§4Ἐὰν διάστημα μὴ πολλαπλάσιον δὶς συντεθῇ, τὸ ὅλον οὔτε πολλαπλάσιον ἔσται, οὔτε ἐπιμόριον.
If an interval which is not multiple is combined twice, the whole will be neither multiple nor superparticular.
Ἔστω γὰρ διάστημα μὴ πολλαπλάσιον, τὸ βγ, καὶ γεγενήσθω ὡς ὁ (p. 27) πρὸς τὸν β ὁ β πρὸς τὸν δ.
For let there be an interval βγ, which is not multiple, and let it be made that as [γ] (p. 27) is to β, so β is to δ.
λέγω ὅτι ὁ δ τοῦ γ οὔτε πολλαπλάσιος οὔτε ἐπιμόριός ἐστιν.
I say that δ is neither multiple nor superparticular of γ.
Ἔστω γὰρ πρῶτον ὁ δ τοῦ γ πολλαπλάσιος.
For let δ first be a multiple of γ.
οὐκοῦν ἐμάθομεν, ὅτι ἐὰν διάστημα δὶς συντεθὲν τὸ ὅλον ποιῇ πολλαπλάσιον, καὶ αὐτὸ πολλαπλάσιόν ἐστιν.
Then we learned that if an interval combined twice makes the whole multiple, it itself is also multiple.
ἔσται ἄρα ὁ β τοῦ πολλαπλάσιος.
Therefore, β will be a multiple of [γ].
οὐκ ἦν δέ.
But it was not.
ἀδύνατον ἄρα τὸν τοῦ εἶναι πολλαπλάσιον.
Therefore, it is impossible for [δ] to be a multiple of [γ].
ἀλλὰ μὴν οὐδʼ ἐπιμόριον.
But indeed, neither is it superparticular.
ἐπιμορίου γὰρ διαστήματος μέσος οὐδεὶς ἀνάλογον ἐμπίπτει.
For between a superparticular interval, no mean proportional falls.
εἰς δὲ τοὺς δγ ἐμπίπτει ὁ β.
And β falls between δ and γ.
ἀδύνατον ἄρα τὸν δ τοῦ γ ἢ πολλαπλάσιον ἢ ἐπιμόριον εἶναι.
Therefore, it is impossible for δ to be either a multiple or a superparticular of γ.
§5Ἐὰν διάστημα δὶς συντεθὲν τὸ ὅλον μὴ ποιῇ πολλαπλάσιον, οὐδʼ αὐτὸ ἔσται πολλαπλάσιον.
If an interval combined twice does not make the whole multiple, it itself will not be multiple either.
Ἔστω γὰρ διάστημα τὸ βγ, καὶ γεγενήσθω ὡς ὁ πρὸς τὸν β ὁ β πρὸς τὸν δ, καὶ μὴ ἔστω ὁ δ τοῦ πολλαπλάσιος.
For let there be an interval βγ, and let it be made that as [γ] is to β, so β is to δ, and let δ not be a multiple of [γ].
λέγω, ὅτι οὐδὲ ὁ β τοῦ γ ἔσται πολλαπλάσιος.
I say that neither will β be a multiple of γ.
εἰ γάρ ἔστιν ὁ β τοῦ (p. 28) πολλαπλάσιος, ἔσται ἄρα ὁ δ τοῦ γ πολλαπλάσιος.
For if β is a multiple of [γ] (p. 28), δ will therefore be a multiple of γ.
οὐκ ἔστι θ ϛ δ δέ.
But it is not (θ, ϛ, δ).
οὐκ ἄρα ὁ β τοῦ ἔσται πολλαπλάσιος.
Therefore, β will not be a multiple of [γ]. 6.
Ϛ. Τὸ διπλάσιον διάστημα ἐκ δύο τῶν μεγίστων ἐπιμορίων συνέστηκεν, ἔκ τε τοῦ ἡμιολίου καὶ ἐκ τοῦ ἐπιτρίτου.
The double interval consists of the two greatest superparticular [intervals], the sesquialter and the sesquitertian.
Ἔστω γὰρ ὁ μὲν β τοῦ δζ ἡμιόλιος, ὁ δὲ δζ τοῦ θ ἐπίτριτος.
For let β be the sesquialter of δζ, and let δζ be the sesquitertian of θ.
φημὶ τὸν β τοῦ θ διπλάσιον εἶναι.
I say that β is the double of θ.
ἀφεῖλον γὰρ ἶσον τῷ θ τὸν ζκ καὶ τῷ δζ τὸν γλ.
For I have subtracted ζκ equal to θ, and γλ equal to δζ.
οὐκοῦν ἐπεὶ ὁ βγ τοῦ δζ ἡμιόλιος, ὁ βλ ἄρα β τοῦ βγ τρίτον μέρος ἐστὶν, τοῦ δὲ δζ ἥμισυ.
Therefore, since βγ is the sesquialter of δζ, βλ is therefore a third part of βγ, and a half of δζ.
πάλιν ἐπεὶ ὁ δζ τοῦ θ ἐπίτριτός ἐστιν, ὁ δκ δ τοῦ μὲν δζ τεταρτημόριον, τοῦ δὲ θ τριτημόριον.
Again, since δζ is the sesquitertian of θ, δκ is a fourth part of δζ, and a third part of θ.
οὐκοῦν ἐπεὶ ὁ δκ τοῦ δζ ἐστι τεταρτημόριον, ὁ δὲ βλ τοῦ δζ ἥμισυ, τοῦ ἄρα βλ ἥμισυ ἔσται ὁ δκ.
Therefore, since δκ is a fourth part of δζ, and βλ is a half of δζ, δκ will therefore be a half of βλ.
ἦν δὲ ὁ βλ τοῦ βγ τρίτον μέρος· ὁ ἄρα δκ τοῦ βγ ἕκτον γ ζ μέρος ἐστίν.
But βλ was a third part of βγ; therefore, δκ is a sixth part of βγ (γ, ζ).
ἦν δὲ ὁ δκ τοῦ θ τρίτον μέρος· ιβ η Ϛ ὁ ἄρα β τοῦ θ διπλάσιός ἐστιν.
And δκ was a third part of θ; (ιβ, η, Ϛ) therefore, β is the double of θ.
Ἄλλως. Ἔστω γὰρ ὁ μὲν α τοῦ β ἡμιόλιος, ὁ δὲ β τοῦ ἐπίτριτος.
Alternatively: For let α be the sesquialter of β, and let β be the sesquitertian of γ.
λέγω ὅτι (p. 29) ὁ α τοῦ γ ἐστι διπλάσιος.
I say that (p. 29) α is the double of γ.
Ἐπεὶ γὰρ ἡμιόλιός ἐστιν ὁ α τοῦ β, ὁ α ἄρα ἔχει τὸν β καὶ τὸ ἥμισυ αὐτοῦ.
For since α is the sesquialter of β, α therefore contains β and a half of it.
δύο β ἄρα οἱ α ἶσοί εἰσι τρισὶ τοῖς β.
Therefore, two α are equal to three β.
πάλιν ἐπεὶ ὁ β τοῦ γ ἐστιν ἐπίτριτος, ὁ β ἄρα ἔχει τὸν γ καὶ τὸ τρίτον αὐτοῦ.
Again, since β is the sesquitertian of γ, β therefore contains γ and a third part of it.
τρεῖς ἄρα οἱ β ἶσοί εἰσι τέτταρσι τοῖς γ.
Therefore, three β are equal to four γ.
τρεῖς δὲ οἱ ν ἶσοί εἰσι δυσὶ τοῖς α.
And three β are equal to two α.
δύο ἄρα οἱ α ἶσοί εἰσι τέτταρσι ιβ η Ϛ τοῖς γ.
Therefore, two α are equal to (ιβ, η, Ϛ) four γ.
ἄρα ὁ α ἶσός ἐστι δυσὶ τοῖς γ· διπλάσιος ἄρα ἐστὶν ὁ α τοῦ γ.
Therefore, α is equal to two γ; α is therefore the double of γ.