§7Ἐκ τοῦ διπλασίου διαστήματος καὶ ἡμιολίου τριπλάσιον διάστημα γίνεται.
From a double interval and a sesquialter, a triple interval is produced.
Ἔστω γὰρ ὁ μὲν α τοῦ β διπλάσιος, ὁ δὲ β τοῦ γ ἡμιόλιος.
For let α be the double of β, and β be the sesquialter of γ.
λέγω ὅτι ὁ α τοῦ γ ἐστι τριπλάσιος.
I say that α is the triple of γ.
Ἐπεὶ γὰρ ὁ α τοῦ β ἐστι διπλάσιος, ὁ α ἄρα ἶσός ἐστι δυσὶ τοῖς β.
For since α is the double of β, α is therefore equal to two β.
πάλιν ἐπεὶ ὁ β ιβ Ϛ δ τοῦ γ ἐστιν ἡμιόλιος, ἄρα ὁ ἔχει τὸν γ καὶ τὸ ἥμισυ αὐτοῦ.
Again, since β is (ιβ, Ϛ, δ) the sesquialter of γ, therefore [β] contains γ and a half of it.
δύο ἄρα οἱ β ἶσοί εἰσι τρισὶ τοῖς γ.
Therefore, two β are equal to three γ.
δύο δὲ οἱ β ἶσοί εἰσι τῷ α.
And two β are equal to α.
καὶ ὁ α ἄρα ἶσός ἐστι τρισὶ τοῖς γ.
Therefore, α also is equal to three γ.
τριπλάσιος ἄρα ἐστὶν ὁ α τοῦ γ (p. 30).
Therefore, α is the triple of γ (p. 30).
§8Ἐὰν ἀπὸ ἡμιολίου διαστήματος ἐπίτριτον διάστημα ἀφαιρεθῇ, τὸ λοιπὸν καταλείπεται ἐπόγδοον.
If a sesquitertian interval is subtracted from a sesquialter interval, the remainder is left as a sesquioctave.
Ἔστω γὰρ ὁ μὲν α τοῦ β ἡμιόλιος, ὁ δὲ γ τοῦ β ἐπίτριτος.
For let α be the sesquialter of β, and let γ be the sesquitertian of β.
λέγω ὅτι ὁ α τοῦ ἐστὶν ἐπόγδοος.
I say that α is the sesquioctave of γ.
Ἐπεὶ γὰρ ὁ α τοῦ β ἐστὶν ἡμιόλιος, 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.
ὀκτὼ ἄρα οἱ α ἶσοί εἰσι δώδεκα τοῖς β.
Therefore, eight α are equal to twelve β.
πάλιν ἐπεὶ ὁ τοῦ ἐστὶν ἐπίτριτος, ὁ γ ἄρα ἔχει θ Ϛ η τὸν β καὶ τὸ τρίτον αὐτοῦ.
Again, since γ is the sesquitertian of [β], γ therefore contains (θ, Ϛ, η) β and a third part of it.
ἐννέα ἄρα οἱ γ ἶσοί εἰσι δώδεκα τοῖς β.
Therefore, nine γ are equal to twelve β.
δώδεκα δὲ οἱ β ἶσοί εἰσιν ὀκτὼ τοῖς α· ὀκτὼ ἄρα οἱ α ἶσοί εἰσιν ἐννέα τοῖς γ.
And twelve β are equal to eight α; therefore, eight α are equal to nine γ.
ὁ α ἄρα ἶσός ἐστι τῷ γ καὶ τῷ ὀγδόῳ αὐτοῦ, ἄρα ὁ α τοῦ ἐστὶν ἐπόγδοος.
Therefore, α is equal to γ and an eighth part of it. Therefore, α is the sesquioctave of γ.
§9Τὰ ἓξ ἐπόγδοα διαστήματα μείζονά ἐστι διαστήματος ἑνὸς διπλασίου.
Six sesquioctave intervals are greater than one double interval.
Ἐστω γὰρ εἶς ἀριθμὸς ὁ α.
For let α be some one number.
καὶ τοῦ μὲν α ἐπόγδοος ἔστω ὁ β, τοῦ δὲ β (p. 31) ἐπόγδοος ὁ γ, τοῦ δὲ ἐπόγδοος ὁ δ, τοῦ δὲ δ ἐπόγδοος ὁ ε, τοῦ ε ἐπόγδοος ὁ ζ, τοῦ ζ ἐπόγδοος ὁ η.
And let β be the sesquioctave of α, and let γ be the sesquioctave of β (p. 31), and let δ be the sesquioctave of [γ], and let ε be the sesquioctave of δ, and let ζ be the sesquioctave of ε, and let η be the sesquioctave of ζ.
λέγω, ὅτι ὁ η τοῦ α μείζων ἐστὶν ἢ διπλάσιος.
I say that η is greater than the double of α.
Ἐπεὶ ἐμάθομεν εὑρεῖν ἑπτὰ ἀριθμοὺς ἐπογδόους ἀλλήλων, εὑρήσθωσαν οἱ α β γ δ ε ζ η, καὶ γίνεται ὁ μὲν α κϚ μύρια βρμδ,
ὁ δὲ β κθ μύρια δϠιβ,
ὁ δὲ γ λγ μύρια αψοϚ
ὁ δὲ δ λζ μύρια γϲμη,
ὁ δὲ ε μα μύρια θϠδ,
ὁ δὲ ζ μζ μύρια βτ Ϟβ,
ὁ δὲ η νγ μύρια αυμα καὶ ἔστιν ὁ η τοῦ α μείζων ἢ διπλάσιος.
Since we learned how to find seven numbers in sesquioctave ratio to one another, let α, β, γ, δ, ε, ζ, η be found, and α becomes 262,144, and β becomes 294,912, and γ becomes 331,776, and δ becomes 373,248, and ε becomes 419,904, and ζ becomes 472,392, and η becomes 531,441. And η is greater than the double of α.
§10Τὸ διὰ πασῶν διάστημά ἐστι πολλαπλάσιον.
The octave interval is multiple.
Ἔστω γὰρ νήτη μὲν ὑπερβολαίων (p. 32) ὁ α, μέση γ δὲ ὁ β, προσλαμβανόμενος δὲ ὁ γ.
For let the nete hyperbolaion (p. 32) be α, let the mese (γ) be β, and let the proslambanomenos be γ.
τὸ ἄρα α γ διάστημα δὶς διὰ πασῶν ὂν ἐστὶ σύμφωνον.
Therefore, the αγ interval, being a double octave, is consonant.
ἤτοι οὖν ἐπιμόριόν ἐστιν, ἢ πολλαπλασιον.
It is therefore either superparticular or multiple.
ἐπιμόριον μὲν οὐκ ἔστιν· ἐπιμορίου γὰρ διαστήματος μέσος οὐδεὶς ἀνάλογον ἐμπίπτει.
It is not superparticular; for between a superparticular interval, no mean proportional falls.
πολλαπλάσιον ἄρα ἐστίν.
Therefore, it is multiple.
ἐπεὶ οὖν δύο ἶσα διαστήματα τὰ αβ βγ συντεθέντα ποιεῖ πολλαπλάσιον τὸ ὅλον, καὶ τὸ αβ ἄρα ἐστὶ πολλαπλάσιον.
Since, therefore, two equal intervals αβ and βγ combined make the whole multiple, αβ therefore is also multiple.
§11Τὸ διὰ τεσσάρων διάστημα καὶ τὸ διὰ πέντε ἑκάτερον ἐπιμόριόν ἐστιν.
The fourth interval and the fifth interval are each superparticular.
Ἔστω γὰρ νήτη μὲν συνημμένων ὁ α, μέση δὲ ὁ β, ὑπάτη δὲ μέσων ὁ γ.
For let the nete synemmenon be α, let the mese be β, and let the hypate meson be γ.
τὸ ἄρα α διάστημα δὶς διὰ τεσσάρων ὂν ἐστὶ διάφωνον· οὐκ ἄρα γ ἐστὶ πολλαπλάσιον.
Therefore, the αγ interval, being a double fourth, is dissonant; therefore, it is not multiple.
ἐπεὶ οὖν δύο διαστήματα β ἶσα τὰ αβ β γ συντεθέντα τὸ ὅλον μὴ α ποιεῖ πολλαπλάσιον, οὐδὲ ἄρα τὸ αβ ἐστὶ πολλαπλάσιον.
Since, therefore, two equal (β) intervals αβ and βγ combined do not make the whole (α) multiple, αβ is therefore not multiple either.
καὶ ἔστι σύμφωνον· ἐπιμόριον ἄρα.
And it is consonant; therefore, it is superparticular.
ἡ αὐτὴ δὲ ἀπόδειξις καὶ ἐπὶ ισ ιβ θ τοῦ διὰ πέντε. (p. 33. )
The same proof also applies to the fifth (ισ, ιβ, θ) (p. 33).