§10.prop1.35εὑρεῖν δύο εὐθείας δυνάμει ἀσυμμέτρους ποιούσας τό τε συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων μέσον καὶ τὸ ὑπʼ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπʼ αὐτῶν τετραγώνῳ.
To find two straight lines incommensurable in square making both 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.
Ἐκκείσθωσαν δύο μέσαι δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ μέσον περιέχουσαι, ὥστε τὴν ΑΒ τῆς ΒΓ μεῖζον δύνασθαι τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, καὶ γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ τὰ λοιπὰ γεγονέτω τοῖς ἐπάνω ὁμοίως.
Let two medial straight lines commensurable in square only AB, BC containing a medial area be set out, so that the square on AB is greater than the square on BC by the square on a straight line incommensurable with AB. And let the semicircle ADB be described on AB, and let the rest be done in the same manner as the things above.
καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΒ μήκει, ἀσύμμετρός ἐστι καὶ ἡ ΑΔ τῇ ΔΒ δυνάμει.
And since AZ is incommensurable in length with ZB, AD is also incommensurable in square with DB.
καὶ ἐπεὶ μέσον ἐστὶ τὸ ἀπὸ τῆς ΑΒ, μέσον ἄρα καὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΖ, ΖΒ ἴσον ἐστὶ τῷ ἀφʼ ἑκατέρας τῶν ΒΕ, ΔΖ, ἴση ἄρα ἐστὶν ἡ ΒΕ τῇ ΔΖ· διπλῆ ἄρα ἡ ΒΓ τῆς ΖΔ· ὥστε καὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ διπλάσιόν ἐστι τοῦ ὑπὸ τῶν ΑΒ, ΖΔ. μέσον δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ·
And since the square on AB is medial, therefore the sum of the squares on AD, DB is also medial. And since the rectangle contained by AZ, ZB is equal to the square on each of BE, DZ, therefore BE is equal to DZ; therefore BC is double of ZD; so that the rectangle contained by AB, BC is also double of the rectangle contained by AB, ZD. But the rectangle contained by AB, BC is medial; therefore the rectangle contained by AB, ZD is also medial.
μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ. καί ἐστιν ἴσον τῷ ὑπὸ τῶν ΑΔ, ΔΒ·
And it is equal to the rectangle contained by AD, DB; therefore the rectangle contained by AD, DB is also medial.
μέσον ἄρα καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΒ. καὶ ἐπεὶ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, σύμμετρος δὲ ἡ ΓΒ τῇ ΒΕ, ἀσύμμετρος ἄρα καὶ ἡ ΑΒ τῇ ΒΕ μήκει· ὥστε καὶ τὸ ἀπὸ τῆς ΑΒ τῷ ὑπὸ τῶν ΑΒ, ΒΕ ἀσύμμετρόν ἐστιν.
And since AB is incommensurable in length with BC, and GB is commensurable with BE, therefore AB is also incommensurable in length with BE; so that the square on AB is also incommensurable with the rectangle contained by AB, BE.
ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΒ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΔ, ΔΒ, τῷ δὲ ὑπὸ τῶν ΑΒ, ΒΕ ἴσον ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΖΔ, τουτέστι τὸ ὑπὸ τῶν ΑΔ, ΔΒ· ἀσύμμετρον ἄρα ἐστὶ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΔ, ΔΒ τῷ ὑπὸ τῶν ΑΔ, ΔΒ.
Εὕρηνται ἄρα δύο εὐθεῖαι αἱ ΑΔ, ΔΒ δυνάμει ἀσύμμετροι ποιοῦσαι τό τε συγκείμενον ἐκ τῶν ἀπʼ αὐτῶν μέσον καὶ τὸ ὑπʼ αὐτῶν μέσον καὶ ἔτι ἀσύμμετρον τῷ συγκειμένῳ ἐκ τῶν ἀπʼ αὐτῶν τετραγώνων· ὅπερ ἔδει δεῖξαι.
But the sum of the squares on AD, DB is equal to the square on AB, and the rectangle contained by AB, ZD, that is, the rectangle contained by AD, DB, is equal to the rectangle contained by AB, BE; therefore the sum of the squares on AD, DB is incommensurable with the rectangle contained by AD, DB. Therefore two straight lines AD, DB incommensurable in square have been found making both 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; which was to be proved.
§10.prop1.36ἐὰν δύο ῥηταὶ δυνάμει μόνον σύμμετροι συντεθῶσιν, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο ὀνομάτων.
If two rational straight lines commensurable in square only be added together, the whole is irrational, and let it be called binomial.
Συγκείσθωσαν γὰρ δύο ῥηταὶ δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ· λέγω, ὅτι ὅλη ἡ ΑΓ ἄλογός ἐστιν.
For let two rational straight lines AB, BC commensurable in square only be added together; I say that the whole AC is irrational.
ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει· δυνάμει γὰρ μόνον εἰσὶ σύμμετροι· ὡς δὲ ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως τὸ ὑπὸ τῶν ΑΒΓ πρὸς τὸ ἀπὸ τῆς ΒΓ, ἀσύμμετρον ἄρα ἐστὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ τῷ ἀπὸ τῆς ΒΓ. ἀλλὰ τῷ μὲν ὑπὸ τῶν ΑΒ, ΒΓ σύμμετρόν ἐστι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, τῷ δὲ ἀπὸ τῆς ΒΓ σύμμετρά ἐστι τὰ ἀπὸ τῶν ΑΒ, ΒΓ· αἱ γὰρ ΑΒ, ΒΓ ῥηταί εἰσι δυνάμει μόνον σύμμετροι·
For since AB is incommensurable in length with BC (for they are commensurable in square only), and as AB is to BC, so is the rectangle contained by AB, BC to the square on BC, therefore the rectangle contained by AB, BC is incommensurable with the square on BC. But the rectangle contained by AB, BC is commensurable with twice the rectangle contained by AB, BC, and the square on BC is commensurable with the sum of the squares on AB, BC (for AB, BC are rational straight lines commensurable in square only); therefore twice the rectangle contained by AB, BC is incommensurable with the sum of the squares on AB, BC.
ἀσύμμετρον ἄρα ἐστὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ τοῖς ἀπὸ τῶν ΑΒ, ΒΓ. καὶ συνθέντι τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ μετὰ τῶν ἀπὸ τῶν ΑΒ, ΒΓ, τουτέστι τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ συγκειμένῳ ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ συγκείμενον ἐκ τῶν ἀπὸ τῶν ΑΒ, ΒΓ· ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ· ὥστε καὶ ἡ ΑΓ ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο ὀνομάτων· ὅπερ ἔδει δεῖξαι.
And on combining twice the rectangle contained by AB, BC with the sum of the squares on AB, BC, that is, the square on AC, is incommensurable with the sum of the squares on AB, BC. But the sum of the squares on AB, BC is rational; therefore the square on AC is irrational; so that AC is also irrational, and let it be called binomial; which was to be proved.
§10.prop1.37ἐὰν δύο μέσαι δυνάμει μόνον σύμμετροι συντεθῶσι ῥητὸν περιέχουσαι, ἡ ὅλη ἄλογός ἐστιν, καλείσθω δὲ ἐκ δύο μέσων πρώτη.
If two medial straight lines commensurable in square only containing a rational area be added together, the whole is irrational, and let it be called the first bimedial.
Συγκείσθωσαν γὰρ δύο μέσαι δυνάμει μόνον σύμμετροι αἱ ΑΒ, ΒΓ ῥητὸν περιέχουσαι· λέγω, ὅτι ὅλη ἡ ΑΓ ἄλογός ἐστιν.
For let two medial straight lines commensurable in square only AB, BC containing a rational area be added together; I say that the whole AC is irrational.
ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΑΒ τῇ ΒΓ μήκει, καὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ ἄρα ἀσύμμετρά ἐστι τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ·
For since AB is incommensurable in length with BC, therefore the sum of the squares on AB, BC is also incommensurable with twice the rectangle contained by AB, BC.
καὶ συνθέντι τὰ ἀπὸ τῶν ΑΒ, ΒΓ μετὰ τοῦ δὶς ὑπὸ τῶν ΑΒ, ΒΓ, ὅπερ ἐστὶ τὸ ἀπὸ τῆς ΑΓ, ἀσύμμετρόν ἐστι τῷ ὑπὸ τῶν ΑΒ, ΒΓ. ῥητὸν δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΓ·
And on combining the sum of the squares on AB, BC with twice the rectangle contained by AB, BC, which is the square on AC, is incommensurable with the rectangle contained by AB, BC.
ὑπόκεινται γὰρ αἱ ΑΒ, ΒΓ ῥητὸν περιέχουσαι· ἄλογον ἄρα τὸ ἀπὸ τῆς ΑΓ· ἄλογος ἄρα ἡ ΑΓ, καλείσθω δὲ ἐκ δύο μέσων πρώτη· ὅπερ ἔδει δεῖξαι.
But the rectangle contained by AB, BC is rational; for AB, BC are by hypothesis set out containing a rational area; therefore the square on AC is irrational; therefore AC is irrational, and let it be called the first bimedial; which was to be proved.