Book V, Proposition 10

Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less.

 Τῶν πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον ἐκεῖνο μεῖζόν ἐστιν: πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλαττόν ἐστιν. Ἐχέτω γὰρ τὸ Α πρὸς τὸ Γ μείζονα λόγον ἤπερ τὸ Β πρὸς τὸ Γ: λέγω, ὅτι μεῖζόν ἐστι τὸ Α τοῦ Β. Εἰ γὰρ μή, ἤτοι ἴσον ἐστὶ τὸ Α τῷ Β ἢ ἔλασσον. ἴσον μὲν οὖν οὔκ ἐστι τὸ Α τῷ Β: ἑκάτερον γὰρ ἂν τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ: οὐκ ἄρα ἴσον ἐστὶ τὸ Α τῷ Β. οὐδὲ μὴν ἔλασσόν ἐστι τὸ Α τοῦ Β: τὸ Α γὰρ ἂν πρὸς τὸ Γ ἐλάσσονα λόγον εἶχεν ἤπερ τὸ Β πρὸς τὸ Γ. οὐκ ἔχει δέ: οὐκ ἄρα ἔλασσόν ἐστι τὸ Α τοῦ Β. ἐδείχθη δὲ οὐδὲ ἴσον: μεῖζον ἄρα ἐστὶ τὸ Α τοῦ Β. Ἐχέτω δὴ πάλιν τὸ Γ πρὸς τὸ Β μείζονα λόγον ἤπερ τὸ Γ πρὸς τὸ Α: λέγω, ὅτι ἔλασσόν ἐστι τὸ Β τοῦ Α. Εἰ γὰρ μή, ἤτοι ἴσον ἐστὶν ἢ μεῖζον. ἴσον μὲν οὖν οὔκ ἐστι τὸ Β τῷ Α: τὸ Γ γὰρ ἂν πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ: οὐκ ἄρα ἴσον ἐστὶ τὸ Α τῷ Β. οὐδὲ μὴν μεῖζόν ἐστι τὸ Β τοῦ Α: τὸ Γ γὰρ ἂν πρὸς τὸ Β ἐλάσσονα λόγον εἶχεν ἤπερ πρὸς τὸ Α. οὐκ ἔχει δέ: οὐκ ἄρα μεῖζόν ἐστι τὸ Β τοῦ Α. ἐδείχθη δέ, ὅτι οὐδὲ ἴσον: ἔλαττον ἄρα ἐστὶ τὸ Β τοῦ Α. Τῶν ἄρα πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον μεῖζόν ἐστιν: καὶ πρὸς ὃ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλαττόν ἐστιν: ὅπερ ἔδει δεῖξαι. Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less. For let A have to C a greater ratio than B has to C; I say that A is greater than B. For, if not, A is either equal to B or less. Now A is not equal to B; for in that case each of the magnitudes A, B would have had the same ratio to C; [V. 7] but they have not; therefore A is not equal to B. Nor again is A less than B; for in that case A would have had to C a less ratio than B has to C; [V. 8] but it has not; therefore A is not less than B. But it was proved not to be equal either; therefore A is greater than B. Again, let C have to B a greater ratio than C has to A; I say that B is less than A. For, if not, it is either equal or greater. Now B is not equal to A; for in that case C would have had the same ratio to each of the magnitudes A, B; [V. 7] but it has not; therefore A is not equal to B. Nor again is B greater than A; for in that case C would have had to B a less ratio than it has to A; [V. 8] but it has not; therefore B is not greater than A. But it was proved that it is not equal either; therefore B is less than A.