§10.prop2.50εὑρεῖν τὴν ἐκ δύο ὀνομάτων τρίτην.
To find the third 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.
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.
ἐκκείσθω δέ τις καὶ ἄλλος μὴ τετράγωνος ἀριθμὸς ὁ Δ, καὶ πρὸς ἑκάτερον τῶν ΒΑ, ΑΓ λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
And let another number D, which is not square, also be set out, and let it not have to each of BA, AC the ratio which a square number has to a square number.
καὶ ἐκκείσθω τις ῥητὴ εὐθεῖα ἡ Ε, καὶ γεγονέτω ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ·
And let a rational straight line E be set out, and let it be made that, as D is to AB, so is the square on E to the square on ZH.
σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς Ε τῷ ἀπὸ τῆς ΖΗ. καί ἐστι ῥητὴ ἡ Ε·
Therefore the square on E is commensurable with the square on ZH.
ῥητὴ ἄρα ἐστὶ καὶ ἡ ΖΗ. καὶ ἐπεὶ ὁ Δ πρὸς τὸν ΑΒ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
And E is rational; therefore ZH is also rational. And since D has not to AB the ratio which a square number has to a square number, therefore the square on E also has not to the square on ZH the ratio which a square number has to a square number.
ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ ΖΗ μήκει.
Therefore E is incommensurable in length with ZH.
γεγονέτω δὴ πάλιν ὡς ὁ ΒΑ ἀριθμὸς πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ·
Let it then again be made that, as the number BA is to AC, so is the square on ZH to the square on HT.
σύμμετρον ἄρα ἐστὶ τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΗΘ. ῥητὴ δὲ ἡ ΖΗ·
Therefore the square on ZH is commensurable with the square on HT.
ῥητὴ ἄρα καὶ ἡ ΗΘ. καὶ ἐπεὶ ὁ ΒΑ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὅν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΘΗ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
And ZH is rational; therefore HT is also rational. And since the number BA has not to AC the ratio which a square number has to a square number, neither has the square on ZH to the square on TH the ratio which a square number has to a square number.
ἀσύμμετρος ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΘ μήκει.
Therefore ZH is incommensurable in length with HT.
αἱ ΖΗ, ΗΘ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἡ ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστίν.
Therefore ZH, HT are rational straight lines commensurable in square only; therefore ZT is a binomial straight line.
λέγω δή, ὅτι καὶ τρίτη.
I say then that it is also a third binomial.
ἐπεὶ γάρ ἐστιν ὡς ὁ Δ πρὸς τὸν ΑΒ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΖΗ, ὡς δὲ ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, διʼ ἴσου ἄρα ἐστὶν ὡς ὁ Δ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς ΗΘ. ὁ δὲ Δ πρὸς τὸν ΑΓ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· οὐδὲ τὸ ἀπὸ τῆς Ε ἄρα πρὸς τὸ ἀπὸ τῆς ΗΘ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
For since as D is to AB, so is the square on E to the square on ZH, and as BA is to AC, so is the square on ZH to the square on HT, therefore, ex aequali, as D is to AC, so is the square on E to the square on HT. But D has not to AC the ratio which a square number has to a square number; therefore neither has the square on E to the square on HT the ratio which a square number has to a square number.
ἀσύμμετρος ἄρα ἐστὶν ἡ Ε τῇ ΗΘ μήκει.
Therefore E is incommensurable in length with HT.
καὶ ἐπεί ἐστιν ὡς ὁ ΒΑ πρὸς τὸν ΑΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς ΗΘ, μεῖζον ἄρα τὸ ἀπὸ τῆς ΖΗ τοῦ ἀπὸ τῆς ΗΘ. ἔστω οὖν τῷ ἀπὸ τῆς ΖΗ ἴσα τὰ ἀπὸ τῶν ΗΘ, Κ· ἀναστρέψαντι ἄρα ὡς ὁ ΑΒ πρὸς τὸν ΒΓ, οὕτως τὸ ἀπὸ τῆς ΖΗ πρὸς τὸ ἀπὸ τῆς Κ. ὁ δὲ ΑΒ πρὸς τὸν ΒΓ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
And since as BA is to AC, so is the square on ZH to the square on HT, therefore the square on ZH is greater than the square on HT. Let then the squares on HT, K be equal to the square on ZH; therefore, by conversion, as AB is to BC, so is the square on ZH to the square on K.
καὶ τὸ ἀπὸ τῆς ΖΗ ἄρα πρὸς τὸ ἀπὸ τῆς Κ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· σύμμετρος ἄρα ἡ ΖΗ τῇ Κ μήκει.
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 K the ratio which a square number has to a square number; therefore ZH is commensurable in length with K.
ἡ ΖΗ ἄρα τῆς ΗΘ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ.
Therefore ZH is greater in square than HT by the square on a straight line commensurable in length with itself.
καί εἰσιν αἱ ΖΗ, ΗΘ ῥηταὶ δυνάμει μόνον σύμμετροι, καὶ οὐδετέρα αὐτῶν σύμμετρός ἐστι τῇ Ε μήκει.
And ZH, HT are rational straight lines commensurable in square only, and neither of them is commensurable in length with E.
ἡ ΖΘ ἄρα ἐκ δύο ὀνομάτων ἐστὶ τρίτη· ὅπερ ἔδει δεῖξαι.
Therefore ZT is a third binomial straight line; which was to be proved.