Page 334 - Euclid's Elements of Geometry

ST EW iþ.

ELEMENTS BOOK 10

ΖΗ. καὶ ἐστι ῥητὴ ἡ ΕΖ· ῥητὴ ἄρα καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ber. Thus, the (square) on EF also has to the (square)
ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς on FG the ratio which (some) number (has) to (some)
πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς number. Hence, the (square) on EF is commensurable
τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς with the (square) on FG [Prop. 10.6]. And EF is ra-
τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ tional. Thus, FG (is) also rational. And since BA does
μήκει. αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· not have to AC the ratio which (some) square number
ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ. λέγω, ὅτι καὶ πρώτη. (has) to (some) square number, thus the (square) on EF
does not have to the (square) on FG the ratio which
(some) square number (has) to (some) square number
either. Thus, EF is incommensurable in length with FG
[Prop 10.9]. EF and FG are thus rational (straight-lines
which are) commensurable in square only. Thus, EG is
a binomial (straight-line) [Prop. 10.36]. I say that (it is)
also a first (binomial straight-line).
∆ Θ D H

Ε Ζ Η E F G

Α Γ Β A C B

᾿Επεὶ γάρ ἐστιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ For since as the number BA is to AC, so the (square)
ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, μείζων δὲ ὁ ΒΑ τοῦ ΑΓ, on EF (is) to the (square) on FG, and BA (is) greater
μεῖζον ἄρα καὶ τὸ ἀπὸ τῆς ΕΖ τοῦ ἀπὸ τῆς ΖΗ. ἔστω οὖν τῷ than AC, the (square) on EF (is) thus also greater than
ἀπὸ τῆς ΕΖ ἴσα τὰ ἀπὸ τῶν ΖΗ, Θ. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ the (square) on FG [Prop. 5.14]. Therefore, let (the sum
πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, of) the (squares) on FG and H be equal to the (square)
ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ on EF. And since as BA is to AC, so the (square)
ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον on EF (is) to the (square) on FG, thus, via conver-
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καὶ sion, as AB is to BC, so the (square) on EF (is) to the
τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν (square) on H [Prop. 5.19 corr.]. And AB has to BC
τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. σύμμετρος the ratio which (some) square number (has) to (some)
ἄρα ἐστὶν ἡ ΕΖ τῇ Θ μήκει· ἡ ΕΖ ἄρα τῆς ΖΗ μεῖζον δύναται square number. Thus, the (square) on EF also has to
τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί εἰσι ῥηταὶ αἱ ΕΖ, ΖΗ, καὶ the (square) on H the ratio which (some) square number
σύμμετρος ἡ ΕΖ τῇ Δ μήκει. (has) to (some) square number. Thus, EF is commensu-
᾿Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πρώτη· ὅπερ ἔδει rable in length with H [Prop. 10.9]. Thus, the square on
δεῖξαι. EF is greater than (the square on) FG by the (square)
on (some straight-line) commensurable (in length) with
(EF). And EF and FG are rational (straight-lines). And †
mjþ
EF (is) commensurable in length with D.
Thus, EG is a first binomial (straight-line) [Def. 10.5].
(Which is) the very thing it was required to show.
√
′ 2
† If the rational straight-line has unit length then the length of a first binomial straight-line is k + k 1 − k . This, and the first apotome, whose
√
2
2
length is k − k 1 − k ′ 2 [Prop. 10.85], are the roots of x − 2 k x + k k ′ 2 = 0.
Proposition 49
.
Εὑρεῖν τὴν ἐκ δύο ὀνομάτων δευτέραν. To find a second binomial (straight-line).

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