§10.prop1.10τῇ προτεθείσῃ εὐθείᾳ προσευρεῖν δύο εὐθείας ἀσυμμέτρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ δυνάμει.
To find two straight lines incommensurable with a given straight line, the one in length only, and the other in square also.
ἔστω ἡ προτεθεῖσα εὐθεῖα ἡ Α· δεῖ δὴ τῇ Α προσευρεῖν δύο εὐθείας ἀσυμμέτρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ δυνάμει.
Let the given straight line be A; it is indeed required to find two straight lines incommensurable with A, the one in length only, and the other in square also.
Ἐκκείσθωσαν γὰρ δύο ἀριθμοὶ οἱ Β, Γ πρὸς ἀλλήλους λόγον μὴ ἔχοντες, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, τουτέστι μὴ ὅμοιοι ἐπίπεδοι, καὶ γεγονέτω ὡς ὁ Β πρὸς τὸν Γ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Δ τετράγωνον· ἐμάθομεν γάρ·
For let two numbers B, Γ be set out which do not have to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers, and let it be made as B to Γ, so the square on A to the square on Δ; for we have learned [how to do this].
σύμμετρον ἄρα τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς Δ. καὶ ἐπεὶ ὁ Β πρὸς τὸν Γ λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδʼ ἄρα τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Δ λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ Δ μήκει.
Therefore the square on A is commensurable with the square on Δ. And since B does not have to Γ the ratio which a square number has to a square number, neither therefore does the square on A have to the square on Δ the ratio which a square number has to a square number; therefore A is incommensurable in length with Δ.
εἰλήφθω τῶν Α, Δ μέση ἀνάλογον ἡ Ε· ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Δ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Ε. ἀσύμμετρος δέ ἐστιν ἡ Α τῇ Δ μήκει· ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς Α τετράγωνον τῷ ἀπὸ τῆς Ε τετραγώνῳ· ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ Ε δυνάμει.
Let a mean proportional E be taken between A, Δ; therefore, as A is to Δ, so is the square on A to the square on E. But A is incommensurable in length with Δ; therefore the square on A is also incommensurable with the square on E; therefore A is incommensurable in square with E.
τῇ ἄρα προτεθείσῃ εὐθείᾳ τῇ Α προσεύρηνται δύο εὐθεῖαι ἀσύμμετροι αἱ Δ, Ε, μήκει μὲν μόνον ἡ Δ, δυνάμει δὲ καὶ μήκει δηλαδὴ ἡ Ε.
Therefore, with the given straight line A, two incommensurable straight lines Δ, E have been found, Δ indeed in length only, and E in square and of course in length too.
§10.prop1.11ἐὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ πρῶτον τῷ δευτέρῳ σύμμετρον ᾖ, καὶ τὸ τρίτον τῷ τετάρτῳ σύμμετρον ἔσται· κἂν τὸ πρῶτον τῷ δευτέρῳ ἀσύμμετρον ᾖ, καὶ τὸ τρίτον τῷ τετάρτῳ ἀσύμμετρον ἔσται.
If four magnitudes be proportional, and the first be commensurable with the second, the third will also be commensurable with the fourth; and if the first be incommensurable with the second, the third will also be incommensurable with the fourth.
ἔστωσαν τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, τὸ Α δὲ τῷ Β σύμμετρον ἔστω· λέγω, ὅτι καὶ τὸ Γ τῷ Δ σύμμετρον ἔσται.
Let four proportional magnitudes be A, B, Γ, Δ, so that as A is to B, so is Γ to Δ, and let A be commensurable with B; I say that Γ will also be commensurable with Δ.
ἐπεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Β, τὸ Α ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
For since A is commensurable with B, therefore A has to B the ratio which a number has to a number.
καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ· καὶ τὸ Γ ἄρα πρὸς τὸ Δ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν· σύμμετρον ἄρα ἐστὶ τὸ Γ τῷ Δ.
ἀλλὰ δὴ τὸ Α τῷ Β ἀσύμμετρον ἔστω· λέγω, ὅτι καὶ τὸ Γ τῷ Δ ἀσύμμετρον ἔσται.
And as A is to B, so is Γ to Δ; therefore Γ also has to Δ the ratio which a number has to a number; therefore Γ is commensurable with Δ. But now let A be incommensurable with B; I say that Γ will also be incommensurable with Δ.
ἐπεὶ γὰρ ἀσύμμετρόν ἐστι τὸ Α τῷ Β, τὸ Α ἄρα πρὸς τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
For since A is incommensurable with B, therefore A does not have to B the ratio which a number has to a number.
καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ· οὐδὲ τὸ Γ ἄρα πρὸς τὸ Δ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν· ἀσύμμετρον ἄρα ἐστὶ τὸ Γ τῷ Δ.
ἐὰν ἄρα τέσσαρα μεγέθη, καὶ τὰ ἑξῆς.
And as A is to B, so is Γ to Δ; therefore Γ also does not have to Δ the ratio which a number has to a number; therefore Γ is incommensurable with Δ. Therefore, if four magnitudes, and the rest.