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ST	EW      iþ.






                                                                                           ELEMENTS BOOK 10



            τετράγωνον ἀριθμόν· καὶ τὸ ἀπὸ τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ have to AC the ratio which (some) square number (has)
            τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον to (some) square number.  Thus, the (square) on E
            ἀριθμόν· σύμμετρος ἄρα [ἐστὶν] ἡ ΖΗ τῇ Κ μήκει. ἡ ΖΗ ἄρα does not have to the (square) on GH the ratio which
            τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. καί εἰσιν (some) square number (has) to (some) square number
            αἱ ΖΗ, ΗΘ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα either. Thus, E is incommensurable in length with GH
            αὐτῶν σύμμετρός ἐστι τῇ Ε μήκει.                    [Prop. 10.9]. And since as BA is to AC, so the (square)
               ῾Η ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη· ὅπερ ἔδει δεῖξαι. on FG (is) to the (square) on GH, the (square) on FG
                                                                (is) thus greater than the (square) on GH [Prop. 5.14].
                                                                Therefore, let (the sum of) the (squares) on GH and
                                                                K be equal to the (square) on FG. Thus, via conver-
                                                                sion, as AB [is] to BC, so the (square) on FG (is)
                                                                to the (square) on K [Prop. 5.19 corr.]. And 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 K the ratio which (some) square
                                                                number (has) to (some) square number. Thus, FG [is]
                                                                commensurable in length with K [Prop. 10.9]. Thus,
                                                                the square on FG is greater than (the square on) GH
                                                                by the (square) on (some straight-line) commensurable
                                   naþ                          and neither of them is commensurable in length with E.
                                                                (in length) with (FG). And FG and GH are rational
                                                                (straight-lines which are) commensurable in square only,
                                                                   Thus, FH is a third binomial (straight-line) [Def.
                                                                     †
                                                                10.7]. (Which is) the very thing it was required to show.
            †  If the rational straight-line has unit length then the length of a third binomial straight-line is k  1/2 (1 +  √ 1 − k ). This, and the third apotome,
                                                                                          ′ 2
                               √
                                                             2
                                    ′ 2
            whose length is k 1/2  (1 −  1 − k ) [Prop. 10.87], are the roots of x − 2 k  1/2  x + k k ′ 2  = 0.
                                                                                 Proposition 51
                                      .
               Εὑρεῖν τὴν ἐκ δύο ὀνομάτων τετάρτην.                To find a fourth 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
            πρὸς τὸν ΒΓ λόγον μὴ ἔχειν μήτε μὴν πρὸς τὸν ΑΓ, such that AB does not have to BC, or to AC either,
            ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.  καὶ the ratio which (some) square number (has) to (some)
            ἐκκείσθω ῥητὴ ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω μήκει ἡ ΕΖ· square number [Prop. 10.28 lem. I]. And let the rational
            ῥητὴ ἄρα ἐστὶ καὶ ἡ ΕΖ. καὶ γεγονέτω ὡς ὁ ΒΑ ἀριθμὸς (straight-line) D be laid down. And let EF be com-
            πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ· mensurable in length with D. Thus, EF is also a ratio-
            σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΖΗ· ῥητὴ ἄρα nal (straight-line). And let it have been contrived that
            ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, as the number BA (is) to AC, so the (square) on EF
            ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ  (is) to the (square) on FG [Prop. 10.6 corr.]. Thus, the
            ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος  (square) on EF is commensurable with the (square) on
            ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ  FG [Prop. 10.6]. Thus, FG is also a rational (straight-
            ΕΖ τῇ ΖΗ μήκει. αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον line). And since BA does not have to AC the ratio which
            σύμμετροι· ὥστε ἡ ΕΗ ἐκ δύο ὀνομάτων ἐστίν. λέγω δή, (some) square number (has) to (some) square number,

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