§10.prop2.48εὑρεῖν τὴν ἐκ δύο ὀνομάτων πρώτην.
To find the first binomial straight line.
Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, πρὸς δὲ τὸν ΓΑ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω τις ῥητὴ ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω μήκει ἡ ΕΖ. ῥητὴ ἄρα ἐστὶ καὶ ἡ ΕΖ. καὶ γεγονέτω ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ. ὁ δὲ ΑΒ πρὸς τὸν ΑΓ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν· καὶ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν·
Let two numbers AC, CB be set out such that the number AB composed of them has to BC the ratio which a square number has to a square number, but to CA has not the ratio which a square number has to a square number; and let a rational straight line D be set out, and let EZ be commensurable in length with D. Therefore EZ is also rational. And let it be made that, as the number BA is to AC, so is the square on EZ to the square on ZH. But AB has to AC the ratio which a number has to a number; therefore the square on EZ also has to the square on ZH the ratio which a number has to a number; so that the square on EZ is commensurable with the square on ZH.
ὥστε σύμμετρόν ἐστι τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΖΗ. καί ἐστι ῥητὴ ἡ ΕΖ·
And EZ is rational; therefore ZH is also rational.
ῥητὴ ἄρα καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
And since the number BA has not to AC the ratio which a square number has to a square number, therefore the square on EZ also has not to the square on ZH the ratio which a square number has to a square number; therefore EZ is incommensurable in length with ZH.
ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ μήκει. αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ.
λέγω, ὅτι καὶ πρώτη.
Therefore EZ, ZH are rational straight lines commensurable in square only; therefore EH is a binomial straight line. I say that it is also a first binomial.
ἐπεὶ γάρ ἐστιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, μείζων δὲ ὁ ΒΑ τοῦ ΑΓ, μεῖζον ἄρα καὶ τὸ ἀπὸ τῆς ΕΖ τοῦ ἀπὸ τῆς ΖΗ. ἔστω οὖν τῷ ἀπὸ τῆς ΕΖ ἴσα τὰ ἀπὸ τῶν ΖΗ, Θ. καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ, ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς Θ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· καὶ τὸ ἀπὸ τῆς ΕΖ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
For since as the number BA is to AC, so is the square on EZ to the square on ZH, and BA is greater than AC, therefore the square on EZ is also greater than the square on ZH. Let then the squares on ZH, T be equal to the square on EZ. And since as BA is to AC, so is the square on EZ to the square on ZH, therefore, by conversion, as AB is to BC, so is the square on EZ to the square on T. But AB has to BC the ratio which a square number has to a square number; therefore the square on EZ also has to the square on T the ratio which a square number has to a square number.
σύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ Θ μήκει· ἡ ΕΖ ἄρα τῆς ΖΗ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
Therefore EZ is commensurable in length with T; therefore EZ is greater in square than ZH by the square on a straight line commensurable in length with itself.
καί εἰσι ῥηταὶ αἱ ΕΖ, ΖΗ, καὶ σύμμετρος ἡ ΕΖ τῇ Δ μήκει.
And EZ, ZH are rational straight lines, and EZ is commensurable in length with D.
ἡ ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ πρώτη· ὅπερ ἔδει δεῖξαι.
Therefore EH is a first binomial straight line; which was to be proved.
§10.prop2.49εὑρεῖν τὴν ἐκ δύο ὀνομάτων δευτέραν.
To find the second binomial straight line.
Ἐκκείσθωσαν δύο ἀριθμοὶ οἱ ΑΓ, ΓΒ, ὥστε τὸν συγκείμενον ἐξ αὐτῶν τὸν ΑΒ πρὸς μὲν τὸν ΒΓ λόγον ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, πρὸς δὲ τὸν ΑΓ λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ἐκκείσθω ῥητὴ ἡ Δ, καὶ τῇ Δ σύμμετρος ἔστω ἡ ΕΖ μήκει·
Let two numbers AC, CB be set out such that the number AB composed of them has to BC the ratio which a square number has to a square number, but to AC has not the ratio which a square number has to a square number; and let a rational straight line D be set out, and let EZ be commensurable in length with D; therefore EZ is rational.
ῥητὴ ἄρα ἐστὶν ἡ ΕΖ. γεγονέτω δὴ καὶ ὡς ὁ ΓΑ ἀριθμὸς πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ·
And let it be made that, as the number GA is to AB, so is the square on EZ to the square on ZH; therefore the square on EZ is commensurable with the square on ZH.
σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΖΗ. ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ ΓΑ ἀριθμὸς πρὸς τὸν ΑΒ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΕΖ πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
Therefore ZH is also rational. And since the number GA has not to AB the ratio which a square number has to a square number, therefore the square on EZ also has not to the square on ZH the ratio which a square number has to a square number.
ἀσύμμετρος ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ μήκει· αἱ ΕΖ, ΖΗ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἐκ δύο ἄρα ὀνομάτων ἐστὶν ἡ ΕΗ.
δεικτέον δή, ὅτι καὶ δευτέρα.
Therefore EZ is incommensurable in length with ZH; therefore EZ, ZH are rational straight lines commensurable in square only; therefore EH is a binomial straight line. It must now be shown that it is also a second binomial.
ἐπεὶ γὰρ ἀνάπαλίν ἐστιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΗΖ πρὸς τὸ ἀπὸ τῆς ΖΕ, μείζων δὲ ὁ ΒΑ τοῦ ΑΓ, μεῖζον ἄρα τὸ ἀπὸ τῆς ΗΖ τοῦ ἀπὸ τῆς ΖΕ. ἔστω τῷ ἀπὸ τῆς ΗΖ ἴσα τὰ ἀπὸ τῶν ΕΖ, Θ· ἀναστρέψαντι ἄρα ἐστὶν ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Θ. ἀλλʼ ὁ ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· καὶ τὸ ἀπὸ τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ τῆς Θ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
For since, inversely, as the number BA is to AC, so is the square on HZ to the square on ZE, and BA is greater than AC, therefore the square on HZ is also greater than the square on ZE. Let then the squares on EZ, T be equal to the square on HZ; therefore, by conversion, as AB is to BC, so is the square on ZH to the square on T. But AB has to BC the ratio which a square number has to a square number; therefore the square on ZH also has to the square on T the ratio which a square number has to a square number.
σύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ Θ μήκει· ὥστε ἡ ΖΗ τῆς ΖΕ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
Therefore ZH is commensurable in length with T; so that ZH is greater in square than ZE by the square on a straight line commensurable in length with itself.
καί εἰσι ῥηταὶ αἱ ΖΗ, ΖΕ δυνάμει μόνον σύμμετροι, καὶ τὸ ΕΖ ἔλασσον ὄνομα τῇ ἐκκειμένῃ ῥητῇ σύμμετρόν ἐστι τῇ Δ μήκει.
And ZH, ZE are rational straight lines commensurable in square only, and EZ, which is the lesser term, is commensurable in length with the set out rational straight line D.
ἡ ΕΗ ἄρα ἐκ δύο ὀνομάτων ἐστὶ δευτέρα· ὅπερ ἔδει δεῖξαι.
Therefore EH is a second binomial straight line; which was to be proved.