§5.prop.13ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα λόγον ἔχῃ ἢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον μείζονα λόγον ἕξει ἢ πέμπτον πρὸς ἕκτον.
If a first magnitude has to a second the same ratio as a third to a fourth, and the third has to the fourth a greater ratio than a fifth to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth.
πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, τρίτον δὲ τὸ Γ πρὸς τέταρτον τὸ Δ μείζονα λόγον ἐχέτω ἢ πέμπτον τὸ Ε πρὸς ἕκτον τὸ Ζ. λέγω, ὅτι καὶ πρῶτον τὸ Α πρὸς δεύτερον τὸ Β μείζονα λόγον ἕξει ἤπερ πέμπτον τὸ Ε πρὸς ἕκτον τὸ Ζ.
ἐπεὶ γὰρ ἔστι τινὰ τῶν μὲν Γ, Ε ἰσάκις πολλαπλάσια, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια, καὶ τὸ μὲν τοῦ Γ πολλαπλάσιον τοῦ τοῦ Δ πολλαπλασίου ὑπερέχει, τὸ δὲ τοῦ Ε πολλαπλάσιον τοῦ τοῦ Ζ πολλαπλασίου οὐχ ὑπερέχει, εἰλήφθω, καὶ ἔστω τῶν μὲν Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, Λ, ὥστε τὸ μὲν Η τοῦ Κ ὑπερέχειν, τὸ δὲ Θ τοῦ Λ μὴ ὑπερέχειν· καὶ ὁσαπλάσιον μέν ἐστι τὸ Η τοῦ Γ, τοσαυταπλάσιον ἔστω καὶ τὸ Μ τοῦ Α, ὁσαπλάσιον δὲ τὸ Κ τοῦ Δ, τοσαυταπλάσιον ἔστω καὶ τὸ Ν τοῦ Β.
καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Μ, Η, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Ν, Κ, εἰ ἄρα ὑπερέχει τὸ Μ τοῦ Ν, ὑπερέχει καὶ τὸ Η τοῦ Κ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον.
For let a first magnitude A have to a second B the same ratio as a third Γ to a fourth Δ, and let the third Γ have to the fourth Δ a greater ratio than a fifth E to a sixth Z. I say that the first A will also have to the second B a greater ratio than the fifth E to the sixth Z. For since there are some equal multiples of Γ, E, and other, as it may happen, equal multiples of Δ, Z, such that the multiple of Γ exceeds the multiple of Δ, but the multiple of E does not exceed the multiple of Z, let them be taken, and let H, Θ be equal multiples of Γ, E, and K, Λ other, as it may happen, equal multiples of Δ, Z, so that H exceeds K, but Θ does not exceed Λ; and let M be the same multiple of A as H is of Γ, and let N be the same multiple of B as K is of Δ. And since as A is to B, so is Γ to Δ, and there have been taken equal multiples M, H of A, Γ, and other, as it may happen, equal multiples N, K of B, Δ, if therefore M exceeds N, H also exceeds K, and if equal, equal, and if less, less.
ὑπερέχει δὲ τὸ Η τοῦ Κ· ὑπερέχει ἄρα καὶ τὸ Μ τοῦ Ν. τὸ δὲ Θ τοῦ Λ οὐχ ὑπερέχει·
But H exceeds K; therefore M also exceeds N.
καί ἐστι τὰ μὲν Μ, Θ τῶν Α, Ε ἰσάκις πολλαπλάσια, τὰ δὲ Ν, Λ τῶν Β, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια· τὸ ἄρα Α πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Ε πρὸς τὸ Ζ.
ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα λόγον ἔχῃ ἢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον μείζονα λόγον ἕξει ἢ πέμπτον πρὸς ἕκτον· ὅπερ ἔδει δεῖξαι.
But Θ does not exceed Λ; and M, Θ are equal multiples of A, E, and N, Λ other, as it may happen, equal multiples of B, Z; therefore A has to B a greater ratio than E has to Z. Therefore, if a first magnitude has to a second the same ratio as a third to a fourth, and the third has to the fourth a greater ratio than a fifth to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth; which was to be proved.
§5.prop.14ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον.
If a first magnitude has to a second the same ratio as a third to a fourth, and the first is greater than the third, the second will also be greater than the fourth, and if equal, equal, and if less, less.
πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, μεῖζον δὲ ἔστω τὸ Α τοῦ Γ· λέγω, ὅτι καὶ τὸ Β τοῦ Δ μεῖζόν ἐστιν.
For let a first magnitude A have to a second B the same ratio as a third Γ to a fourth Δ, and let A be greater than Γ; I say that B is also greater than Δ.
ἐπεὶ γὰρ τὸ Α τοῦ Γ μεῖζόν ἐστιν, ἄλλο δέ, ὃ ἔτυχεν, τὸ Β, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ·
For since A is greater than Γ, and B is some other magnitude as it may happen, A therefore has to B a greater ratio than Γ has to B. And as A is to B, so is Γ to Δ; therefore Γ also has to Δ a greater ratio than Γ has to B.
καὶ τὸ Γ ἄρα πρὸς τὸ Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλασσόν ἐστιν· ἔλασσον ἄρα τὸ Δ τοῦ Β·
But that to which the same magnitude has a greater ratio is less; therefore Δ is less than B; so that B is greater than Δ.
ὥστε μεῖζόν ἐστι τὸ Β τοῦ Δ.
ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Β τῷ Δ, κἂν ἔλασσον ᾖ τὸ Α τοῦ Γ, ἔλασσον ἔσται καὶ τὸ Β τοῦ Δ.
ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον· ὅπερ ἔδει δεῖξαι.
Similarly indeed we will show that even if A is equal to Γ, B will also be equal to Δ, and if A is less than Γ, B will also be less than Δ. Therefore, if a first magnitude has to a second the same ratio as a third to a fourth, and the first is greater than the third, the second will also be greater than the fourth, and if equal, equal, and if less, less; which was to be proved.