Page 335 - Euclid's Elements of Geometry

ST EW iþ.

Θ ∆ H D ELEMENTS BOOK 10

Ε Ζ Η E F G

Α Γ Β A C B

᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν Let the two numbers AC and CB be laid down such
συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον that their sum AB has to BC the ratio which (some)
ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, square number (has) to (some) square number, and does
πρὸς δὲ τὸν ΑΓ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς not have to AC the ratio which (some) square number
πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω ῥητὴ ἡ Δ, καὶ τῇ Δ (has) to (some) square number [Prop. 10.28 lem. I]. And
σύμμετρος ἔστω ἡ ΕΖ μήκει· ῥητὴ ἄρα ἐστὶν ἡ ΕΖ. γεγονέτω let the rational (straight-line) D be laid down. And let
δὴ καὶ ὡς ὁ ΓΑ ἀριθμὸς πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς ΕΖ EF be commensurable in length with D. EF is thus a
πρὸς τὸ ἀπὸ τῆς ΖΗ· σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΖ τῷ rational (straight-line). So, let it also have been contrived
ἀπὸ τῆς ΖΗ. ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΓΑ ἀριθμὸς that as the number CA (is) to AB, so the (square) on EF
πρὸς τὸν ΑΒ λὸγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the
τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς (square) on EF is commensurable with the (square) on
ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον FG [Prop. 10.6]. Thus, FG is also a rational (straight-
ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ μήκει· αἱ ΕΖ, line). And since the number CA does not have to AB
ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα the ratio which (some) square number (has) to (some)
ὀνομάτων ἐστὶν ἡ ΕΗ. δεικτέον δή, ὅτι καὶ δευτέρα. square number, the (square) on EF does not have to the
᾿Επεὶ γὰρ ἀνάπαλίν ἐστιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, (square) on FG the ratio which (some) square number
οὕτως τὸ ἀπὸ τῆς ΗΖ πρὸς τὸ ἀπὸ τῆς ΖΕ, μείζων δὲ ὁ ΒΑ (has) to (some) square number either. Thus, EF is in-
τοῦ ΑΓ, μεῖζον ἄρα [καὶ] τὸ ἀπὸ τῆς ΗΖ τοῦ ἀπὸ τῆς ΖΕ. commensurable in length with FG [Prop. 10.9]. EF and
ἔστω τῷ ἀπὸ τῆς ΗΖ ἴσα τὰ ἀπὸ τῶν ΕΖ, Θ· ἀναστρέψαντι FG are thus rational (straight-lines which are) commen-
ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ surable in square only. Thus, EG is a binomial (straight-
πρὸς τὸ ἀπὸ τῆς Θ. ἀλλ᾿ ὁ ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν line) [Prop. 10.36]. So, we must show that (it is) also a
τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· καὶ τὸ ἀπὸ second (binomial straight-line).
τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος For since, inversely, as the number BA is to AC, so
ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. σύμμετρος ἄρα ἐστὶν ἡ the (square) on GF (is) to the (square) on FE [Prop. 5.7
ΖΗ τῇ Θ μήκει· ὥστε ἡ ΖΗ τῆς ΖΕ μεῖζον δύναται τῷ ἀπὸ corr.], and BA (is) greater than AC, the (square) on
συμμέτρου ἑαυτῇ. καί εἰσι ῥηταὶ αἱ ΖΗ, ΖΕ δυνάμει μόνον GF (is) thus [also] greater than the (square) on FE
σύμμετροι, καὶ τὸ ΕΖ ἔλασσον ὄνομα τῇ ἐκκειμένῃ ῥητῇ [Prop. 5.14]. Let (the sum of) the (squares) on EF and
σύμμετρόν ἐστι τῇ Δ μήκει. H be equal to the (square) on GF. Thus, via conver-
῾Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ δευτέρα· ὅπερ ἔδει sion, as AB is to BC, so the (square) on FG (is) to the
δεῖξαι. (square) on H [Prop. 5.19 corr.]. But, AB has to BC
the ratio which (some) square number (has) to (some)
square number. Thus, the (square) on FG also has to
the (square) on H the ratio which (some) square number
(has) to (some) square number. Thus, FG is commensu-
rable in length with H [Prop. 10.9]. Hence, the square on
FG is greater than (the square on) FE by the (square) on
(some straight-line) commensurable in length with (FG).
And FG and FE are rational (straight-lines which are)
commensurable in square only. And the lesser term EF
is commensurable in length with the rational (straight-
line) D (previously) laid down.
Thus, EG is a second binomial (straight-line) [Def.
†
10.6]. (Which is) the very thing it was required to show.
√
† If the rational straight-line has unit length then the length of a second binomial straight-line is k/ 1 − k ′ 2 + k. This, and the second apotome,

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