Euclid · Elements §5.prop.17-5.prop.18

Proportionality in Separando and Componendo

Passage 78 of 316 · Greek

Summary

Proposition 17 proves that if compounded magnitudes are proportional, they will also be proportional when separated. Proposition 18 proves the converse, showing that if separated magnitudes are proportional, they will also be proportional when compounded, using proof by contradiction.

§5.prop.17ἐὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται.

If compounded magnitudes are proportional, they will also be proportional separated.

ἔστω συγκείμενα μεγέθη ἀνάλογον τὰ ΑΒ, ΒΕ, ΓΔ, ΔΖ, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ· λέγω, ὅτι καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΔΖ. εἰλήφθω γὰρ τῶν μὲν ΑΕ, ΕΒ, ΓΖ, ΖΔ ἰσάκις πολλαπλάσια τὰ ΗΘ, ΘΚ, ΛΜ, ΜΝ, τῶν δὲ ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ ΚΞ, ΝΠ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΘΚ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΗΚ τοῦ ΑΒ. ἰσάκις δέ ἐστι πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΛΜ τοῦ ΓΖ· ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΜ τοῦ ΓΖ. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΜΝ τοῦ ΖΔ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΛΝ τοῦ ΓΔ. ἰσάκις δὲ ἦν πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΗΚ τοῦ ΑΒ·

Let compounded magnitudes be proportional, AB, BE, ΓΔ, ΔΖ, so that as AB is to BE, so is ΓΔ to ΔΖ; I say that they will also be proportional separated, so that as AE is to EB, so is ΓΖ to ΔΖ. For let equal multiples HΘ, ΘK, ΛM, MN be taken of AE, EB, ΓΖ, ΖΔ, and other, as it may happen, equal multiples KΞ, NΠ of EB, ΖΔ. And since HΘ is an equal multiple of AE and ΘK of EB, therefore HΘ is an equal multiple of AE and HK of AB. And HΘ is an equal multiple of AE and ΛM of ΓΖ; therefore HK is an equal multiple of AB and ΛM of ΓΖ. Again, since ΛM is an equal multiple of ΓΖ and MN of ΖΔ, therefore ΛM is an equal multiple of ΓΖ and ΛN of ΓΔ. And ΛM was an equal multiple of ΓΖ and HK of AB; therefore HK is an equal multiple of AB and ΛN of ΓΔ.

ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΝ τοῦ ΓΔ. τὰ ΗΚ, ΛΝ ἄρα τῶν ΑΒ, ΓΔ ἰσάκις ἐστὶ πολλαπλάσια.

Therefore HK, ΛN are equal multiples of AB, ΓΔ.

πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΘΚ τοῦ ΕΒ καὶ τὸ ΜΝ τοῦ ΖΔ, ἔστι δὲ καὶ τὸ ΚΞ τοῦ ΕΒ ἰσάκις πολλαπλάσιον καὶ τὸ ΝΠ τοῦ ΖΔ, καὶ συντεθὲν τὸ ΘΞ τοῦ ΕΒ ἰσάκις ἐστὶ πολλαπλάσιον καὶ τὸ ΜΠ τοῦ ΖΔ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, καὶ εἴληπται τῶν μὲν ΑΒ, ΓΔ ἰσάκις πολλαπλάσια τὰ ΗΚ, ΛΝ, τῶν δὲ ΕΒ, ΖΔ ἰσάκις πολλαπλάσια τὰ ΘΞ, ΜΠ, εἰ ἄρα ὑπερέχει τὸ ΗΚ τοῦ ΘΞ, ὑπερέχει καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον.

Again, since ΘK is an equal multiple of EB and MN of ΖΔ, and KΞ is also an equal multiple of EB and NΠ of ΖΔ, therefore, combined, ΘΞ is also an equal multiple of EB and MΠ of ΖΔ. And since as AB is to BE, so is ΓΔ to ΔΖ, and equal multiples HK, ΛN have been taken of AB, ΓΔ, and equal multiples ΘΞ, MΠ of EB, ΖΔ, therefore, if HK exceeds ΘΞ, ΛN also exceeds MΠ, and if equal, equal, and if less, less.

ὑπερεχέτω δὴ τὸ ΗΚ τοῦ ΘΞ, καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΘΚ ὑπερέχει ἄρα καὶ τὸ ΗΘ τοῦ ΚΞ. ἀλλὰ εἰ ὑπερεῖχε τὸ ΗΚ τοῦ ΘΞ, ὑπερεῖχε καὶ τὸ ΛΝ τοῦ ΜΠ· ὑπερέχει ἄρα καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΜΝ ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ· ὥστε εἰ ὑπερέχει τὸ ΗΘ τοῦ ΚΞ, ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ ΗΘ τῷ ΚΞ, ἴσον ἔσται καὶ τὸ ΛΜ τῷ ΝΠ, κἂν ἔλαττον, ἔλαττον.

Let HK then exceed ΘΞ; and, HK and ΘΞ having ΘK subtracted from them in common, HΘ also exceeds KΞ. But if HK exceeded ΘΞ, ΛN also exceeded MΠ; therefore ΛN also exceeds MΠ, and, MN being subtracted in common, ΛM also exceeds NΠ; so that if HΘ exceeds KΞ, ΛM also exceeds NΠ. In like manner we shall also prove that, even if HΘ is equal to KΞ, ΛM will also be equal to NΠ, and if less, less.

καί ἐστι τὰ μὲν ΗΘ, ΛΜ τῶν ΑΕ, ΓΖ ἰσάκις πολλαπλάσια, τὰ δὲ ΚΞ, ΝΠ τῶν ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια·

And HΘ, ΛM are equal multiples of AE, ΓΖ, and KΞ, NΠ other, as it may happen, equal multiples of EB, ΖΔ; therefore, as AE is to EB, so is ΓΖ to ΖΔ.

ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. ἐὰν ἄρα συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται· ὅπερ ἔδει δεῖξαι.

Therefore, if compounded magnitudes are proportional, they will also be proportional separated; which was to be proved.

§5.prop.18ἐὰν διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα ἀνάλογον ἔσται.

If separated magnitudes are proportional, they will also be proportional compounded.

ἔστω διῃρημένα μεγέθη ἀνάλογον τὰ ΑΕ, ΕΒ, ΓΖ, ΖΔ, ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ· λέγω, ὅτι καὶ συντεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΖΔ. εἰ γὰρ μή ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, ἔσται ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ ἤτοι πρὸς ἔλασσόν τι τοῦ ΔΖ ἢ πρὸς μεῖζον.

Let separated magnitudes be proportional, AE, EB, ΓΖ, ΖΔ, so that as AE is to EB, so is ΓΖ to ΖΔ; I say that they will also be proportional compounded, so that as AB is to BE, so is ΓΔ to ΖΔ. For if as AB is to BE, so is not ΓΔ to ΔΖ, then as AB is to BE, so will ΓΔ be either to some magnitude less than ΔΖ or to some magnitude greater.

ἔστω πρότερον πρὸς ἔλασσον τὸ ΔΗ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΗ, συγκείμενα μεγέθη ἀνάλογόν ἐστιν· ὥστε καὶ διαιρεθέντα ἀνάλογον ἔσται.

Let it first be to a less magnitude ΔH. And since as AB is to BE, so is ΓΔ to ΔH, the compounded magnitudes are proportional; so that they will also be proportional separated.

ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΗ πρὸς τὸ ΗΔ. ὑπόκειται δὲ καὶ ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. καὶ ὡς ἄρα τὸ ΓΗ πρὸς τὸ ΗΔ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. μεῖζον δὲ τὸ πρῶτον τὸ ΓΗ τοῦ τρίτου τοῦ ΓΖ·

Therefore, as AE is to EB, so is ΓH to HΔ. But it is also assumed that as AE is to EB, so is ΓΖ to ΖΔ. Therefore, as ΓH is to HΔ, so also is ΓΖ to ΖΔ. But the first ΓH is greater than the third ΓΖ; therefore the second HΔ is also greater than the fourth ΖΔ.

μεῖζον ἄρα καὶ τὸ δεύτερον τὸ ΗΔ τοῦ τετάρτου τοῦ ΖΔ. ἀλλὰ καὶ ἔλαττον· ὅπερ ἐστὶν ἀδύνατον·

But it is also less; which is impossible.

οὐκ ἄρα ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς ἔλασσον τοῦ ΖΔ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ πρὸς μεῖζον· πρὸς αὐτὸ ἄρα.

Therefore, as AB is to BE, so is not ΓΔ to a magnitude less than ΖΔ. In like manner we shall also prove that neither is it to a greater; therefore it is to itself.

ἐὰν ἄρα διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα ἀνάλογον ἔσται· ὅπερ ἔδει δεῖξαι.

Therefore, if separated magnitudes are proportional, they will also be proportional compounded; which was to be proved.

Notes

5.prop.17συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται — For the nominative neuter plural noun phrase συγκείμενα μεγέθη (compounded magnitudes), the verb ᾖ in the conditional clause is in the third-person singular (present subjunctive), and the verb ἔσται in the consequent clause is also in the third-person singular (future). This follows the Greek grammatical rule where a neuter plural subject takes a singular verb. The participle διαιρεθέντα (separated) is also neuter plural nominative, functioning as the grammatical subject for the singular verb ἔσται.
5.prop.18ἤτοι πρὸς ἔλασσόν τι τοῦ ΔΖ ἢ πρὸς μεῖζον — The correlative conjunctions ἤτοι ... ἤ (either ... or) present two mutually exclusive possibilities for the ratio's standard (either to something less than ΔΖ or to something greater). The comparative adjectives ἔλασσον (less) and μεῖζον (greater) take the genitive of comparison τοῦ ΔΖ (than ΔΖ) as their referent.

Cite this passage

Euclid, Elements §5.prop.17-5.prop.18. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:5.prop.17-5.prop.18

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