§9.prop.20οἱ πρῶτοι ἀριθμοὶ πλείους εἰσὶ παντὸς τοῦ προτεθέντος πλήθους πρώτων ἀριθμῶν.
Prime numbers are more than any assigned multitude of prime numbers.
ἔστωσαν οἱ προτεθέντες πρῶτοι ἀριθμοὶ οἱ Α, Β, Γ· λέγω, ὅτι τῶν Α, Β, Γ πλείους εἰσὶ πρῶτοι ἀριθμοί.
Let the assigned prime numbers be A, B, Γ; I say that there are more prime numbers than A, B, Γ.
εἰλήφθω γὰρ ὁ ὑπὸ τῶν Α, Β, Γ ἐλάχιστος μετρούμενος καὶ ἔστω ὁ ΔΕ, καὶ προσκείσθω τῷ ΔΕ μονὰς ἡ ΔΖ. ὁ δὴ ΕΖ ἤτοι πρῶτός ἐστιν ἢ οὔ.
For let the least number measured by A, B, Γ be taken, and let it be ΔE, and let the unit ΔZ be added to ΔE. Then EZ is either prime or not.
ἔστω πρότερον πρῶτος· εὑρημένοι ἄρα εἰσὶ πρῶτοι ἀριθμοὶ οἱ Α, Β, Γ, ΕΖ πλείους τῶν Α, Β, Γ.
ἀλλὰ δὴ μὴ ἔστω ὁ ΕΖ πρῶτος· ὑπὸ πρώτου ἄρα τινὸς ἀριθμοῦ μετρεῖται.
Let it first be prime; therefore prime numbers A, B, Γ, EZ have been found, more than A, B, Γ. But indeed let EZ not be prime; therefore it is measured by some prime number.
μετρείσθω ὑπὸ πρώτου τοῦ Η· λέγω, ὅτι ὁ Η οὐδενὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός.
Let it be measured by the prime H; I say that H is not the same as any of A, B, Γ.
εἰ γὰρ δυνατόν, ἔστω.
For, if possible, let it be.
οἱ δὲ Α, Β, Γ τὸν ΔΕ μετροῦσιν· καὶ ὁ Η ἄρα τὸν ΔΕ μετρήσει.
But A, B, Γ measure ΔE; therefore H will also measure ΔE.
μετρεῖ δὲ καὶ τὸν ΕΖ· καὶ λοιπὴν τὴν ΔΖ μονάδα μετρήσει ὁ Η ἀριθμὸς ὤν· ὅπερ ἄτοπον.
And it also measures EZ; therefore H, being a number, will also measure the remaining unit ΔZ; which is absurd.
οὐκ ἄρα ὁ Η ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός.
Therefore H is not the same as any of A, B, Γ.
καὶ ὑπόκειται πρῶτος.
And it is assumed to be prime.
εὑρημένοι ἄρα εἰσὶ πρῶτοι ἀριθμοὶ πλείους τοῦ προτεθέντος πλήθους τῶν Α, Β, Γ οἱ Α, Β, Γ, Η· ὅπερ ἔδει δεῖξαι.
Therefore prime numbers A, B, Γ, H have been found, more than the assigned multitude of A, B, Γ; which it was required to prove.
§9.prop.21ἐὰν ἄρτιοι ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, ὁ ὅλος ἄρτιός ἐστιν.
If any number of even numbers be added together, the whole is even.
Συγκείσθωσαν γὰρ ἄρτιοι ἀριθμοὶ ὁποσοιοῦν οἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ· λέγω, ὅτι ὅλος ὁ ΑΕ ἄρτιός ἐστιν.
For let any number of even numbers, AB, BΓ, ΓΔ, ΔE, be added together; I say that the whole AE is even.
ἐπεὶ γὰρ ἕκαστος τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ ἄρτιός ἐστιν, ἔχει μέρος ἥμισυ· ὥστε καὶ ὅλος ὁ ΑΕ ἔχει μέρος ἥμισυ.
For, since each of AB, BΓ, ΓΔ, ΔE is even, it has a half part; so that the whole AE also has a half part.
ἄρτιος δὲ ἀριθμός ἐστιν ὁ δίχα διαιρούμενος· ἄρτιος ἄρα ἐστὶν ὁ ΑΕ· ὅπερ ἔδει δεῖξαι.
And an even number is that which is divided into two equal parts; therefore AE is even; which it was required to prove.
§9.prop.22ἐὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ πλῆθος αὐτῶν ἄρτιον ᾖ, ὁ ὅλος ἄρτιος ἔσται.
If any number of odd numbers be added together, and their multitude be even, the whole will be even.
Συγκείσθωσαν γὰρ περισσοὶ ἀριθμοὶ ὁσοιδηποτοῦν ἄρτιοι τὸ πλῆθος οἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ· λέγω, ὅτι ὅλος ὁ ΑΕ ἄρτιός ἐστιν.
For let any number of odd numbers, even in multitude, AB, BΓ, ΓΔ, ΔE, be added together; I say that the whole AE is even.
ἐπεὶ γὰρ ἕκαστος τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ περιττός ἐστιν, ἀφαιρεθείσης μονάδος ἀφʼ ἑκάστου ἕκαστος τῶν λοιπῶν ἄρτιος ἔσται· ὥστε καὶ ὁ συγκείμενος ἐξ αὐτῶν ἄρτιος ἔσται.
For, since each of AB, BΓ, ΓΔ, ΔE is odd, if a unit is subtracted from each, each of the remainders will be even; so that the number composed of them will also be even.
ἔστι δὲ καὶ τὸ πλῆθος τῶν μονάδων ἄρτιον.
But the multitude of the units is also even.
καὶ ὅλος ἄρα ὁ ΑΕ ἄρτιός ἐστιν· ὅπερ ἔδει δεῖξαι.
Therefore the whole AE is also even; which it was required to prove.
§9.prop.23ἐὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ πλῆθος αὐτῶν περισσὸν ᾖ, καὶ ὁ ὅλος περισσὸς ἔσται.
If any number of odd numbers be added together, and their multitude be odd, the whole will also be odd.
Συγκείσθωσαν γὰρ ὁποσοιοῦν περισσοὶ ἀριθμοί, ὧν τὸ πλῆθος περισσὸν ἔστω, οἱ ΑΒ, ΒΓ, ΓΔ· λέγω, ὅτι καὶ ὅλος ὁ ΑΔ περισσός ἐστιν.
For let any number of odd numbers be added together, of which let the multitude be odd, namely AB, BΓ, ΓΔ; I say that the whole AΔ is also odd.
ἀφῃρήσθω ἀπὸ τοῦ ΓΔ μονὰς ἡ ΔΕ· λοιπὸς ἄρα ὁ ΓΕ ἄρτιός ἐστιν.
For let the unit ΔE be subtracted from ΓΔ; therefore the remainder ΓE is even.
ἔστι δὲ καὶ ὁ ΓΑ ἄρτιος· καὶ ὅλος ἄρα ὁ ΑΕ ἄρτιός ἐστιν.
But ΓA is also even; therefore the whole AE is also even.
καί ἐστι μονὰς ἡ ΔΕ. περισσὸς ἄρα ἐστὶν ὁ ΑΔ· ὅπερ ἔδει δεῖξαι.
And ΔE is a unit; therefore AΔ is odd; which it was required to prove.
§9.prop.24ἐὰν ἀπὸ ἀρτίου ἀριθμοῦ ἄρτιος ἀφαιρεθῇ, ὁ λοιπὸς ἄρτιος ἔσται.
If an even number be subtracted from an even number, the remainder will be even.
ἀπὸ γὰρ ἀρτίου τοῦ ΑΒ ἄρτιος ἀφῃρήσθω ὁ ΒΓ· λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν.
For let the even number BΓ be subtracted from the even number AB; I say that the remainder ΓA is even.
ἐπεὶ γὰρ ὁ ΑΒ ἄρτιός ἐστιν, ἔχει μέρος ἥμισυ.
For, since AB is even, it has a half part.
διὰ τὰ αὐτὰ δὴ καὶ ὁ ΒΓ ἔχει μέρος ἥμισυ· ὥστε καὶ λοιπὸς ἄρτιος ἐστὶν ὁ ΑΓ· ὅπερ ἔδει δεῖξαι.
For the same reasons indeed BΓ also has a half part; so that the remainder AΓ is also even; which it was required to prove.
§9.prop.25ἐὰν ἀπὸ ἀρτίου ἀριθμοῦ περισσὸς ἀφαιρεθῇ, ὁ λοιπὸς περισσὸς ἔσται.
If an odd number be subtracted from an even number, the remainder will be odd.
ἀπὸ γὰρ ἀρτίου τοῦ ΑΒ περισσὸς ἀφῃρήσθω ὁ ΒΓ· λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ περισσός ἐστιν.
For let the odd number BΓ be subtracted from the even number AB; I say that the remainder ΓA is odd.
ἀφῃρήσθω γὰρ ἀπὸ τοῦ ΒΓ μονὰς ἡ ΓΔ· ὁ ΔΒ ἄρα ἄρτιός ἐστιν.
For let the unit ΓΔ be subtracted from BΓ; therefore ΔB is even.
ἔστι δὲ καὶ ὁ ΑΒ ἄρτιος· καὶ λοιπὸς ἄρα ὁ ΑΔ ἄρτιός ἐστιν.
But AB is also even; therefore the remainder AΔ is also even.
καί ἐστι μονὰς ἡ ΓΔ· ὁ ΓΑ ἄρα περισσός ἐστιν· ὅπερ ἔδει δεῖξαι.
And ΓΔ is a unit; therefore ΓA is odd; which it was required to prove.
§9.prop.26ἐὰν ἀπὸ περισσοῦ ἀριθμοῦ περισσὸς ἀφαιρεθῇ, ὁ λοιπὸς ἄρτιος ἔσται.
If an odd number be subtracted from an odd number, the remainder will be even.
ἀπὸ γὰρ περισσοῦ τοῦ ΑΒ περισσὸς ἀφῃρήσθω ὁ ΒΓ· λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν.
For let the odd number BΓ be subtracted from the odd number AB; I say that the remainder ΓA is even.
ἐπεὶ γὰρ ὁ ΑΒ περισσός ἐστιν, ἀφῃρήσθω μονὰς ἡ ΒΔ· λοιπὸς ἄρα ὁ ΑΔ ἄρτιός ἐστιν.
For, since AB is odd, let the unit BΔ be subtracted; therefore the remainder AΔ is even.
διὰ τὰ αὐτὰ δὴ καὶ ὁ ΓΔ ἄρτιός ἐστιν· ὥστε καὶ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν· ὅπερ ἔδει δεῖξαι.
For the same reasons indeed ΓΔ is also even; so that the remainder ΓA is also even; which it was required to prove.