10. Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure.
10. σύμμετρα μεγέθη λέγεται τὰ τῷ αὐτῷ μέτρῳ μετρούμενα, ἀσύμμετρα δέ, ὧν μηδὲν ἐνδέχεται κοινὸν μέτρον γενέσθαι.
Straight lines are commensurable in square when the squares on them are measured by the same area, and incommensurable in square when the squares on them cannot possibly have any area as a common measure.
εὐθεῖαι δυνάμει σύμμετροί εἰσιν, ὅταν τὰ ἀπʼ αὐτῶν τετράγωνα τῷ αὐτῷ χωρίῳ μετρῆται, ἀσύμμετροι δέ, ὅταν τοῖς ἀπʼ αὐτῶν τετραγώνοις μηδὲν ἐνδέχηται χωρίον κοινὸν μέτρον γενέσθαι.
With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable respectively, some in length only, and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square or in square only, rational, but those which are incommensurable with it irrational.
τούτων ὑποκειμένων δείκνυται, ὅτι τῇ προτεθείσῃ εὐθείᾳ ὑπάρχουσιν εὐθεῖαι πλήθει ἄπειροι σύμμετροί τε καὶ ἀσύμμετροι αἱ μὲν μήκει μόνον, αἱ δὲ καὶ δυνάμει. καλείσθω οὖν ἡ μὲν προτεθεῖσα εὐθεῖα ῥητή, καὶ αἱ ταύτῃ σύμμετροι εἴτε μήκει καὶ δυνάμει εἴτε δυνάμει μόνον ῥηταί, αἱ δὲ ταύτῃ ἀσύμμετροι ἄλογοι καλείσθωσαν.
And let the square on the assigned straight line be called rational and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them.
καὶ τὸ μὲν ἀπὸ τῆς προτεθείσης εὐθείας τετράγωνον ῥητόν, καὶ τὰ τούτῳ σύμμετρα ῥητά, τὰ δὲ τούτῳ ἀσύμμετρα ἄλογα καλείσθω, καὶ αἱ δυνάμεναι αὐτὰ ἄλογοι, εἰ μὲν τετράγωνα εἴη, αὐταὶ αἱ πλευραί, εἰ δὲ ἕτερά τινα εὐθύγραμμα, αἱ ἴσα αὐτοῖς τετράγωνα ἀναγράφουσαι. δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους.
Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the lesser magnitude set out.
ἔστω δύο μεγέθη ἄνισα τὰ ΑΒ, Γ, ὧν μεῖζον τὸ ΑΒ· λέγω, ὅτι, ἐὰν ἀπὸ τοῦ ΑΒ ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ Γ μεγέθους.
Let AB, C be two unequal magnitudes of which AB is the greater: I say that, if from AB there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C. For C if multiplied will sometime be greater than AB. [cf. ]
τὸ Γ γὰρ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ ΑΒ μεῖζον. πεπολλαπλασιάσθω, καὶ ἔστω τὸ ΔΕ τοῦ μὲν Γ πολλαπλάσιον, τοῦ δὲ ΑΒ μεῖζον, καὶ διῃρήσθω τὸ ΔΕ εἰς τὰ τῷ Γ ἴσα τὰ ΔΖ, ΖΗ, ΗΕ, καὶ ἀφῃρήσθω ἀπὸ μὲν τοῦ ΑΒ μεῖζον ἢ τὸ ἥμισυ τὸ ΒΘ, ἀπὸ δὲ τοῦ ΑΘ μεῖζον ἢ τὸ ἥμισυ τὸ ΘΚ, καὶ τοῦτο ἀεὶ γιγνέσθω, ἕως ἂν αἱ ἐν τῷ ΑΒ διαιρέσεις ἰσοπληθεῖς γένωνται ταῖς ἐν τῷ ΔΕ διαιρέσεσιν.
Let it be multiplied, and let DE be a multiple of C, and greater than. AB; let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE. Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, GE.
ἔστωσαν οὖν αἱ ΑΚ, ΚΘ, ΘΒ διαιρέσεις ἰσοπληθεῖς οὖσαι ταῖς ΔΖ, ΖΗ, ΗΕ· καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΔΕ τοῦ ΑΒ, καὶ ἀφῄρηται ἀπὸ μὲν τοῦ ΔΕ ἔλασσον τοῦ ἡμίσεος τὸ ΕΗ, ἀπὸ δὲ τοῦ ΑΒ μεῖζον ἢ τὸ ἥμισυ τὸ ΒΘ, λοιπὸν ἄρα τὸ ΗΔ λοιποῦ τοῦ ΘΑ μεῖζόν ἐστιν.
Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.
καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΗΔ τοῦ ΘΑ, καὶ ἀφῄρηται τοῦ μὲν ΗΔ ἥμισυ τὸ ΗΖ, τοῦ δὲ ΘΑ μεῖζον ἢ τὸ ἥμισυ τὸ ΘΚ, λοιπὸν ἄρα τὸ ΔΖ λοιποῦ τοῦ ΑΚ μεῖζόν ἐστιν. ἴσον δὲ τὸ ΔΖ τῷ Γ·
And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK.
καὶ τὸ Γ ἄρα τοῦ ΑΚ μεῖζόν ἐστιν. ἔλασσον ἄρα τὸ ΑΚ τοῦ Γ. καταλείπεται ἄρα ἀπὸ τοῦ ΑΒ μεγέθους τὸ ΑΚ μέγεθος ἔλασσον ὂν τοῦ ἐκκειμένου ἐλάσσονος μεγέθους τοῦ Γ·
But DF is equal to C; therefore C is also greater than AK. Therefore AK is less than C.
ὅπερ ἔδει δεῖξαι. ¯ὁμοίως δὲ δειχθήσεται, κἂν ἡμίση ᾖ τὰ ἀφαιρούμενα.
Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C. Q. E. D. And the theorem can be similarly proved even if the parts subtracted be halves.
ἐὰν δύο μεγεθῶν [ἐκκειμένων] ἀνίσων ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος τὸ καταλειπόμενον μηδέποτε καταμετρῇ τὸ πρὸ ἑαυτοῦ, ἀσύμμετρα ἔσται τὰ μεγέθη.
If, when the less of two unequal magnitudes is continually subtracted in turn from the greater, that which is left never measures the one before it, the magnitudes will be incommensurable.
δύο γὰρ μεγεθῶν ὄντων ἀνίσων τῶν ΑΒ, ΓΔ καὶ ἐλάσσονος τοῦ ΑΒ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος τὸ περιλειπόμενον μηδέποτε καταμετρείτω τὸ πρὸ ἑαυτοῦ·
For, there being two unequal magnitudes AB, CD, and AB being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it; I say that the magnitudes AB, CD are incommensurable.
λέγω, ὅτι ἀσύμμετρά ἐστι τὰ ΑΒ, ΓΔ μεγέθη. εἰ γάρ ἐστι σύμμετρα, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Ε·
For, if they are commensurable, some magnitude will measure them.
καὶ τὸ μὲν ΑΒ τὸ ΖΔ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΓΖ, τὸ δὲ ΓΖ τὸ ΒΗ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΑΗ, καὶ τοῦτο ἀεὶ γινέσθω, ἕως οὗ λειφθῇ τι μέγεθος, ὅ ἐστιν ἔλασσον τοῦ Ε.
Let a magnitude measure them, if possible, and let it be E; let AB, measuring FD, leave CF less than itself, let CF measuring BG, leave AG less than itself, and let this process be repeated continually, until there is left some magnitude which is less than E.
γεγονέτω, καὶ λελείφθω τὸ ΑΗ ἔλασσον τοῦ Ε. ἐπεὶ οὖν τὸ Ε τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΔΖ μετρεῖ, καὶ τὸ Ε ἄρα τὸ ΖΔ μετρήσει. μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ·
Suppose this done, and let there be left AG less than E.
καὶ λοιπὸν ἄρα τὸ ΓΖ μετρήσει.
Then, since E measures AB, while AB measures DF, therefore E will also measure FD.
ἀλλὰ τὸ ΓΖ τὸ ΒΗ μετρεῖ· καὶ τὸ Ε ἄρα τὸ ΒΗ μετρεῖ. μετρεῖ δὲ καὶ ὅλον τὸ ΑΒ·
But it measures the whole CD also; therefore it will also measure the remainder CF.
καὶ λοιπὸν ἄρα τὸ ΑΗ μετρήσει, τὸ μεῖζον τὸ ἔλασσον. ὅπερ ἐστὶν ἀδύνατον.
But CF measures BG; therefore E also measures BG.
οὐκ ἄρα τὰ ΑΒ, ΓΔ μεγέθη μετρήσει τι μέγεθος·
But it measures the whole AB also; therefore it will also measure the remainder AG, the greater the less: which is impossible.
ἀσύμμετρα ἄρα ἐστὶ τὰ ΑΒ, ΓΔ μεγέθη. ἐὰν ἄρα δύο μεγεθῶν ἀνίσων, καὶ τὰ ἑξῆς.
Therefore no magnitude will measure the magnitudes AB, CD; therefore the magnitudes AB, CD are incommensurable. Therefore etc. []
δύο μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν. ἔστω τὰ δοθέντα δύο μεγέθη σύμμετρα τὰ ΑΒ, ΓΔ, ὧν ἔλασσον τὸ ΑΒ·
Given two commensurable magnitudes, to find their greatest common measure.
δεῖ δὴ τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν.
Let the two given commensurable magnitudes be AB, CD of which AB is the less; thus it is required to find the greatest common measure of AB, CD.
τὸ ΑΒ γὰρ μέγεθος ἤτοι μετρεῖ τὸ ΓΔ ἢ οὔ. εἰ μὲν οὖν μετρεῖ, μετρεῖ δὲ καὶ ἑαυτό, τὸ ΑΒ ἄρα τῶν ΑΒ, ΓΔ κοινὸν μέτρον ἐστίν·
Now the magnitude AB either measures CD or it does not.
καὶ φανερόν, ὅτι καὶ μέγιστον.
If then it measures it—and it measures itself also—AB is a common measure of AB, CD.
μεῖζον γὰρ τοῦ ΑΒ μεγέθους τὸ ΑΒ οὐ μετρήσει. μὴ μετρείτω δὴ τὸ ΑΒ τὸ ΓΔ.
And it is manifest that it is also the greatest; for a greater magnitude than the magnitude AB will not measure AB. Next, let AB not measure CD.
καὶ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, τὸ περιλειπόμενον μετρήσει ποτὲ τὸ πρὸ ἑαυτοῦ διὰ τὸ μὴ εἶναι ἀσύμμετρα τὰ ΑΒ, ΓΔ·
Then, if the less be continually subtracted in turn from the greater, that which is left over will sometime measure the one before it, because AB, CD are not incommensurable; [cf. ] let AB, measuring ED, leave EC less than itself, let EC, measuring FB, leave AF less than itself, and let AF measure CE.
καὶ τὸ μὲν ΑΒ τὸ ΕΔ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΕΓ, τὸ δὲ ΕΓ τὸ ΖΒ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΑΖ, τὸ δὲ ΑΖ τὸ ΓΕ μετρείτω. ἐπεὶ οὖν τὸ ΑΖ τὸ ΓΕ μετρεῖ, ἀλλὰ τὸ ΓΕ τὸ ΖΒ μετρεῖ, καὶ τὸ ΑΖ ἄρα τὸ ΖΒ μετρήσει.
Since, then, AF measures CE, while CE measures FB, therefore AF will also measure FB.
μετρεῖ δὲ καὶ ἑαυτό· καὶ ὅλον ἄρα τὸ ΑΒ μετρήσει τὸ ΑΖ. ἀλλὰ τὸ ΑΒ τὸ ΔΕ μετρεῖ·
But it measures itself also; therefore AF will also measure the whole AB.
καὶ τὸ ΑΖ ἄρα τὸ ΕΔ μετρήσει. μετρεῖ δὲ καὶ τὸ ΓΕ·
But AB measures DE; therefore AF will also measure ED.
καὶ ὅλον ἄρα τὸ ΓΔ μετρεῖ·
But it measures CE also; therefore it also measures the whole CD.
τὸ ΑΖ ἄρα τῶν ΑΒ, ΓΔ κοινὸν μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον.
Therefore AF is a common measure of AB, CD. I say next that it is also the greatest.
εἰ γὰρ μή, ἔσται τι μέγεθος μεῖζον τοῦ ΑΖ, ὃ μετρήσει τὰ ΑΒ, ΓΔ.
For, if not, there will be some magnitude greater than AF which will measure AB, CD. Let it be G.
ἔστω τὸ Η. ἐπεὶ οὖν τὸ Η τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΕΔ μετρεῖ, καὶ τὸ Η ἄρα τὸ ΕΔ μετρήσει.
Since then G measures AB, while AB measures ED, therefore G will also measure ED.
μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ· καὶ λοιπὸν ἄρα τὸ ΓΕ μετρήσει τὸ Η.
But it measures the whole CD also; therefore G will also measure the remainder CE.
ἀλλὰ τὸ ΓΕ τὸ ΖΒ μετρεῖ· καὶ τὸ Η ἄρα τὸ ΖΒ μετρήσει.
But CE measures FB; therefore G will also measure FB.
μετρεῖ δὲ καὶ ὅλον τὸ ΑΒ, καὶ λοιπὸν τὸ ΑΖ μετρήσει, τὸ μεῖζον τὸ ἔλασσον·
But it measures the whole AB also, and it will therefore measure the remainder AF, the greater the less: which is impossible.
ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα μεῖζόν τι μέγεθος τοῦ ΑΖ τὰ ΑΒ, ΓΔ μετρήσει·
Therefore no magnitude greater than AF will measure AB, CD; therefore AF is the greatest common measure of AB, CD.
τὸ ΑΖ ἄρα τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν μέτρον ἐστίν. δύο ἄρα μεγεθῶν συμμέτρων δοθέντων τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν μέτρον ηὕρηται·
Therefore the greatest common measure of the two given commensurable magnitudes AB, CD has been found. Q. E. D.
ὅπερ ἔδει δεῖξαι.
Porism. From this it is manifest that, if a magnitude measure two magnitudes, it will also measure their greatest common measure.
Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν μέγεθος δύο μεγέθη μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει. τριῶν μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν.
Given three commensurable magnitudes, to find their greatest common measure.
ἔστω τὰ δοθέντα τρία μεγέθη σύμμετρα τὰ Α, Β, Γ·
Let A, B, C be the three given commensurable magnitudes; thus it is required to find the greatest common measure of A, B, C.
δεῖ δὴ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν.
Let the greatest common measure of the two magnitudes A, B be taken, and let it be D; [] then D either measures C, or does not measure it. First, let it measure it.
εἰλήφθω γὰρ δύο τῶν Α, Β τὸ μέγιστον κοινὸν μέτρον, καὶ ἔστω τὸ Δ· τὸ δὴ Δ τὸ Γ ἤτοι μετρεῖ ἢ οὔ [μετρεῖ]. μετρείτω πρότερον.
Since then D measures C, while it also measures A, B, therefore D is a common measure of A, B, C.
ἐπεὶ οὖν τὸ Δ τὸ Γ μετρεῖ, μετρεῖ δὲ καὶ τὰ Α, Β, τὸ Δ ἄρα τὰ Α, Β, Γ μετρεῖ· τὸ Δ ἄρα τῶν Α, Β, Γ κοινὸν μέτρον ἐστίν.
And it is manifest that it is also the greatest; for a greater magnitude than the magnitude D does not measure A, B. Next, let D not measure C.
καὶ φανερόν, ὅτι καὶ μέγιστον· μεῖζον γὰρ τοῦ Δ μεγέθους τὰ Α, Β οὐ μετρεῖ. μὴ μετρείτω δὴ τὸ Δ τὸ Γ.
I say first that C, D are commensurable.
λέγω πρῶτον, ὅτι σύμμετρά ἐστι τὰ Γ, Δ.
For, since A, B, C are commensurable, some magnitude will measure them, and this will of course measure A, B also; so that it will also measure the greatest common measure of A, B, namely D. []
ἐπεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, Γ, μετρήσει τι αὐτὰ μέγεθος, ὃ δηλαδὴ καὶ τὰ Α, Β μετρήσει· ὥστε καὶ τὸ τῶν Α, Β μέγιστον κοινὸν μέτρον τὸ Δ μετρήσει. μετρεῖ δὲ καὶ τὸ Γ·
But it also measures C; so that the said magnitude will measure C, D; therefore C, D are commensurable.
ὥστε τὸ εἰρημένον μέγεθος μετρήσει τὰ Γ, Δ· σύμμετρα ἄρα ἐστὶ τὰ Γ, Δ.
Now let their greatest common measure be taken, and let it be E. []
εἰλήφθω οὖν αὐτῶν τὸ μέγιστον κοινὸν μέτρον, καὶ ἔστω τὸ Ε.
Since then E measures D, while D measures A, B, therefore E will also measure A, B.
ἐπεὶ οὖν τὸ Ε τὸ Δ μετρεῖ, ἀλλὰ τὸ Δ τὰ Α, Β μετρεῖ, καὶ τὸ Ε ἄρα τὰ Α, Β μετρήσει.
But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C.
μετρεῖ δὲ καὶ τὸ Γ. τὸ Ε ἄρα τὰ Α, Β, Γ μετρεῖ·
I say next that it is also the greatest.
τὸ Ε ἄρα τῶν Α, Β, Γ κοινόν ἐστι μέτρον.
For, if possible, let there be some magnitude F greater than E, and let it measure A, B, C.
λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ δυνατόν, ἔστω τι τοῦ Ε μεῖζον μέγεθος τὸ Ζ, καὶ μετρείτω τὰ Α, Β, Γ.
Now, since F measures A, B, C, it will also measure A, B, and will measure the greatest common measure of A, B. []
καὶ ἐπεὶ τὸ Ζ τὰ Α, Β, Γ μετρεῖ, καὶ τὰ Α, Β ἄρα μετρήσει καὶ τὸ τῶν Α, Β μέγιστον κοινὸν μέτρον μετρήσει.
But the greatest common measure of A, B is D; therefore F measures D.
τὸ δὲ τῶν Α, Β μέγιστον κοινὸν μέτρον ἐστὶ τὸ Δ·
But it measures C also; therefore F measures C, D; therefore F will also measure the greatest common measure of C, D. []
τὸ Ζ ἄρα τὸ Δ μετρεῖ. μετρεῖ δὲ καὶ τὸ Γ· τὸ Ζ ἄρα τὰ Γ, Δ μετρεῖ·
But that is E; therefore F will measure E, the greater the less: which is impossible.
καὶ τὸ τῶν Γ, Δ ἄρα μέγιστον κοινὸν μέτρον μετρήσει τὸ Ζ. ἔστι δὲ τὸ ε·
Therefore no magnitude greater than the magnitude E will measure A, B, C; therefore E is the greatest common measure of A, B, C if D do not measure C, and, if it measure it, D is itself the greatest common measure. Therefore the greatest common measure of the three given commensurable magnitudes has been found.
τὸ Ζ ἄρα τὸ Ε μετρήσει, τὸ μεῖζον τὸ ἔλασσον· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα μεῖζόν τι τοῦ Ε μεγέθους [μέγεθος] τὰ Α, Β, Γ μετρεῖ· τὸ Ε ἄρα τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον ἐστίν, ἐὰν μὴ μετρῇ τὸ Δ τὸ Γ, ἐὰν δὲ μετρῇ, αὐτὸ τὸ Δ.
Porism. From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure.
τριῶν ἄρα μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον κοινὸν μέτρον ηὕρηται [ὅπερ ἔδει δεῖξαι].
Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended. Q. E. D.
Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν μέγεθος τρία μεγέθη μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει.
Commensurable magnitudes have to one another the ratio which a number has to a number.
ὁμοίως δὴ καὶ ἐπὶ πλειόνων τὸ μέγιστον κοινὸν μέτρον ληφθήσεται, καὶ τὸ πόρισμα προχωρήσει. ὅπερ ἔδει δεῖξαι.
Let A, B be commensurable magnitudes; I say that A has to B the ratio which a number has to a number.
τὰ σύμμετρα μεγέθη πρὸς ἄλληλα λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἔστω σύμμετρα μεγέθη τὰ Α, Β·
For, since A, B are commensurable, some magnitude will measure them. Let it measure them, and let it be C.
λέγω, ὅτι τὸ Α πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
And, as many times as C measures A, so many units let there be in D; and, as many times as C measures B, so many units let there be in E.
ἐπεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, μετρήσει τι αὐτὰ μέγεθος. μετρείτω, καὶ ἔστω τὸ Γ.
Since then C measures A according to the units in D, while the unit also measures D according to the units in it, therefore the unit measures the number D the same number of times as the magnitude C measures A; therefore, as C is to A, so is the unit to D; [] therefore, inversely, as A is to C, so is D to the unit. [cf. ]
καὶ ὁσάκις τὸ Γ τὸ Α μετρεῖ τοσαῦται μονάδες ἔστωσαν ἐν τῷ Δ, ὁσάκις δὲ τὸ Γ τὸ Β μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν τὸ Γ τὸ Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας, μετρεῖ δὲ καὶ ἡ μονὰς τὸν Δ κατὰ τὰς ἐν αὐτῷ μονάδας, ἰσάκις ἄρα ἡ μονὰς τὸν Δ μετρεῖ ἀριθμὸν καὶ τὸ Γ μέγεθος τὸ Α·
Again, since C measures B according to the units in E, while the unit also measures E according to the units in it, therefore the unit measures E the same number of times as C measures B; therefore, as C is to B, so is the unit to E.
ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ· ἀνάπαλιν ἄρα, ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὴν μονάδα. πάλιν ἐπεὶ τὸ Γ τὸ Β μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, μετρεῖ δὲ καὶ ἡ μονὰς τὸν Ε κατὰ τὰς ἐν αὐτῷ μονάδας, ἰσάκις ἄρα ἡ μονὰς τὸν Ε μετρεῖ καὶ τὸ Γ τὸ Β·
But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to B, so is the number D to E. []
ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Β, οὕτως ἡ μονὰς πρὸς τὸν Ε.
Therefore the commensurable magnitudes A, B have to one another the ratio which the number D has to the number E. Q. E. D.
ἐδείχθη δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, ὁ Δ πρὸς τὴν μονάδα·
If two magnitudes have to one another the ratio which a number has to a number, the magnitudes will be commensurable.
διʼ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Δ ἀριθμὸς πρὸς τὸν Ε. τὰ ἄρα σύμμετρα μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον ἔχει, ὃν ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε·
For let the two magnitudes A, B have to one another the ratio which the number D has to the number E; I say that the magnitudes A, B are commensurable.
ὅπερ ἔδει δεῖξαι. ἐὰν δύο μεγέθη πρὸς ἄλληλα λόγον ἔχῃ, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρα ἔσται τὰ μεγέθη.
For let A be divided into as many equal parts as there are units in D, and let C be equal to one of them; and let F be made up of as many magnitudes equal to C as there are units in E.
δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον ἐχέτω, ὃν ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε· λέγω, ὅτι σύμμετρά ἐστι τὰ Α, Β μεγέθη.
Since then there are in A as many magnitudes equal to C as there are units in D, whatever part the unit is of D, the same part is C of A also; therefore, as C is to A, so is the unit to D. [] But the unit measures the number D; therefore C also measures A.
ὅσαι γάρ εἰσιν ἐν τῷ Δ μονάδες, εἰς τοσαῦτα ἴσα διῃρήσθω τὸ Α, καὶ ἑνὶ αὐτῶν ἴσον ἔστω τὸ Γ· ὅσαι δέ εἰσιν ἐν τῷ Ε μονάδες, ἐκ τοσούτων μεγεθῶν ἴσων τῷ Γ συγκείσθω τὸ Ζ.
And since, as C is to A, so is the unit to D, therefore, inversely, as A is to C, so is the number D to the unit. [cf. ]
ἐπεὶ οὖν, ὅσαι εἰσὶν ἐν τῷ Δ μονάδες, τοσαῦτά εἰσι καὶ ἐν τῷ Α μεγέθη ἴσα τῷ Γ, ὃ ἄρα μέρος ἐστὶν ἡ μονὰς τοῦ Δ, τὸ αὐτὸ μέρος ἐστὶ καὶ τὸ Γ τοῦ Α·
Again, since there are in F as many magnitudes equal to C as there are units in E, therefore, as C is to F, so is the unit to E. []
ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ. μετρεῖ δὲ ἡ μονὰς τὸν Δ ἀριθμόν· μετρεῖ ἄρα καὶ τὸ Γ τὸ Α.
But it was also proved that, as A is to C, so is D to the unit; therefore, ex aequali, as A is to F, so is D to E. [] But, as D is to E, so is A to B; therefore also, as A is to B, so is it to F also. []
καὶ ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ [ἀριθμόν], ἀνάπαλιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ ἀριθμὸς πρὸς τὴν μονάδα.
Therefore A has the same ratio to each of the magnitudes B, F; therefore B is equal to F. [] But C measures F; therefore it measures B also.
πάλιν ἐπεί, ὅσαι εἰσὶν ἐν τῷ Ε μονάδες, τοσαῦτά εἰσι καὶ ἐν τῷ Ζ ἴσα τῷ Γ, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ, οὕτως ἡ μονὰς πρὸς τὸν Ε [ἀριθμόν].
Further it measures A also; therefore C measures A, B. Therefore A is commensurable with B.
ἐδείχθη δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὴν μονάδα·
Therefore etc.
διʼ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Ζ, οὕτως ὁ Δ πρὸς τὸν Ε.
Porism. From this it is manifest that, if there be two numbers, as D, E, and a straight line, as A, it is possible to make a straight line [F] such that the given straight line is to it as the number D is to the number E.
ἀλλʼ ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ἐστὶ τὸ Α πρὸς τὸ Β· καὶ ὡς ἄρα τὸ Α πρὸς τὸ Β, οὕτως καὶ πρὸς τὸ Ζ. τὸ Α ἄρα πρὸς ἑκάτερον τῶν Β, Ζ τὸν αὐτὸν ἔχει λόγον·
And, if a mean proportional be also taken between A, F, as B,
ἴσον ἄρα ἐστὶ τὸ Β τῷ Ζ. μετρεῖ δὲ τὸ Γ τὸ Ζ·
as A is to F, so will the square on A be to the square on B, that is, as the first is to the third, so is the figure on the first to that which is similar and similarly described on the second. []
μετρεῖ ἄρα καὶ τὸ Β. ἀλλὰ μὴν καὶ τὸ Α· τὸ Γ ἄρα τὰ Α, Β μετρεῖ. σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Β. ἐὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς.
But, as A is to F, so is the number D to the number E; therefore it has been contrived that, as the number D is to the number E, so also is the figure on the straight line A to the figure on the straight line B. Q. E. D. Incommensurable magnitudes have not to one another the ratio which a number has to a number.
Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν ὦσι δύο ἀριθμοί, ὡς οἱ Δ, Ε, καὶ εὐθεῖα, ὡς ἡ Α, δύνατόν ἐστι ποιῆσαι ὡς ὁ Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν, οὕτως τὴν εὐθεῖαν πρὸς εὐθεῖαν.
Let A, B be incommensurable magnitudes; I say that A has not to B the ratio which a number has to a number.
ἐὰν δὲ καὶ τῶν Α, Ζ μέση ἀνάλογον ληφθῇ, ὡς ἡ Β, ἔσται ὡς ἡ Α πρὸς τὴν Ζ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Β, τουτέστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον.
For, if A has to B the ratio which a number has to a number, A will be commensurable with B. [] But it is not; therefore A has not to B the ratio which a number has to a number.
ἀλλʼ ὡς ἡ Α πρὸς τὴν Ζ, οὕτως ἐστὶν ὁ Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν·
Therefore etc.
γέγονεν ἄρα καὶ ὡς ὁ Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν, οὕτως τὸ ἀπὸ τῆς Α εὐθείας πρὸς τὸ ἀπὸ τῆς Β εὐθείας·
If two magnitudes have not to one another the ratio which a number has to a number, the magnitudes will be incommensurable.
ὅπερ ἔδει δεῖξαι.
For let the two magnitudes A, B not have to one another the ratio which a number has to a number; I say that the magnitudes A, B are incommensurable.
τὰ ἀσύμμετρα μεγέθη πρὸς ἄλληλα λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἔστω ἀσύμμετρα μεγέθη τὰ Α, Β· λέγω, ὅτι τὸ Α πρὸς τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
For, if they are commensurable, A will have to B the ratio which a number has to a number. []
εἰ γὰρ ἔχει τὸ Α πρὸς τὸ Β λόγον, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρον ἔσται τὸ Α τῷ Β.
But it has not; therefore the magnitudes A, B are incommensurable.
οὐκ ἔστι δέ·
Therefore etc.
οὐκ ἄρα τὸ Α πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
The squares on straight lines commensurable in length have to one another the ratio which a square number has to a square number; and squares which have to one another the ratio which a square number has to a square number will also have their sides commensurable in length. But the squares on straight lines incommensurable in length have not to one another the ratio which a square number has to a square number; and squares which have not to one another the ratio which a square number has to a square number will not have their sides commensurable in length either.
τὰ ἄρα ἀσύμμετρα μεγέθη πρὸς ἄλληλα λόγον οὐκ ἔχει, καὶ τὰ ἑξῆς. ἐὰν δύο μεγέθη πρὸς ἄλληλα λόγον μὴ ἔχῃ, ὃν ἀριθμὸς πρὸς ἀριθμόν, ἀσύμμετρα ἔσται τὰ μεγέθη. δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον μὴ ἐχέτω, ὃν ἀριθμὸς πρὸς ἀριθμόν· λέγω, ὅτι ἀσύμμετρά ἐστι τὰ Α, Β μεγέθη. εἰ γὰρ ἔσται σύμμετρα, τὸ Α πρὸς τὸ Β λόγον ἕξει, ὃν ἀριθμὸς πρὸς ἀριθμόν. οὐκ ἔχει δέ. ἀσύμμετρα ἄρα ἐστὶ τὰ Α, Β μεγέθη. ἐὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς.
For let A, B be commensurable in length; I say that the square on A has to the square on B the ratio which a square number has to a square number.
τὰ ἀπὸ τῶν μήκει συμμέτρων εὐθειῶν τετράγωνα πρὸς ἄλληλα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
For, since A is commensurable in length with B, therefore A has to B the ratio which a number has to a number. [] Let it have to it the ratio which C has to D.
καὶ τὰ τετράγωνα τὰ πρὸς ἄλληλα λόγον ἔχοντα, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ τὰς πλευρὰς ἕξει μήκει συμμέτρους.
Since then, as A is to B, so is C to D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, for similar figures are in the duplicate ratio of their corresponding sides; [] and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, for between two square numbers there is one mean proportional number, and the square number has to the square number the ratio duplicate of that which the side has to the side; [] therefore also, as the square on A is to the square on B, so is the square on C to the square on D.
τὰ δὲ ἀπὸ τῶν μήκει ἀσυμμέτρων εὐθειῶν τετράγωνα πρὸς ἄλληλα λόγον οὐκ ἔχει, ὅνπερ τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· καὶ τὰ τετράγωνα τὰ πρὸς ἄλληλα λόγον μὴ ἔχοντα, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, οὐδὲ τὰς πλευρὰς ἕξει μήκει συμμέτρους. ἔστωσαν γὰρ αἱ Α, Β μήκει σύμμετροι· λέγω, ὅτι τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. ἐπεὶ γὰρ σύμμετρός ἐστιν ἡ Α τῇ Β μήκει, ἡ Α ἄρα πρὸς τὴν Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
Next, as the square on A is to the square on B, so let the square on C be to the square on D; I say that A is commensurable in length with B.
ἐχέτω, ὃν ὁ Γ πρὸς τὸν Δ.
For since, as the square on A is to the square on B, so is the square on C to the square on D, while the ratio of the square on A to the square on B is duplicate of the ratio of A to B, and the ratio of the square on C to the square on D is duplicate of the ratio of C to D, therefore also, as A is to B, so is C to D. Therefore A has to B the ratio which the number C has to the number D; therefore A is commensurable in length with B. []
ἐπεὶ οὖν ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ Γ πρὸς τὸν Δ, ἀλλὰ τοῦ μὲν τῆς Α πρὸς τὴν Β λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τῆς Α τετραγώνου πρὸς τὸ ἀπὸ τῆς Β τετράγωνον· τὰ γὰρ ὅμοια σχήματα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν· τοῦ δὲ τοῦ Γ [ἀριθμοῦ] πρὸς τὸν Δ [ἀριθμὸν] λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τοῦ Γ τετραγώνου πρὸς τὸν ἀπὸ τοῦ Δ τετράγωνον·
Next, let A be incommensurable in length with B; I say that the square on A has not to the square on B the ratio which a square number has to a square number.
δύο γὰρ τετραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καὶ ὁ τετράγωνος πρὸς τὸν τετράγωνον [ἀριθμὸν] διπλασίονα λόγον ἔχει, ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν·
For, if the square on A has to the square on B the ratio which a square number has to a square number, A will be commensurable with B.
ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος [ἀριθμὸς] πρὸς τὸν ἀπὸ τοῦ Δ [ἀριθμοῦ] τετράγωνον [ἀριθμόν].
But it is not; therefore the square on A has not to the square on B the ratio which a square number has to a square number.
ἀλλὰ δὴ ἔστω ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος πρὸς τὸν ἀπὸ τοῦ Δ [τετράγωνον]·
Again, let the square on A not have to the square on B the ratio which a square number has to a square number; I say that A is incommensurable in length with B. For, if A is commensurable with B, the square on A will have to the square on B the ratio which a square number has to a square number. But it has not; therefore A is not commensurable in length with B. Therefore etc.
λέγω, ὅτι σύμμετρός ἐστιν ἡ Α τῇ Β μήκει.
Porism. And it is manifest from what has been proved that straight lines commensurable in length are always commensurable in square also, but those commensurable in square are not always commensurable in length also.
ἐπεὶ γάρ ἐστιν ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον], οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος πρὸς τὸν ἀπὸ τοῦ Δ [τετράγωνον], ἀλλʼ ὁ μὲν τοῦ ἀπὸ τῆς Α τετραγώνου πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] λόγος διπλασίων ἐστὶ τοῦ τῆς Α πρὸς τὴν Β λόγου, ὁ δὲ τοῦ ἀπὸ τοῦ Γ [ἀριθμοῦ] τετραγώνου [ἀριθμοῦ] πρὸς τὸν ἀπὸ τοῦ Δ [ἀριθμοῦ] τετράγωνον [ἀριθμὸν] λόγος διπλασίων ἐστὶ τοῦ τοῦ Γ [ἀριθμοῦ] πρὸς τὸν Δ [ἀριθμὸν] λόγου, ἔστιν ἄρα καὶ ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ Γ [ἀριθμὸς] πρὸς τὸν Δ [ἀριθμόν].
[Lemma. It has been proved in the arithmetical books that similar plane numbers have to one another the ratio which a square number has to a square number, [] and that, if two numbers have to one another the ratio which a square number has to a square number, they are similar plane numbers. [Converse of ]
ἡ Α ἄρα πρὸς τὴν Β, λόγον ἔχει, ὃν ἀριθμὸς ὁ Γ πρὸς ἀριθμὸν τὸν Δ· σύμμετρος ἄρα ἐστὶν ἡ Α τῇ Β μήκει. ἀλλὰ δὴ ἀσύμμετρος ἔστω ἡ Α τῇ Β μήκει·
And it is manifest from these propositions that numbers which are not similar plane numbers, that is, those which have not their sides proportional, have not to one another the ratio which a square number has to a square number.
λέγω, ὅτι τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. εἰ γὰρ ἔχει τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, σύμμετρος ἔσται ἡ Α τῇ Β.
For, if they have, they will be similar plane numbers: which is contrary to the hypothesis.
οὐκ ἔστι δέ·
Therefore numbers which are not similar plane numbers have not to one another the ratio which a square number has to a square number.]
οὐκ ἄρα τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
To find two straight lines incommensurable, the one in length only, and the other in square also, with an assigned straight line.
πάλιν δὴ τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν·
Let A be the assigned straight line; thus it is required to find two straight lines incommensurable, the one in length only, and the other in square also, with A.
λέγω, ὅτι ἀσύμμετρός ἐστιν ἡ Α τῇ Β μήκει.
Let two numbers B, C be set out which have not 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 contrived that, as B is to C, so is the square on A to the square on D —for we have learnt how to do this— [] therefore the square on A is commensurable with the square on D. []
εἰ γάρ ἐστι σύμμετρος ἡ Α τῇ Β, ἕξει τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Β λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. οὐκ ἔχει δέ· οὐκ ἄρα σύμμετρός ἐστιν ἡ Α τῇ Β μήκει. τὰ ἄρα ἀπὸ τῶν μήκει συμμέτρων, καὶ τὰ ἑξῆς.
And, since B has not to C the ratio which a square number has to a square number, therefore neither has the square on A to the square on D the ratio which a square number has to a square number; therefore A is incommensurable in length with D. []
Πόρισμα καὶ φανερὸν ἐκ τῶν δεδειγμένων ἔσται, ὅτι αἱ μήκει σύμμετροι πάντως καὶ δυνάμει, αἱ δὲ δυνάμει οὐ πάντως καὶ μήκει [εἴπερ τὰ ἀπὸ τῶν μήκει συμμέτρων εὐθειῶν τετράγωνα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, τὰ δὲ λόγον ἔχοντα, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρά ἐστιν.
Let E be taken a mean proportional between A, D; therefore, as A is to D, so is the square on A to the square on E. [] But A is incommensurable in length with D; therefore the square on A is also incommensurable with the square on E; [] therefore A is incommensurable in square with E.
ὥστε αἱ μήκει σύμμετροι εὐθεῖαι οὐ μόνον [εἰσὶ] μήκει σύμμετροι, ἀλλὰ καὶ δυνάμει.
Therefore two straight lines D, E have been found incommensurable, D in length only, and E in square and of course in length also, with the assigned straight line A.]
πάλιν ἐπεί, ὅσα τετράγωνα πρὸς ἄλληλα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, μήκει ἐδείχθη σύμμετρα καὶ δυνάμει ὄντα σύμμετρα τῷ τὰ τετράγωνα λόγον ἔχειν, ὃν ἀριθμὸς πρὸς ἀριθμόν, ὅσα ἄρα τετράγωνα λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ἀλλὰ ἁπλῶς, ὃν ἀριθμὸς πρὸς ἀριθμόν, σύμμετρα μὲν ἔσται αὐτὰ τὰ τετράγωνα δυνάμει, οὐκέτι δὲ καὶ μήκει·
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.
ὥστε τὰ μὲν μήκει σύμμετρα πάντως καὶ δυνάμει, τὰ δὲ δυνάμει οὐ πάντως καὶ μήκει, εἰ μὴ καὶ λόγον ἔχοιεν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
Let A, B, C, D be four magnitudes in proportion, so that, as A is to B, so is C to D, and let A be commensurable with B; I say that C will also be commensurable with D. For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. []
λέγω δή, ὅτι [καὶ] αἱ μήκει ἀσύμμετροι οὐ πάντως καὶ δυνάμει, ἐπειδήπερ αἱ δυνάμει σύμμετροι δύνανται λόγον μὴ ἔχειν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ διὰ τοῦτο δυνάμει οὖσαι σύμμετροι μήκει εἰσὶν ἀσύμμετροι.
And, as A is to B, so is C to D; therefore C also has to D the ratio which a number has to a number; therefore C is commensurable with D. []
ὥστε οὐχ αἱ τῷ μήκει ἀσύμμετροι πάντως καὶ δυνάμει, ἀλλὰ δύνανται μήκει οὖσαι ἀσύμμετροι δυνάμει εἶναι καὶ ἀσύμμετροι καὶ σύμμετροι.
Next, let A be incommensurable with B; I say that C will also be incommensurable with D.
αἱ δὲ δυνάμει ἀσύμμετροι πάντως καὶ μήκει ἀσύμμετροι·
For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. []
εἰ γὰρ [εἰσι] μήκει σύμμετροι, ἔσονται καὶ δυνάμει σύμμετροι. ὑπόκεινται δὲ καὶ ἀσύμμετροι·
And, as A is to B, so is C to D; therefore neither has C to D the ratio which a number has to a number; therefore C is incommensurable with D. [] Therefore etc. Magnitudes commensurable with the same magnitude are commensurable with one another also.
ὅπερ ἄτοπον. αἱ ἄρα δυνάμει ἀσύμμετροι πάντως καὶ μήκει].
For let each of the magnitudes A, B be commensurable with C; I say that A is also commensurable with B. For, since A is commensurable with C, therefore A has to C the ratio which a number has to a number. [] Let it have the ratio which D has to E.
λῆμμα δέδεικται ἐν τοῖς ἀριθμητικοῖς, ὅτι οἱ ὅμοιοι ἐπίπεδοι ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, καὶ ὅτι, ἐὰν δύο ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχωσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν, ὅμοιοί εἰσιν ἐπίπεδοι.
Again, since C is commensurable with B, therefore C has to B the ratio which a number has to a number. [] Let it have the ratio which F has to G.
καὶ δῆλον ἐκ τούτων, ὅτι οἱ μὴ ὅμοιοι ἐπίπεδοι ἀριθμοί, τουτέστιν οἱ μὴ ἀνάλογον ἔχοντες τὰς πλευράς, πρὸς ἀλλήλους λόγον οὐκ ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
And, given any number of ratios we please, namely the ratio which D has to E and that which F has to G, let the numbers H, K, L be taken continuously in the given ratios; [cf. ] so that, as D is to E, so is H to K, and, as F is to G, so is K to L.
εἰ γὰρ ἕξουσιν, ὅμοιοι ἐπίπεδοι ἔσονται· ὅπερ οὐχ ὑπόκειται. οἱ ἄρα μὴ ὅμοιοι ἐπίπεδοι πρὸς ἀλλήλους λόγον οὐκ ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
Since, then, as A is to C, so is D to E, while, as D is to E, so is H to K, therefore also, as A is to C, so is H to K. []
τῇ προτεθείσῃ εὐθείᾳ προσευρεῖν δύο εὐθείας ἀσυμμέτρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ δυνάμει. ἔστω ἡ προτεθεῖσα εὐθεῖα ἡ Α·
Again, since, as C is to B, so is F to G, while, as F is to G, so is K to L, therefore also, as C is to B, so is K to L. [] But also, as A is to C, so is H to K; therefore, ex aequali, as A is to B, so is H to L. []
δεῖ δὴ τῇ Α προσευρεῖν δύο εὐθείας ἀσυμμέτρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ δυνάμει.
Therefore A has to B the ratio which a number has to a number; therefore A is commensurable with B. [] Therefore etc. Q. E. D.
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·
If two magnitudes be commensurable, and the one of them be incommensurable with any magnitude, the remaining one will also be incommensurable with the same.
ἐμάθομεν γάρ· σύμμετρον ἄρα τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς Δ.
Let A, B be two commensurable magnitudes, and let one of them, A, be incommensurable with any other magnitude C; I say that the remaining one, B, will also be incommensurable with C.
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·
For, if B is commensurable with C, while A is also commensurable with B, A is also commensurable with C. []
ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ Δ μήκει. εἰλήφθω τῶν Α, Δ μέση ἀνάλογον ἡ Ε·
But it is also incommensurable with it: which is impossible.
ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Δ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Ε.
Therefore B is not commensurable with C; therefore it is incommensurable with it.
ἀσύμμετρος δέ ἐστιν ἡ Α τῇ Δ μήκει·
Therefore etc.
ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ τῆς Α τετράγωνον τῷ ἀπὸ τῆς Ε τετραγώνῳ·
Lemma. Given two unequal straight lines, to find by what square the square on the greater is greater than the square on the less.
ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ Ε δυνάμει.
Let AB, C be the given two unequal straight lines, and let AB be the greater of them; thus it is required to find by what square the square on AB is greater than the square on C.
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 [ὅπερ ἔδει δεῖξαι].
Let the semicircle ADB be described on AB, and let AD be fitted into it equal to C; [] let DB be joined.
ἐὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ πρῶτον τῷ δευτέρῳ σύμμετρον ᾖ, καὶ τὸ τρίτον τῷ τετάρτῳ σύμμετρον ἔσται·
It is then manifest that the angle ADB is right, [] and that the square on AB is greater than the square on AD, that is, C, by the square on DB. []
κἂν τὸ πρῶτον τῷ δευτέρῳ ἀσύμμετρον ᾖ, καὶ τὸ τρίτον τῷ τετάρτῳ ἀσύμμετρον ἔσται.
Similarly also, if two straight lines be given, the straight line the square on which is equal to the sum of the squares on them is found in this manner.
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 AD, DB be the given two straight lines, and let it be required to find the straight line the square on which is equal to the sum of the squares on them.
ἐπεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Β, τὸ Α ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ·
Let them be placed so as to contain a right angle, that formed by AD, DB; and let AB be joined.
καὶ τὸ Γ ἄρα πρὸς τὸ Δ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν· σύμμετρον ἄρα ἐστὶ τὸ Γ τῷ Δ.
It is again manifest that the straight line the square on which is equal to the sum of the squares on AD, DB is AB. [] Q. E. D.
ἀλλὰ δὴ τὸ Α τῷ Β ἀσύμμετρον ἔστω· λέγω, ὅτι καὶ τὸ Γ τῷ Δ ἀσύμμετρον ἔσται.
If four straight lines be proportional, and the square on the first be greater than the square on the second by the square on a straight line commensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line commensurable with the third.
ἐπεὶ γὰρ ἀσύμμετρόν ἐστι τὸ Α τῷ Β, τὸ Α ἄρα πρὸς τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. 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, if the square on the first be greater than the square on the second by the square on a straight line incommensurable with the first, the square on the third will also be greater than the square on the fourth by the square on a straight line in- commensurable with the third.
τὰ τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ σύμμετρα. ἑκάτερον γὰρ τῶν Α, Β τῷ Γ ἔστω σύμμετρον. λέγω, ὅτι καὶ τὸ Α τῷ Β ἐστι σύμμετρον. ἐπεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Γ, τὸ Α ἄρα πρὸς τὸ Γ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν.
Let A, B, C, D be four straight lines in proportion, so that, as A is to B, so is C to D; and let the square on A be greater than the square on B by the square on E, and let the square on C be greater than the square on D by the square on F; I say that, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F.
ἐχέτω, ὃν ὁ Δ πρὸς τὸν Ε. πάλιν, ἐπεὶ σύμμετρόν ἐστι τὸ Γ τῷ Β, τὸ Γ ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ Ζ πρὸς τὸν Η. καὶ λόγων δοθέντων ὁποσωνοῦν τοῦ τε, ὃν ἔχει ὁ Δ πρὸς τὸν Ε, καὶ ὁ Ζ πρὸς τὸν Η εἰλήφθωσαν ἀριθμοὶ ἑξῆς ἐν τοῖς δοθεῖσι λόγοις οἱ Θ, Κ, Λ·
For since, as A is to B, so is C to D, therefore also, as the square on A is to the square on B, so is the square on C to the square on D. []
ὥστε εἶναι ὡς μὲν τὸν Δ πρὸς τὸν Ε, οὕτως τὸν Θ πρὸς τὸν Κ, ὡς δὲ τὸν Ζ πρὸς τὸν Η, οὕτως τὸν Κ πρὸς τὸν Λ.
But the squares on E, B are equal to the square on A, and the squares on D, F are equal to the square on C.
ἐπεὶ οὖν ἐστιν ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὸν Ε, ἀλλʼ ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Θ πρὸς τὸν Κ, ἔστιν ἄρα καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ.
Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D; therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D; [] therefore also, as E is to B, so is F to D; [] therefore, inversely, as B is to E, so is D to F.
πάλιν, ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Β, οὕτως ὁ Ζ πρὸς τὸν Η, ἀλλʼ ὡς ὁ Ζ πρὸς τὸν Η, [οὕτως] ὁ Κ πρὸς τὸν Λ, καὶ ὡς ἄρα τὸ Γ πρὸς τὸ Β, οὕτως ὁ Κ πρὸς τὸν Λ. ἔστι δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ·
But, as A is to B, so also is C to D; therefore, ex aequali, as A is to E, so is C to F. []
διʼ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Θ πρὸς τὸν Λ. τὸ Α ἄρα πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς ὁ Θ πρὸς ἀριθμὸν τὸν Λ·
Therefore, if A is commensurable with E, C is also commensurable with F, and, if A is incommensurable with E, C is also incommensurable with F. []
σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Β. τὰ ἄρα τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ σύμμετρα·
Therefore etc.
ὅπερ ἔδει δεῖξαι.
If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable.
ἐὰν ᾖ δύο μεγέθη σύμμετρα, τὸ δὲ ἕτερον αὐτῶν μεγέθει τινὶ ἀσύμμετρον ᾖ, καὶ τὸ λοιπὸν τῷ αὐτῷ ἀσύμμετρον ἔσται. ἔστω δύο μεγέθη σύμμετρα τὰ Α, Β, τὸ δὲ ἕτερον αὐτῶν τὸ Α ἄλλῳ τινὶ τῷ Γ ἀσύμμετρον ἔστω·
For let the two commensurable magnitudes AB, BC be added together; I say that the whole AC is also commensurable with each of the magnitudes AB, BC.
λέγω, ὅτι καὶ τὸ λοιπὸν τὸ Β τῷ Γ ἀσύμμετρόν ἐστιν.
For, since AB, BC are commensurable, some magnitude will measure them.
εἰ γάρ ἐστι σύμμετρον τὸ Β τῷ Γ, ἀλλὰ καὶ τὸ Α τῷ Β σύμμετρόν ἐστιν, καὶ τὸ Α ἄρα τῷ Γ σύμμετρόν ἐστιν.
Let it measure them, and let it be D.
ἀλλὰ καὶ ἀσύμμετρον·
Since then D measures AB, BC, it will also measure the whole AC.
ὅπερ ἀδύνατον. οὐκ ἄρα σύμμετρόν ἐστι τὸ Β τῷ Γ·
But it measures AB, BC also; therefore D measures AB, BC, AC; therefore AC is commensurable with each of the magnitudes AB, BC. [] Next, let AC be commensurable with AB; I say that AB, BC are also commensurable.
ἀσύμμετρον ἄρα. ἐὰν ἄρα ᾖ δύο μεγέθη σύμμετρα, καὶ τὰ ἑξῆς. λῆμμα δύο δοθεισῶν εὐθειῶν ἀνίσων εὑρεῖν, τίνι μεῖζον δύναται ἡ μείζων τῆς ἐλάσσονος.
For, since AC, AB are commensurable, some magnitude will measure them.
ἔστωσαν αἱ δοθεῖσαι δύο ἄνισοι εὐθεῖαι αἱ ΑΒ, Γ, ὧν μείζων ἔστω ἡ ΑΒ·
Let it measure them, and let it be D.
δεῖ δὴ εὑρεῖν, τίνι μεῖζον δύναται ἡ ΑΒ τῆς Γ.
Since then D measures CA, AB, it will also measure the remainder BC. But it measures AB also; therefore D will measure AB, BC; therefore AB, BC are commensurable. []
γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ εἰς αὐτὸ ἐνηρμόσθω τῇ Γ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΒ.
Therefore etc.
φανερὸν δή, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΑΔΒ γωνία, καὶ ὅτι ἡ ΑΒ τῆς ΑΔ, τουτέστι τῆς Γ, μεῖζον δύναται τῇ ΔΒ.
If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.
ὁμοίως δὲ καὶ δύο δοθεισῶν εὐθειῶν ἡ δυναμένη αὐτὰς εὑρίσκεται οὕτως. ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ, καὶ δέον ἔστω εὑρεῖν τὴν δυναμένην αὐτάς.
For let the two incommensurable magnitudes AB, BC be added together; I say that the whole AC is also incommensurable with each of the magnitudes AB, BC.
κείσθωσαν γάρ, ὥστε ὀρθὴν γωνίαν περιέχειν τὴν ὑπὸ ΑΔ, ΔΒ, καὶ ἐπεζεύχθω ἡ ΑΒ·
For, if CA, AB are not incommensurable, some magnitude will measure them. Let it measure them, if possible, and let it be D.
φανερὸν πάλιν, ὅτι ἡ τὰς ΑΔ, ΔΒ δυναμένη ἐστὶν ἡ ΑΒ· ὅπερ ἔδει δεῖξαι.
Since then D measures CA, AB, therefore it will also measure the remainder BC. But it measures AB also; therefore D measures AB, BC. Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.
ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, δύνηται δὲ ἡ πρώτη τῆς δευτέρας μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ [μήκει], καὶ ἡ τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ συμμέτρου ἑαυτῇ [μήκει].
Therefore no magnitude will measure CA, AB; therefore CA, AB are incommensurable. [] Similarly we can prove that AC, CB are also incommensurable.
καὶ ἐὰν ἡ πρώτη τῆς δευτέρας μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ [μήκει], καὶ ἡ τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ [μήκει].
Therefore AC is incommensurable with each of the magnitudes AB, BC. Next, let AC be incommensurable with one of the magnitudes AB, BC. First, let it be incommensurable with AB; I say that AB, BC are also incommensurable.
ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, Δ, ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, καὶ ἡ Α μὲν τῆς β μεῖζον δυνάσθω τῷ ἀπὸ τῆς Ε, ἡ δὲ Γ τῆς Δ μεῖζον δυνάσθω τῷ ἀπὸ τῆς Ζ·
For, if they are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC. therefore it will also measure the whole AC.
λέγω, ὅτι, εἴτε σύμμετρός ἐστιν ἡ Α τῇ Ε, σύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός ἐστιν ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ.
But it measures AB also; therefore D measures CA, AB. Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.
ἐπεὶ γάρ ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Β, οὕτως τὸ ἀπὸ τῆς Γ πρὸς τὸ ἀπὸ τῆς Δ.
Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable. [] Therefore etc.
ἀλλὰ τῷ μὲν ἀπὸ τῆς Α ἴσα ἐστὶ τὰ ἀπὸ τῶν Ε, Β, τῷ δὲ ἀπὸ τῆς Γ ἴσα ἐστὶ τὰ ἀπὸ τῶν Δ, Ζ.
Lemma. If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application.
ἔστιν ἄρα ὡς τὰ ἀπὸ τῶν Ε, Β πρὸς τὸ ἀπὸ τῆς Β, οὕτως τὰ ἀπὸ τῶν Δ, Ζ πρὸς τὸ ἀπὸ τῆς Δ· διελόντι ἄρα ἐστὶν ὡς τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς Β, οὕτως τὸ ἀπὸ τῆς Ζ πρὸς τὸ ἀπὸ τῆς Δ· ἔστιν ἄρα καὶ ὡς ἡ Ε πρὸς τὴν Β, οὕτως ἡ Ζ πρὸς τὴν Δ·
For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB; I say that AD is equal to the rectangle contained by AC, CB.
ἀνάπαλιν ἄρα ἐστὶν ὡς ἡ Β πρὸς τὴν Ε, οὕτως ἡ Δ πρὸς τὴν Ζ. ἔστι δὲ καὶ ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ·
This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle AC, CD, that is, the rectangle AC, CB.
διʼ ἴσου ἄρα ἐστὶν ὡς ἡ Α πρὸς τὴν Ε, οὕτως ἡ Γ πρὸς τὴν Ζ.
Therefore etc.
εἴτε οὖν σύμμετρός ἐστιν ἡ Α τῇ Ε, σύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός ἐστιν ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ.
If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and deficient by a square figure, and if it divide it into parts which are commensurable in length, then the square on the greater will be greater than the square on the less by the square on a straight line commensurable with the greater.
ἐὰν ἄρα, καὶ τὰ ἑξῆς. ἐὰν δύο μεγέθη σύμμετρα συντεθῇ, καὶ τὸ ὅλον ἑκατέρῳ αὐτῶν σύμμετρον ἔσται· κἂν τὸ ὅλον ἑνὶ αὐτῶν σύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη σύμμετρα ἔσται. Συγκείσθω γὰρ δύο μεγέθη σύμμετρα τὰ ΑΒ, ΒΓ· λέγω, ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἐστι σύμμετρον.