Page 339 - Euclid's Elements of Geometry

ST EW iþ.

ELEMENTS BOOK 10

ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο tio which (some) square number (has) to (some) square
ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ. λέγω δή, ὅτι καὶ πέμπτη. number. Thus, the (square) on EF does not have to the
᾿Επεὶ γάρ ἐστιν ὡς ὁ ΓΑ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ (square) on FG the ratio which (some) square number
τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, ἀνάπαλιν ὡς ὁ ΒΑ πρὸς τὸν (has) to (some) square number either. Thus, EF and
ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΖΕ· μεῖζον FG are rational (straight-lines which are) commensu-
ἄρα τὸ ἀπὸ τῆς ΗΖ τοῦ ἀπὸ τῆς ΖΕ. ἔστω οὖν τῷ ἀπὸ rable in square only [Prop. 10.9]. Thus, EG is a binomial
τῆς ΗΖ ἴσα τὰ ἀπὸ τῶν ΕΖ, Θ· ἀναστρέψαντι ἄρα ἐστὶν ὡς (straight-line) [Prop. 10.36]. So, I say that (it is) also a
ὁ ΑΒ ἀριθμὸς πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΗΖ πρὸς fifth (binomial straight-line).
τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον οὐκ ἔχει, ὃν For since as CA is to AB, so the (square) on EF
τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδ᾿ ἄρα τὸ (is) to the (square) on FG, inversely, as BA (is) to
ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος AC, so the (square) on FG (is) to the (square) on FE
ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἀσύμμετρος ἄρα ἐστὶν [Prop. 5.7 corr.]. Thus, the (square) on GF (is) greater
ἡ ΖΗ τῇ Θ μήκει· ὥστε ἡ ΖΗ τῆς ΖΕ μεῖζον δύναται τῷ than the (square) on FE [Prop. 5.14]. Therefore, let
ἀπὸ ἀσυμμέτρου ἑαυτῇ. καί εἰσιν αἱ ΗΖ, ΖΕ ῥηταὶ δυνάμει (the sum of) the (squares) on EF and H be equal to
μόνον σύμμετροι, καὶ τὸ ΕΖ ἔλαττον ὄνομα σύμμετρόν ἐστι the (square) on GF. Thus, via conversion, as the number
τῇ ἐκκειμένῃ ῥητῇ τῇ Δ μήκει. AB is to BC, so the (square) on GF (is) to the (square)
῾Η ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πέμπτη· ὅπερ ἔδει on H [Prop. 5.19 corr.]. And AB does not have to BC
δεῖξαι. the ratio which (some) square number (has) to (some)
square number. Thus, the (square) on FG does not have
to the (square) on H the ratio which (some) square num-
ber (has) to (some) square number either. Thus, FG is
incommensurable 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) incommensurable
(in length) with (FG). And GF and FE are rational
ngþ the rational (straight-line previously) laid down, D.
(straight-lines which are) commensurable in square only.
And the lesser term EF is commensurable in length with

†
Thus, EG is a fifth binomial (straight-line). (Which
is) the very thing it was required to show.

√
† If the rational straight-line has unit length then the length of a fifth binomial straight-line is k ( 1 + k + 1). This, and the fifth apotome, whose
′
√ 2 √ 2 ′
′
′
length is k ( 1 + k − 1) [Prop. 10.89], are the roots of x − 2 k 1 + k x + k k = 0.
Proposition 53
.
Εὑρεῖν τὴν ἐκ δύο ὀνομάτων ἕκτην. To find a sixth binomial (straight-line).
Ε Κ E K
Ζ Η Θ F G H
∆ D
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 each of them the ratio which
ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἔστω δὲ καὶ ἕτερος (some) square number (has) to (some) square number.
ἀριθμὸς ὁ Δ μὴ τετράγωνος ὢν μηδὲ πρὸς ἑκάτερον And let D also be another number, which is not square,
τῶν ΒΑ, ΑΓ λόγον ἔχων, ὃν τετράγωνος ἀριθμὸς πρὸς and does not have to each of BA and AC the ratio which
τετράγωνον ἀριθμόν· καὶ ἐκκείσθω τις ῥητὴ εὐθεῖα ἡ Ε, (some) square number (has) to (some) square number ei-
καὶ γεγονέτω ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε ther [Prop. 10.28 lem. I]. And let some rational straight-
πρὸς τὸ ἀπὸ τῆς ΖΗ· σύμμετρον ἄρα τὸ ἀπὸ τῆς Ε τῷ ἀπὸ line E be laid down. And let it have been contrived that

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