§10.prop1.40ἐὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦσαι τὸ μὲν συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον, τὸ δʼ ὑπʼ αὐτῶν ῥητόν, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, καλείσθω δὲ ῥητὸν καὶ μέσον δυναμένη.
If two straight lines incommensurable in square be added together making the sum of the squares on them medial, but the rectangle contained by them rational, the whole straight line is irrational, and let it be called that which produces a rational and a medial area.
Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα· λέγω, ὅτι ἄλογός ἐστιν ἡ ΑΓ.
ἐπεὶ γὰρ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ μέσον ἐστίν, τὸ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ῥητόν, ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ· ὥστε καὶ τὸ ἀπὸ τῆς ΑΓ ἀσύμμετρόν ἐστι τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ·
For let two straight lines AB, BC incommensurable in square be added together making the proposed conditions; I say that AC is irrational. For since the sum of the squares on AB, BC is medial, but twice the rectangle contained by AB, BC is rational, therefore the sum of the squares on AB, BC is incommensurable with twice the rectangle contained by AB, BC; so that the square on AC is also incommensurable with twice the rectangle contained by AB, BC. And twice the rectangle contained by AB, BC is rational; therefore the square on AC is irrational.
ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ. ἄλογος ἄρα ἡ ΑΓ, καλείσθω δὲ ῥητὸν καὶ μέσον δυναμένη. ὅπερ ἔδει δεῖξαι.
Therefore AC is irrational, and let it be called that which produces a rational and a medial area; which was to be proved.
§10.prop1.41ἐὰν δύο εὐθεῖαι δυνάμει ἀσύμμετροι συντεθῶσι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπʼ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων, ἡ ὅλη εὐθεῖα ἄλογός ἐστιν, καλείσθω δὲ δύο μέσα δυναμένη.
If two straight lines incommensurable in square be added together making the sum of the squares on them medial and the rectangle contained by them medial, and moreover incommensurable with the sum of the squares on them, the whole straight line is irrational, and let it be called that which produces two medial areas.
Συγκείσθωσαν γὰρ δύο εὐθεῖαι δυνάμει ἀσύμμετροι αἱ ΑΒ, ΒΓ ποιοῦσαι τὰ προκείμενα· λέγω, ὅτι ἡ ΑΓ ἄλογός ἐστιν.
For let two straight lines AB, BC incommensurable in square be added together making the proposed conditions; I say that AC is irrational.
Ἐκκείσθω ῥητὴ ἡ ΔΕ, καὶ παραβεβλήσθω παρὰ τὴν ΔΕ τοῖς μὲν ἀπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΔΖ, τῷ δὲ δὶς ὑπὸ τῶν ΑΒ, ΒΓ ἴσον τὸ ΗΘ· ὅλον ἄρα τὸ ΔΘ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ.
For let the rational straight line DE be set out, and let there be applied to DE the rectangle DZ equal to the sum of the squares on AB, BC, and the rectangle HTh equal to twice the rectangle contained by AB, BC; therefore the whole DTh is equal to the square on AC.
καὶ ἐπεὶ μέσον ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ, καί ἐστιν ἴσον τῷ ΔΖ, μέσον ἄρα ἐστὶ καὶ τὸ ΔΖ. καὶ παρὰ ῥητὴν τὴν ΔΕ παράκειται· ῥητὴ ἄρα ἐστὶν ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΔΕ μήκει.
And since the sum of the squares on AB, BC is medial, and it is equal to DZ, therefore DZ is also medial. And it is applied to the rational straight line DE; therefore DH is a rational straight line and incommensurable in length with DE.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ΗΚ ῥητή ἐστι καὶ ἀσύμμετρος τῇ ΗΖ, τουτέστι τῇ ΔΕ, μήκει.
For the same reasons, HK is also a rational straight line and incommensurable in length with HZ, that is, with DE.
καὶ ἐπεὶ ἀσύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ἀσύμμετρόν ἐστι τὸ ΔΖ τῷ ΗΘ· ὥστε καὶ ἡ ΔΗ τῇ ΗΚ ἀσύμμετρός ἐστιν.
And since the sum of the squares on AB, BC is incommensurable with twice the rectangle contained by AB, BC, DZ is incommensurable with HTh; so that DH is also incommensurable with HK.
καί εἰσι ῥηταί· αἱ ΔΗ, ΗΚ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἄλογος ἄρα ἐστὶν ἡ ΔΚ ἡ καλουμένη ἐκ δύο ὀνομάτων.
And they are rational; therefore DH, HK are rational straight lines commensurable in square only; therefore DK, which is called binomial, is irrational.
ῥητὴ δὲ ἡ ΔΕ· ἄλογον ἄρα ἐστὶ τὸ ΔΘ καὶ ἡ δυναμένη αὐτὸ ἄλογός ἐστιν.
And DE is rational; therefore DTh is irrational, and the straight line which is equal in square to it is irrational.
δύναται δὲ τὸ ΘΔ ἡ ΑΓ· ἄλογος ἄρα ἐστὶν ἡ ΑΓ, καλείσθω δὲ δύο μέσα δυναμένη. ὅπερ ἔδει δεῖξαι. λῆμμα
ὅτι δὲ αἱ εἰρημέναι ἄλογοι μοναχῶς διαιροῦνται εἰς τὰς εὐθείας, ἐξ ὧν σύγκεινται ποιουσῶν τὰ προκείμενα εἴδη, δείξομεν ἤδη προεκθέμενοι λημμάτιον τοιοῦτον· Ἐκκείσθω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω ἡ ὅλη εἰς ἄνισα καθʼ ἑκάτερον τῶν Γ, Δ, ὑποκείσθω δὲ μείζων ἡ ΑΓ τῆς ΔΒ· λέγω, ὅτι τὰ ἀπὸ τῶν ΑΓ, ΓΒ μείζονά ἐστι τῶν ἀπὸ τῶν ΑΔ, ΔΒ.
τετμήσθω γὰρ ἡ ΑΒ δίχα κατὰ τὸ Ε. καὶ ἐπεὶ μείζων ἐστὶν ἡ ΑΓ τῆς ΔΒ, κοινὴ ἀφῃρήσθω ἡ ΔΓ· λοιπὴ ἄρα ἡ ΑΔ λοιπῆς τῆς ΓΒ μείζων ἐστίν.
And AC is equal in square to ThD; therefore AC is irrational, and let it be called that which produces two medial areas; which was to be proved. [Lemma] And that the aforesaid irrational straight lines are divided in one way only into the straight lines of which they are composed and which produce the proposed types, we will now show, after first setting out the following lemma: Let the straight line AB be set out, and let the whole be cut into unequal parts at each of the points C, D, and let AC be assumed greater than DB; I say that the sum of the squares on AC, CB is greater than the sum of the squares on AD, DB. For let AB be bisected at E. And since AC is greater than DB, let the common part DC be subtracted; therefore the remainder AD is greater than the remainder CB.
ἴση δὲ ἡ ΑΕ τῇ ΕΒ· ἐλάττων ἄρα ἡ ΔΕ τῆς ΕΓ· τὰ Γ, Δ ἄρα σημεῖα οὐκ ἴσον ἀπέχουσι τῆς διχοτομίας.
And AE is equal to EB; therefore DE is less than EC; therefore the points C, D do not lie at an equal distance from the point of bisection.
καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΒ, ἀλλὰ μὴν καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΒ μετὰ τοῦ ἀπὸ ΔΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΒ, τὸ ἄρα ὑπὸ τῶν ΑΓ, ΓΒ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΔ, ΔΒ μετὰ τοῦ ἀπὸ τῆς ΔΕ·
And since the rectangle contained by AC, CB with the square on EC is equal to the square on EB, and moreover the rectangle contained by AD, DB with the square on DE is also equal to the square on EB, therefore the rectangle contained by AC, CB with the square on EC is equal to the rectangle contained by AD, DB with the square on DE.
ὧν τὸ ἀπὸ τῆς ΔΕ ἔλασσόν ἐστι τοῦ ἀπὸ τῆς ΕΓ· καὶ λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΓ, ΓΒ ἔλασσόν ἐστι τοῦ ὑπὸ τῶν ΑΔ, ΔΒ. ὥστε καὶ τὸ δὶς ὑπὸ τῶν ΑΓ, ΓΒ ἔλασσόν ἐστι τοῦ δὶς ὑπὸ ΑΔ, ΔΒ. καὶ λοιπὸν ἄρα τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΓ, ΓΒ μεῖζόν ἐστι τοῦ συγκειμένου ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ· ὅπερ ἔδει δεῖξαι.
And of these, the square on DE is less than the square on EC; therefore the remainder, the rectangle contained by AC, CB, is less than the rectangle contained by AD, DB. So that twice the rectangle contained by AC, CB is also less than twice the rectangle contained by AD, DB. And therefore the remainder, the sum of the squares on AC, CB, is greater than the sum of the squares on AD, DB; which was to be proved.