elem.10.18

Book X, Proposition 18

Ἐὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, καὶ εἰς ἀσύμμετρα αὐτὴν διαιρῇ [ μήκει ], ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διαιρεῖ [ μήκει ]. ̓́εστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ ΒΓ, τῷ δὲ τετάρτῳ [ μέρει ] τοῦ ἀπὸ τῆς ἐλάσσονος τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔΓ, ἀσύμμετρος δὲ ἔστω ἡ ΒΔ τῇ ΔΓ μήκει: λέγω, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. Τῶν γὰρ αὐτῶν κατασκευασθέντων τῷ πρότερον ὁμοίως δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. δεικτέον [ οὖν ], ὅτι ἀσύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ μήκει. ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει. ἀλλὰ ἡ ΔΓ σύμμετρός ἐστι συναμφοτέραις ταῖς ΒΖ, ΔΓ: καὶ ἡ ΒΓ ἄρα ἀσύμμετρός ἐστι συναμφοτέραις ταῖς ΒΖ, ΔΓ. ὥστε καὶ λοιπῇ τῇ ΖΔ ἀσύμμετρός ἐστιν ἡ ΒΓ μήκει. καὶ ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ: ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. Δυνάσθω δὴ πάλιν ἡ ΒΓ τῆς Α μεῖζον τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει. Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. ἀλλὰ ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει: ὥστε καὶ λοιπῇ συναμφοτέρῳ τῇ ΒΖ, ΔΓ ἀσύμμετρός ἐστιν ἡ ΒΓ. ἀλλὰ συναμφότερος ἡ ΒΖ, ΔΓ τῇ ΔΓ σύμμετρός ἐστι μήκει: καὶ ἡ ΒΓ ἄρα τῇ ΔΓ ἀσύμμετρός ἐστι μήκει: ὥστε καὶ διελόντι ἡ ΒΔ τῇ ΔΓ ἀσύμμετρός ἐστι μήκει. Ἐὰν ἄρα ὦσι δύο εὐθεῖαι, καὶ τὰ ἑξῆς. Λῆμμα Ἐπεὶ δέδεικται, ὅτι αἱ μήκει σύμμετροι πάντως καὶ δυνάμει [ εἰσὶ σύμμετροι ], αἱ δὲ δυνάμει οὐ πάντως καὶ μήκει, ἀλλὰ δὴ δύνανται μήκει καὶ σύμμετροι εἶναι καὶ ἀσύμμετροι, φανερόν, ὅτι, ἐὰν τῇ ἐκκειμένῃ ῥητῇ σύμμετρός τις ᾖ μήκει, λέγεται ῥητὴ καὶ σύμμετρος αὐτῇ οὐ μόνον μήκει, ἀλλὰ καὶ δυνάμει, ἐπεὶ αἱ μήκει σύμμετροι πάντως καὶ δυνάμει. ἐὰν δὲ τῇ ἐκκειμένῃ ῥητῇ σύμμετρός τις ᾖ δυνάμει, εἰ μὲν καὶ μήκει, λέγεται καὶ οὕτως ῥητὴ καὶ σύμμετρος αὐτῇ μήκει καὶ δυνάμει: εἰ δὲ τῇ ἐκκειμένῃ πάλιν ῥητῇ σύμμετρός τις οὖσα δυνάμει μήκει αὐτῇ ᾖ ἀσύμμετρος, λέγεται καὶ οὕτως ῥητὴ δυνάμει μόνον σύμμετρος.

If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are incommensurable, the square on the greater will be greater than the square on the less by the square on a straight line incommensurable with the greater. And, if the square on the greater be greater than the square on the less by the square on a straight line incommensurable with the greater, and if there be applied to the greater a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, it divides it into parts which are incommensurable. Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma before X. 17] and let BD be incommensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD. It is to be proved that BC is incommensurable in length with DF. Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16] But DC is commensurable with the sum of BF, DC; [X. 6] therefore BC is also incommensurable with the sum of BF, DC; [X. 13] so that BC is also incommensurable in length with the remainder FD. [X. 16] And the square on BC is greater than the square on A by the square on FD; therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC. Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC, and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure. Let this be the rectangle BD, DC. It is to be proved that BD is incommensurable in length with DC. For, with the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD. But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC; therefore BC is incommensurable in length with FD. so that BC is also incommensurable with the remainder, the sum of BF, DC. [X. 16] But the sum of BF, DC is commensurable in length with DC; [X. 6] therefore BC is also incommensurable in length with DC, [X. 13] so that, separando, BD is also incommensurable in length with DC. [X. 16] Therefore etc. LEMMA. [Since it has been proved that straight lines commensurable in length are always commensurable in square also, while those commensurable in square are not always commensurable in length also, but can of course be either commensurable or incommensurable in length, it is manifest that, if any straight line be commensurable in length with a given rational straight line, it is called rational and commensurable with the other not only in length but in square also, since straight lines commensurable in length are always commensurable in square also.