§9.prop.8ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ μὲν τρίτος ἀπὸ τῆς μονάδος τετράγωνος ἔσται καὶ οἱ ἕνα διαλείποντες, ὁ δὲ τέταρτος κύβος καὶ οἱ δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος κύβος ἅμα καὶ τετράγωνος καὶ οἱ πέντε διαλείποντες.
If any number of numbers starting from a unit be continuously proportional, the third from the unit will be square, and those which leave out one; the fourth will be cube, and all those which leave out two; and the seventh will be both cube and square, and those which leave out five.
ἔστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, Ε, Ζ· λέγω, ὅτι ὁ μὲν τρίτος ἀπὸ τῆς μονάδος ὁ Β τετράγωνός ἐστι καὶ οἱ ἕνα διαλείποντες πάντες, ὁ δὲ τέταρτος ὁ Γ κύβος καὶ οἱ δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος ὁ Ζ κύβος ἅμα καὶ τετράγωνος καὶ οἱ πέντε διαλείποντες πάντες.
Let there be any number of numbers starting from a unit continuously proportional, A, B, Γ, Δ, Ε, Ζ; I say that the third from the unit, B, is square, and all those which leave out one, and the fourth, Γ, is cube, and all those which leave out two, and the seventh, Ζ, is both cube and square, and all those which leave out five.
ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας· καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας.
For since, as the unit is to A, so is A to B, therefore the unit measures the number A as many times as A measures B. But the unit measures the number A according to the units in it; therefore A also measures B according to the units in A.
ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν· τετράγωνος ἄρα ἐστὶν ὁ Β. καὶ ἐπεὶ οἱ Β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Β τετράγωνός ἐστιν, καὶ ὁ Δ ἄρα τετράγωνός ἐστιν.
Therefore A by multiplying itself has made B; therefore B is square. And since B, Γ, Δ are continuously proportional, and B is square, therefore Δ is also square.
διὰ τὰ αὐτὰ δὴ καὶ ὁ Ζ τετράγωνός ἐστιν.
For the same reasons indeed Ζ is also square.
ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ ἕνα διαλείποντες πάντες τετράγωνοί εἰσιν.
Similarly indeed we will show that all those which leave out one are square.
λέγω δή, ὅτι καὶ ὁ τέταρτος ἀπὸ τῆς μονάδος ὁ Γ κύβος ἐστὶ καὶ οἱ δύο διαλείποντες πάντες.
I say indeed that the fourth from the unit, Γ, is also a cube, and all those which leave out two.
ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Γ, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Γ. ἡ δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας· καὶ ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας·
For since, as the unit is to A, so is B to Γ, therefore the unit measures the number A as many times as B measures Γ. But the unit measures the number A according to the units in A; therefore B also measures Γ according to the units in A.
ὁ Α ἄρα τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν.
Therefore A by multiplying B has made Γ.
ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, κύβος ἄρα ἐστὶν ὁ Γ. καὶ ἐπεὶ οἱ Γ, Δ, Ε, Ζ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Γ κύβος ἐστίν, καὶ ὁ Ζ ἄρα κύβος ἐστίν.
Since, therefore, A by multiplying itself has made B, and by multiplying B has made Γ, therefore Γ is a cube. And since Γ, Δ, Ε, Ζ are continuously proportional, and Γ is a cube, therefore Ζ is also a cube.
ἐδείχθη δὲ καὶ τετράγωνος· ὁ ἄρα ἕβδομος ἀπὸ τῆς μονάδος κύβος τέ ἐστι καὶ τετράγωνος.
But it was also proved square; therefore the seventh from the unit is both a cube and a square.
ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ πέντε διαλείποντες πάντες κύβοι τέ εἰσι καὶ τετράγωνοι· ὅπερ ἔδει δεῖξαι.
Similarly indeed we will show that all those which leave out five are both cubes and squares; which it was required to prove.
§9.prop.9ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἑξῆς κατὰ τὸ συνεχὲς ἀριθμοὶ ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα τετράγωνος ᾖ, καὶ οἱ λοιποὶ πάντες τετράγωνοι ἔσονται.
If any number of numbers starting from a unit be continuously proportional, and the one after the unit be square, all the rest will also be square.
καὶ ἐὰν ὁ μετὰ τὴν μονάδα κύβος ᾖ, καὶ οἱ λοιποὶ πάντες κύβοι ἔσονται.
And if the one after the unit be cube, all the rest will also be cube.
ἔστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α τετράγωνος ἔστω· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοι ἔσονται.
Let there be any number of numbers starting from a unit continuously proportional, A, B, Γ, Δ, Ε, Ζ, and let the one after the unit, A, be square; I say that all the rest will also be square.
ὅτι μὲν οὖν ὁ τρίτος ἀπὸ τῆς μονάδος ὁ Β τετράγωνός ἐστι καὶ οἱ ἕνα διαλείποντες πάντες, δέδεικται· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοί εἰσιν.
Now, that the third from the unit, B, is square, and all those which leave out one, has been proved; I say that all the rest are also square.
ἐπεὶ γὰρ οἱ Α, Β, Γ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α τετράγωνος, καὶ ὁ Γ τετράγωνός ἐστιν.
For since A, B, Γ are continuously proportional, and A is square, Γ is also square.
πάλιν, ἐπεὶ οἱ β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Β τετράγωνος, καὶ ὁ Δ τετράγωνός ἐστιν.
Again, since B, Γ, Δ are continuously proportional, and B is square, Δ is also square.
ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοί εἰσιν.
Similarly indeed we will show that all the rest are also square.
ἀλλὰ δὴ ἔστω ὁ Α κύβος· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες κύβοι εἰσίν.
But indeed, let A be a cube; I say that all the rest are also cubes.
ὅτι μὲν οὖν ὁ τέταρτος ἀπὸ τῆς μονάδος ὁ Γ κύβος ἐστὶ καὶ οἱ δύο διαλείποντες πάντες, δέδεικται· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες κύβοι εἰσίν.
Now, that the fourth from the unit, Γ, is a cube, and all those which leave out two, has been proved; I say that all the rest are also cubes.
ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β, ἰσάκις ἄρα ἡ μονὰς τὸν Α μετρεῖ καὶ ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας·
For since, as the unit is to A, so is A to B, therefore the unit measures A as many times as A measures B.
καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας· ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν.
But the unit measures A according to the units in it; therefore A also measures B according to the units in itself; therefore A by multiplying itself has made B.
καί ἐστιν ὁ Α κύβος.
And A is a cube.
ἐὰν δὲ κύβος ἀριθμὸς ἑαυτὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἐστίν· καὶ ὁ Β ἄρα κύβος ἐστίν.
But if a cube number by multiplying itself make some number, the product is a cube; therefore B is also a cube.
καὶ ἐπεὶ τέσσαρες ἀριθμοὶ οἱ Α, Β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α κύβος, καὶ ὁ Δ ἄρα κύβος ἐστίν.
And since four numbers A, B, Γ, Δ are continuously proportional, and A is a cube, therefore Δ is also a cube.
διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε κύβος ἐστίν, καὶ ὁμοίως οἱ λοιποὶ πάντες κύβοι εἰσίν· ὅπερ ἔδει δεῖξαι.
For the same reasons indeed Ε is also a cube, and similarly all the rest are cubes; which it was required to prove.