§5.prop.6ἐὰν δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολλαπλάσια, καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολλαπλάσια, καὶ τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν πολλαπλάσια.
If two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted be equimultiples of the same, then the remainders also are either equal to them or equimultiples of them.
δύο γὰρ μεγέθη τὰ ΑΒ, ΓΔ δύο μεγεθῶν τῶν Ε, Ζ ἰσάκις ἔστω πολλαπλάσια, καὶ ἀφαιρεθέντα τὰ ΑΗ, ΓΘ τῶν αὐτῶν τῶν Ε, Ζ ἰσάκις ἔστω πολλαπλάσια·
For let two magnitudes, AB, ΓΔ, be equimultiples of two magnitudes, E, Z, and let subtracted parts, AH, ΓΘ, be equimultiples of the same, E, Z.
λέγω, ὅτι καὶ λοιπὰ τὰ ΗΒ, ΘΔ τοῖς Ε, Ζ ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν πολλαπλάσια.
I say that the remainders, HB, ΘΔ, also are either equal to E, Z, or equimultiples of them.
ἔστω γὰρ πρότερον τὸ ΗΒ τῷ Ε ἴσον.
For let, first, HB be equal to E.
λέγω, ὅτι καὶ τὸ ΘΔ τῷ Ζ ἴσον ἐστίν.
I say that ΘΔ is also equal to Z.
κείσθω γὰρ τῷ Ζ ἴσον τὸ ΓΚ. ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΗ τοῦ Ε καὶ τὸ ΓΘ τοῦ Ζ, ἴσον δὲ τὸ μὲν ΗΒ τῷ Ε, τὸ δὲ ΚΓ τῷ Ζ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΚΘ τοῦ Ζ. ἰσάκις δὲ ὑπόκειται πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΓΔ τοῦ Ζ·
For let ΓK be laid down equal to Z. Since AH is the same multiple of E that ΓΘ is of Z, and HB is equal to E, and KΓ to Z, therefore AB is the same multiple of E that KΘ is of Z. But AB is supposed to be the same multiple of E that ΓΔ is of Z; therefore KΘ is the same multiple of Z that ΓΔ is of Z.
ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΚΘ τοῦ Ζ καὶ τὸ ΓΔ τοῦ Ζ. ἐπεὶ οὖν ἑκάτερον τῶν ΚΘ, ΓΔ τοῦ Ζ ἰσάκις ἐστὶ πολλαπλάσιον, ἴσον ἄρα ἐστὶ τὸ ΚΘ τῷ ΓΔ. κοινὸν ἀφῃρήσθω τὸ ΓΘ·
Since therefore each of the magnitudes KΘ, ΓΔ is the same multiple of Z, therefore KΘ is equal to ΓΔ. Let the common part, ΓΘ, be subtracted; therefore the remainder, KΓ, is equal to the remainder, ΘΔ.
λοιπὸν ἄρα τὸ ΚΓ λοιπῷ τῷ ΘΔ ἴσον ἐστίν. ἀλλὰ τὸ Ζ τῷ ΚΓ ἐστιν ἴσον· καὶ τὸ ΘΔ ἄρα τῷ Ζ ἴσον ἐστίν.
But Z is equal to KΓ; therefore ΘΔ is also equal to Z.
ὥστε εἰ τὸ ΗΒ τῷ Ε ἴσον ἐστίν, καὶ τὸ ΘΔ ἴσον ἔσται τῷ Ζ.
ὁμοίως δὴ δείξομεν, ὅτι, κἂν πολλαπλάσιον ᾖ τὸ ΗΒ τοῦ Ε, τοσαυταπλάσιον ἔσται καὶ τὸ ΘΔ τοῦ Ζ.
ἐὰν ἄρα δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολλαπλάσια, καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολλαπλάσια, καὶ τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν πολλαπλάσια· ὅπερ ἔδει δεῖξαι.
So that, if HB is equal to E, ΘΔ will also be equal to Z. In the same manner we shall prove that, even if HB be a multiple of E, ΘΔ will also be the same multiple of Z. Therefore, if two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted be equimultiples of the same, then the remainders also are either equal to them or equimultiples of them; which was to be proved.
§5.prop.7τὰ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ αὐτὸ πρὸς τὰ ἴσα.
Equal magnitudes have the same ratio to the same magnitude, and the same has the same ratio to equal magnitudes.
ἔστω ἴσα μεγέθη τὰ Α, Β, ἄλλο δέ τι, ὃ ἔτυχεν, μέγεθος τὸ Γ·
For let A, B be equal magnitudes, and Γ any other arbitrary magnitude.
λέγω, ὅτι ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν ἔχει λόγον, καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β.
εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Δ, Ε, τοῦ δὲ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον τὸ Ζ.
ἐπεὶ οὖν ἰσάκις ἐστὶ πολλαπλάσιον τὸ Δ τοῦ Α καὶ τὸ Ε τοῦ Β, ἴσον δὲ τὸ Α τῷ Β, ἴσον ἄρα καὶ τὸ Δ τῷ Ε. ἄλλο δέ, ὃ ἔτυχεν, τὸ Ζ. εἰ ἄρα ὑπερέχει τὸ Δ τοῦ Ζ, ὑπερέχει καὶ τὸ Ε τοῦ Ζ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον.
I say that each of the magnitudes A, B has the same ratio to Γ, and Γ has the same ratio to each of the magnitudes A, B. For let equimultiples Δ, E be taken of A, B, and another arbitrary multiple, Z, of Γ. Since therefore Δ is the same multiple of A that E is of B, and A is equal to B, therefore Δ is also equal to E.
καί ἐστι τὰ μὲν Δ, Ε τῶν Α, Β ἰσάκις πολλαπλάσια, τὸ δὲ Ζ τοῦ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον· ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Γ.
λέγω, ὅτι καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν ἔχει λόγον.
And Z is another arbitrary magnitude. Therefore, if Δ exceeds Z, E also exceeds Z, and if equal, equal, and if less, less. And Δ, E are equimultiples of A, B, and Z is another arbitrary multiple of Γ; therefore, as A is to Γ, so is B to Γ. I say that Γ also has the same ratio to each of the magnitudes A, B.
τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἴσον ἐστὶ τὸ Δ τῷ Ε· ἄλλο δέ τι τὸ Ζ· εἰ ἄρα ὑπερέχει τὸ Ζ τοῦ Δ, ὑπερέχει καὶ τοῦ Ε, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον.
For with the same construction we shall prove in the same manner that Δ is equal to E; and Z is some other magnitude; therefore, if Z exceeds Δ, it also exceeds E, and if equal, equal, and if less, less.
καί ἐστι τὸ μὲν Ζ τοῦ Γ πολλαπλάσιον, τὰ δὲ Δ, Ε τῶν Α, Β ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως τὸ Γ πρὸς τὸ Β.
τὰ ἴσα ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ αὐτὸ πρὸς τὰ ἴσα.
And Z is a multiple of Γ, and Δ, E are other arbitrary equimultiples of A, B; therefore, as Γ is to A, so is Γ to B. Therefore, equal magnitudes have the same ratio to the same magnitude, and the same has the same ratio to equal magnitudes.
Πόρισμα
ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν μεγέθη τινὰ ἀνάλογον ᾖ, καὶ ἀνάπαλιν ἀνάλογον ἔσται. ὅπερ ἔδει δεῖξαι.
Corollary From this it is manifest that, if any magnitudes be proportional, they will also be proportional inversely; which was to be proved.