Page 338 - Euclid's Elements of Geometry

ST EW iþ.

ELEMENTS BOOK 10

ὅτι καὶ τετάρτη. the (square) on EF does not have to the (square) on FG
᾿Επεὶ γάρ ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ the ratio which (some) square number (has) to (some)
τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ [μείζων δὲ ὁ ΒΑ τοῦ ΑΓ], square number either. Thus, EF is incommensurable
μεῖζον ἄρα τὸ ἀπὸ τῆς ΕΖ τοῦ ἀπὸ τῆς ΖΗ. ἔστω οὖν τῷ in length with FG [Prop. 10.9]. Thus, EF and FG
ἀπὸ τῆς ΕΖ ἴσα τὰ ἀπὸ τῶν ΖΗ, Θ· ἀναστρέψαντι ἄρα ὡς are rational (straight-lines which are) commensurable
ὁ ΑΒ ἀριθμὸς πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς in square only. Hence, EG is a binomial (straight-line)
τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν [Prop. 10.36]. So, I say that (it is) also a fourth (binomial
τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδ᾿ ἄρα τὸ straight-line).
ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος For since as BA is to AC, so the (square) on EF (is) to
ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν the (square) on FG [and BA (is) greater than AC], the
ἡ ΕΖ τῇ Θ μήκει· ἡ ΕΖ ἄρα τῆς ΗΖ μεῖζον δύναται τῷ ἀπὸ (square) on EF (is) thus greater than the (square) on FG
ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΕΖ, ΖΗ ῥηταὶ δυνάμει μόνον [Prop. 5.14]. Therefore, let (the sum of) the squares on
σύμμετροι, καὶ ἡ ΕΖ τῇ Δ σύμμετρός ἐστι μήκει. FG and H be equal to the (square) on EF. Thus, via con-
῾Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τετάρτη· ὅπερ ἔδει version, as the number AB (is) to BC, so the (square) on
δεῖξαι. EF (is) to the (square) on H [Prop. 5.19 corr.]. And AB
does not have to BC the ratio which (some) square num-
ber (has) to (some) square number. Thus, the (square) on
EF does not have to the (square) on H the ratio which
(some) square number (has) to (some) square number
either. Thus, EF is incommensurable in length with H
[Prop. 10.9]. Thus, the square on EF is greater than (the
square on) GF by the (square) on (some straight-line) in-
nbþ square only. And EF is commensurable in length with D.
commensurable (in length) with (EF). And EF and FG
are rational (straight-lines which are) commensurable in

Thus, EG is a fourth binomial (straight-line) [Def.
†
10.8]. (Which is) the very thing it was required to show.
√
† If the rational straight-line has unit length then the length of a fourth binomial straight-line is k (1 + 1/ 1 + k ). This, and the fourth apotome,
′
√ 2 2 ′ ′
′
whose length is k (1 − 1/ 1 + k ) [Prop. 10.88], are the roots of x − 2 k x + k k /(1 + k ) = 0.
.
Proposition 52
Εὑρεῖν τὴν ἐκ δύο ὀνομάτων πέμπτην. To find a fifth binomial straight-line.
Ε Ζ Η E F G

∆ Θ D H

Α Γ Β A C B
᾿Εκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν ΑΒ Let the two numbers AC and CB be laid down such
πρὸς ἑκάτερον αὐτῶν λόγον μὴ ἔχειν, ὃν τετράγωνος that AB does not have to either of them the ratio which
ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω ῥητή τις (some) square number (has) to (some) square number
εὐθεῖα ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω [μήκει] ἡ ΕΖ· ῥητὴ [Prop. 10.38 lem.]. And let some rational straight-line
ἄρα ἡ ΕΖ. καὶ γεγονέτω ὡς ὁ ΓΑ πρὸς τὸν ΑΒ, οὕτως τὸ D be laid down. And let EF be commensurable [in
ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ. ὁ δὲ ΓΑ πρὸς τὸν ΑΒ length] with D. Thus, EF (is) a rational (straight-
λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον line). And let it have been contrived that as CA (is) to
ἀριθμόν· οὑδὲ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον AB, so the (square) on EF (is) to the (square) on FG
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. αἱ [Prop. 10.6 corr.]. And CA does not have to AB the ra-

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