B are cubes, A, B are similar solids. Therefore between A, B two mean proportional numbers fall; so that also between Δ, Γ two mean proportional numbers will fall.
καί ἐστι κύβος ὁ Δ· κύβος ἄρα καὶ ὁ Γ· ὅπερ ἔδει δεῖξαι.
And Δ is a cube; therefore Γ is also a cube; which it was required to prove.
§9.prop.5ἐὰν κύβος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας κύβον ποιῇ, καὶ ὁ πολλαπλασιασθεὶς κύβος ἔσται.
If a cube number by multiplying some number make a cube, the multiplied number will also be a cube.
κύβος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλαπλασιάσας κύβον τὸν Γ ποιείτω· λέγω, ὅτι ὁ Β κύβος ἐστίν.
For let a cube number A by multiplying some number B make a cube Γ; I say that B is a cube.
ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· κύβος ἄρα ἐστίν ὁ Δ. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Δ, Γ κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν.
For let A by multiplying itself make Δ; therefore Δ is a cube. And since A by multiplying itself has made Δ, and by multiplying B has made Γ, therefore, as A is to B, so is Δ to Γ. And since Δ, Γ are cubes, they are similar solids.
τῶν Δ, Γ ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί.
Therefore between Δ, Γ two mean proportional numbers fall.
καί ἐστιν ὡς ὁ Δ πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν Β· καὶ τῶν α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί.
And, as Δ is to Γ, so is A to B; therefore also between A, B two mean proportional numbers fall.
καί ἐστι κύβος ὁ Α· κύβος ἄρα ἐστὶ καὶ ὁ Β· ὅπερ ἔδει δεῖξαι.
And A is a cube; therefore B is also a cube; which it was required to prove.
§9.prop.6ἐὰν ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ αὐτὸς κύβος ἔσται.
If a number by multiplying itself make a cube, it itself will also be a cube.
ἀριθμὸς γὰρ ὁ Α ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β ποιείτω· λέγω, ὅτι καὶ ὁ Α κύβος ἐστίν.
For let a number A by multiplying itself make a cube B; I say that A is also a cube.
ʽὁ γὰρ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω.
For let A by multiplying B make Γ.
ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Γ ἄρα κύβος ἐστίν.
Since, therefore, A by multiplying itself has made B, and by multiplying B has made Γ, therefore Γ is a cube.
καὶ ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας.
And since A by multiplying itself has made B, therefore A measures B according to the units in itself.
μετρεῖ δὲ καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας.
But the unit also measures A according to the units in it.
ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β. καὶ ἐπεὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας.
Therefore, as the unit is to A, so is A to B. And since A by multiplying B has made Γ, therefore B measures Γ according to the units in A.
μετρεῖ δὲ καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας.
But the unit also measures A according to the units in it.
ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Γ. ἀλλʼ ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Β, Γ κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν.
Therefore, as the unit is to A, so is B to Γ. But as the unit is to A, so is A to B; therefore also, as A is to B, so is B to Γ. And since B, Γ are cubes, they are similar solids.
τῶν Β, Γ ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί.
Therefore between B, Γ two mean proportional numbers exist.
καί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β. καὶ τῶν Α, Β ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί.
And, as B is to Γ, so is A to B. Therefore also between A, B two mean proportional numbers exist.
καί ἐστι κύβος ὁ Β· κύβος ἄρα ἐστὶ καὶ ὁ Α· ὅπερ ἔδει δεῖξαι.
And B is a cube; therefore A is also a cube; which it was required to prove.
§9.prop.7ἐὰν σύνθετος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος στερεὸς ἔσται.
If a composite number by multiplying some number make some number, the product will be a solid.
σύνθετος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ στερεός ἐστιν.
For let a composite number A by multiplying some number B make Γ; I say that Γ is a solid.
ἐπεὶ γὰρ ὁ Α σύνθετός ἐστιν, ὑπὸ ἀριθμοῦ τινος μετρηθήσεται.
For since A is composite, it will be measured by some number.
μετρείσθω ὑπὸ τοῦ Δ, καὶ ὁσάκις ὁ Δ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν ὁ Δ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, ὁ Ε ἄρα τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν.
Let it be measured by Δ, and let there be as many units in Ε as the times Δ measures A. Since, therefore, Δ measures A according to the units in Ε, therefore Ε by multiplying Δ has made A.
καὶ ἐπεὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ δὲ Α ἐστιν ὁ ἐκ τῶν Δ, Ε, ὁ ἄρα ἐκ τῶν Δ, Ε τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν.
And since A by multiplying B has made Γ, and A is the product of Δ, Ε, therefore the product of Δ, Ε by multiplying B has made Γ.
ὁ Γ ἄρα στερεός ἐστιν, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Δ, Ε, Β· ὅπερ ἔδει δεῖξαι.
Therefore Γ is a solid, and its sides are Δ, Ε, B; which it was required to prove.