---
title: "Elements · Στοιχεῖα"
author: "Euclid"
translator: "Thomas Little Heath"
languages: [grc, en]
shelf: "Classical Greek"
section: 29 of 49
url: https://scriptura.si/works/euclid-elements/29/
cite: "Euclid, tr. Thomas Little Heath, Elements · Στοιχεῖα, section 29. Scriptura (scriptura.si), https://scriptura.si/works/euclid-elements/29/"
---

# Elements · Στοιχεῖα

*Euclid, tr. Thomas Little Heath*

The thirteen books of geometry and number theory that taught mathematics for two thousand years, from the definition of a point to the five regular solids.

> Greek: ed. Johan Ludvig Heiberg, from the Perseus Digital Library (canonical-greekLit, tlg1799.tlg001, CC BY-SA 4.0).

> English: Thomas Little Heath, from the Perseus Digital Library (tlg1799.tlg001.perseus-eng2). Notes are not included.

> Rows follow the references the two texts share (books, chapters and sections), cut where both reach the end of a sentence.

## 29. Book IX · Propositions · 1–15

Source: https://scriptura.si/works/euclid-elements/29/ · cite a paragraph as https://scriptura.si/works/euclid-elements/29/#p-N

[¶0] 1. If two similar plane numbers by multiplying one another make some number, the product will be square. Let A, B be two similar plane numbers, and let A by multiplying B make C; I say that C is square. For let A by multiplying itself make D. Therefore D is square. Since then A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [] And, since A, B are similar plane numbers, therefore one mean proportional number falls between A, B. [] But, if numbers fall between two numbers in continued proportion, as many as fall between them, so many also fall between those which have the same ratio; [] so that one mean proportional number falls between D, C also. And D is square; therefore C is also square. [] Q. E. D.

> Greek: 1. ἐὰν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος τετράγωνος ἔσται. ἔστωσαν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ τετράγωνός ἐστιν. ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω. ὁ Δ ἄρα τετράγωνός ἐστιν. ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί, τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. ἐὰν δὲ δύο ἀριθμῶν μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν ἀριθμοί, ὅσοι εἰς αὐτοὺς ἐμπίπτουσι, τοσοῦτοι καὶ εἰς τοὺς τὸν αὐτὸν λόγον ἔχοντας· ὥστε καὶ τῶν Δ, Γ εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. καί ἐστι τετράγωνος ὁ Δ· τετράγωνος ἄρα καὶ ὁ Γ· ὅπερ ἔδει δεῖξαι.

[¶1] 2. If two numbers by multiplying one another make a square number, they are similar plane numbers. Let A, B be two numbers, and let A by multiplying B make the square number C; I say that A, B are similar plane numbers. For let A by multiplying itself make D; therefore D is square. Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [] And, since D is square, and C is so also, therefore D, C are similar plane numbers. Therefore one mean proportional number falls between D, C. [] And, as D is to C, so is A to B; therefore one mean proportional number falls between A, B also. [] But, if one mean proportional number fall between two numbers, they are similar plane numbers; [] therefore A, B are similar plane numbers. Q. E. D.

> Greek: 2. ἐὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσι τετράγωνον, ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί. ἔστωσαν δύο ἀριθμοὶ οἱ Α, Β, καὶ ὁ Α τὸν Β πολλαπλασιάσας τετράγωνον τὸν Γ ποιείτω· λέγω, ὅτι οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί. ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· ὁ Δ ἄρα τετράγωνός ἐστιν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ ὁ Δ τετράγωνός ἐστιν, ἀλλὰ καὶ ὁ Γ, οἱ Δ, Γ ἄρα ὅμοιοι ἐπίπεδοί εἰσιν. τῶν Δ, Γ ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει. καί ἐστιν ὡς ὁ Δ πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν Β· καὶ τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει. ἐὰν δὲ δύο ἀριθμῶν εἷς μέσος ἀνάλογον ἐμπίπτῃ, ὅμοιοι ἐπίπεδοί εἰσιν [οἱ] ἀριθμοί· οἱ ἄρα Α, Β ὅμοιοί εἰσιν ἐπίπεδοι· ὅπερ ἔδει δεῖξαι.

[¶2] 3. If a cube number by multiplying itself make some number, the product will be cube. For let the cube number A by multiplying itself make B; I say that B is cube. For let C, the side of A, be taken, and let C by multiplying itself make D. It is then manifest that C by multiplying D has made A. Now, since C by multiplying itself has made D, therefore C measures D according to the units in itself. But further the unit also measures C according to the units in it; therefore, as the unit is to C, so is C to D. [] Again, since C by multiplying D has made A, therefore D measures A according to the units in C. But the unit also measures C according to the units in it; therefore, as the unit is to C, so is D to A. But, as the unit is to C, so is C to D; therefore also, as the unit is to C, so is C to D, and D to A. Therefore between the unit and the number A two mean proportional numbers C, D have fallen in continued proportion. Again, since A by multiplying itself has made B, therefore A measures B according to the units in itself. But the unit also measures A according to the units in it; therefore, as the unit is to A, so is A to B. [] But between the unit and A two mean proportional numbers have fallen; therefore two mean proportional numbers will also fall between A, B. [] But, if two mean proportional numbers fall between two numbers, and the first be cube, the second will also be cube. [] And A is cube; therefore B is also cube. Q. E. D.

> Greek: 3. ἐὰν κύβος ἀριθμὸς ἑαυτὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἔσται. κύβος γὰρ ἀριθμὸς ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β ποιείτω· λέγω, ὅτι ὁ Β κύβος ἐστίν. εἰλήφθω γὰρ τοῦ Α πλευρὰ ὁ Γ, καὶ ὁ Γ ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω. φανερὸν δή ἐστιν, ὅτι ὁ Γ τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ ἐπεὶ ὁ Γ ἑαυτὸν πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ Γ ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας. ἀλλὰ μὴν καὶ ἡ μονὰς τὸν Γ μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας· ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Δ. πάλιν, ἐπεὶ ὁ Γ τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν, ὁ Δ ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Γ κατὰ τὰς ἐν αὐτῷ μονάδας· ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Δ πρὸς τὸν Α. ἀλλʼ ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Δ· καὶ ὡς ἄρα ἡ μονὰς πρὸς τὸν Γ, οὕτως ὁ Γ πρὸς τὸν Δ καὶ ὁ Δ πρὸς τὸν Α. τῆς ἄρα μονάδος καὶ τοῦ Α ἀριθμοῦ δύο μέσοι ἀνάλογον κατὰ τὸ συνεχὲς ἐμπεπτώκασιν ἀριθμοὶ οἱ Γ, Δ. πάλιν, ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας· ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς τὸν Β. τῆς δὲ μονάδος καὶ τοῦ Α δύο μέσοι ἀνάλογον ἐμπεπτώκασιν ἀριθμοί· καὶ τῶν Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπεσοῦνται ἀριθμοί. ἐὰν δὲ δύο ἀριθμῶν δύο μέσοι ἀνάλογον ἐμπίπτωσιν, ὁ δὲ πρῶτος κύβος ᾖ, καὶ ὁ δεύτερος κύβος ἔσται. καί ἐστιν ὁ Α κύβος· καὶ ὁ Β ἄρα κύβος ἐστίν· ὅπερ ἔδει δεῖξαι.

[¶3] 4. If a cube number by multiplying a cube number make some number, the product will be cube. For let the cube number A by multiplying the cube number B make C; I say that C is cube. For let A by multiplying itself make D; therefore D is cube. [] And, since A by multiplying itself has made D, and by multiplying B has made C therefore, as A is to B, so is D to C. [] And, since A, B are cube numbers, A, B are similar solid numbers. Therefore two mean proportional numbers fall between A, B; [] so that two mean proportional numbers will fall between D, C also. [] And D is cube; therefore C is also cube [] Q. E. D.

> Greek: 4. ἐὰν κύβος ἀριθμὸς κύβον ἀριθμὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἔσται. κύβος γὰρ ἀριθμὸς ὁ Α κύβον ἀριθμὸν τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ κύβος ἐστίν. ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· ὁ Δ ἄρα κύβος ἐστίν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Α, Β κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν οἱ Α, Β. τῶν Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί· ὥστε καὶ τῶν Δ, Γ δύο μέσοι ἀνάλογον ἐμπεσοῦνται ἀριθμοί. καί ἐστι κύβος ὁ Δ· κύβος ἄρα καὶ ὁ Γ· ὅπερ ἔδει δεῖξαι.

[¶4] 5. If a cube number by multiplying any number make a cube number, the multiplied number will also be cube. For let the cube number A by multiplying any number B make the cube number C; I say that B is cube. For let A by multiplying itself make D; therefore D is cube. [] Now, since A by multiplying itself has made D, and by multiplying B has made C, therefore, as A is to B, so is D to C. [] And since D, C are cube, they are similar solid numbers. Therefore two mean proportional numbers fall between D, C. [] And, as D is to C, so is A to B; therefore two mean proportional numbers fall between A, B also. [] And A is cube; therefore B is also cube. [] If a number by multiplying itself make a cube number, it will itself also be cube.

> Greek: 5. ἐὰν κύβος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας κύβον ποιῇ, καὶ ὁ πολλαπλασιασθεὶς κύβος ἔσται. κύβος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλαπλασιάσας κύβον τὸν Γ ποιείτω· λέγω, ὅτι ὁ Β κύβος ἐστίν. ὁ γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· κύβος ἄρα ἐστίν ὁ Δ. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Δ, Γ κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν. τῶν Δ, Γ ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. καί ἐστιν ὡς ὁ Δ πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν Β· καὶ τῶν α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. καί ἐστι κύβος ὁ Α· κύβος ἄρα ἐστὶ καὶ ὁ Β· ὅπερ ἔδει δεῖξαι.

[¶5] 6. For let the number A by multiplying itself make the cube number B; I say that A is also cube. For let A by multiplying B make C. Since, then, A by multiplying itself has made B, and by multiplying B has made C, therefore C is cube. And, since A by multiplying itself has made B, therefore A measures B according to the units in itself. But the unit also measures A according to the units in it. Therefore, as the unit is to A, so is A to B. [] And, since A by multiplying B has made C, therefore B measures C according to the units in A. But the unit also measures A according to the units in it. Therefore, as the unit is to A, so is B to C. [] But, as the unit is to A, so is A to B; therefore also, as A is to B, so is B to C. And, since B, C are cube, they are similar solid numbers. Therefore there are two mean proportional numbers between B, C. [] And, as B is to C, so is A to B. Therefore there are two mean proportional numbers between A, B also. [] And B is cube; therefore A is also cube. [cf. ] Q. E. D.

> Greek: 6. ἐὰν ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ αὐτὸς κύβος ἔσται. ἀριθμὸς γὰρ ὁ Α ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β ποιείτω· λέγω, ὅτι καὶ ὁ Α κύβος ἐστίν. ʽὁ γὰρ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω. ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Γ ἄρα κύβος ἐστίν. καὶ ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας. ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β. καὶ ἐπεὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας. ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Γ. ἀλλʼ ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Β, Γ κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν. τῶν Β, Γ ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί. καί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β. καὶ τῶν Α, Β ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί. καί ἐστι κύβος ὁ Β· κύβος ἄρα ἐστὶ καὶ ὁ Α· ὅπερ ἔδει δεῖξαι.

[¶6] 7. If a composite number by multiplying any number make some number, the product will be solid. For let the composite number A by multiplying any number B make C; I say that C is solid. For, since A is composite, it will be measured by some number. [] Let it be measured by D; and, as many times as D measures A, so many units let there be in E. Since then D measures A according to the units in E, therefore E by multiplying D has made A. [] And, since A by multiplying B has made C, and A is the product of D, E, therefore the product of D, E by multiplying B has made C. Therefore C is solid, and D, E, B are its sides. Q. E. D.

> Greek: 7. ἐὰν σύνθετος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος στερεὸς ἔσται. σύνθετος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ στερεός ἐστιν. ἐπεὶ γὰρ ὁ Α σύνθετός ἐστιν, ὑπὸ ἀριθμοῦ τινος μετρηθήσεται. μετρείσθω ὑπὸ τοῦ Δ, καὶ ὁσάκις ὁ Δ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν ὁ Δ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, ὁ Ε ἄρα τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ ἐπεὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ δὲ Α ἐστιν ὁ ἐκ τῶν Δ, Ε, ὁ ἄρα ἐκ τῶν Δ, Ε τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. ὁ Γ ἄρα στερεός ἐστιν, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Δ, Ε, Β· ὅπερ ἔδει δεῖξαι.

[¶7] 8. If as many numbers as we please beginning from an unit be in continued proportion, the third from the unit will be square, as will also those which successively leave out one; the fourth will be cube, as will also all those which leave out two; and the seventh will be at once cube and square, as will also those which leave out five. Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion; I say that B, the third from the unit, is square, as are also all those which leave out one; C, the fourth, is cube, as are also all those which leave out two; and F, the seventh, is at once cube and square, as are also all those which leave out five. For since, as the unit is to A, so is A to B, therefore the unit measures the number A the same number of times that A measures B. [] But the unit measures the number A according to the units in it; therefore A also measures B according to the units in A. Therefore A by multiplying itself has made B; therefore B is square. And, since B, C, D are in continued proportion, and B is square, therefore D is also square. [] For the same reason F is also square. Similarly we can prove that all those which leave out one are square. I say next that C, the fourth from the unit, is cube, as are also all those which leave out two. For since, as the unit is to A, so is B to C, therefore the unit measures the number A the same number of times that B measures C. But the unit measures the number A according to the units in A; therefore B also measures C according to the units in A. Therefore A by multiplying B has made C. Since then A by multiplying itself has made B, and by multiplying B has made C, therefore C is cube. And, since C, D, E, F are in continued proportion, and C is cube, therefore F is also cube. [] But it was also proved square; therefore the seventh from the unit is both cube and square. Similarly we can prove that all the numbers which leave out five are also both cube and square. Q. E. D.

> Greek: 8. ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ μὲν τρίτος ἀπὸ τῆς μονάδος τετράγωνος ἔσται καὶ οἱ ἕνα διαλείποντες, ὁ δὲ τέταρτος κύβος καὶ οἱ δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος κύβος ἅμα καὶ τετράγωνος καὶ οἱ πέντε διαλείποντες. ἔστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, Ε, Ζ· λέγω, ὅτι ὁ μὲν τρίτος ἀπὸ τῆς μονάδος ὁ Β τετράγωνός ἐστι καὶ οἱ ἕνα διαλείποντες πάντες, ὁ δὲ τέταρτος ὁ Γ κύβος καὶ οἱ δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος ὁ Ζ κύβος ἅμα καὶ τετράγωνος καὶ οἱ πέντε διαλείποντες πάντες. ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας· καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν· τετράγωνος ἄρα ἐστὶν ὁ Β. καὶ ἐπεὶ οἱ Β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Β τετράγωνός ἐστιν, καὶ ὁ Δ ἄρα τετράγωνός ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ζ τετράγωνός ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ ἕνα διαλείποντες πάντες τετράγωνοί εἰσιν. λέγω δή, ὅτι καὶ ὁ τέταρτος ἀπὸ τῆς μονάδος ὁ Γ κύβος ἐστὶ καὶ οἱ δύο διαλείποντες πάντες. ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Γ, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Γ. ἡ δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας· καὶ ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας· ὁ Α ἄρα τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, κύβος ἄρα ἐστὶν ὁ Γ. καὶ ἐπεὶ οἱ Γ, Δ, Ε, Ζ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Γ κύβος ἐστίν, καὶ ὁ Ζ ἄρα κύβος ἐστίν. ἐδείχθη δὲ καὶ τετράγωνος· ὁ ἄρα ἕβδομος ἀπὸ τῆς μονάδος κύβος τέ ἐστι καὶ τετράγωνος. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ πέντε διαλείποντες πάντες κύβοι τέ εἰσι καὶ τετράγωνοι· ὅπερ ἔδει δεῖξαι.

[¶8] 9. If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be square, all the rest will also be square. And, if the number after the unit be cube, all the rest will also be cube. Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, be square; I say that all the rest will also be square. Now it has been proved that B, the third from the unit, is square, as are also all those which leave out one; [] I say that all the rest are also square. For, since A, B, C are in continued proportion, and A is square, therefore C is also square. [] Again, since B, C, D are in continued proportion, and B is square, D is also square. [] Similarly we can prove that all the rest are also square. Next, let A be cube; I say that all the rest are also cube. Now it has been proved that C, the fourth from the unit, is cube, as also are all those which leave out two; [] I say that all the rest are also cube. For, since, as the unit is to A, so is A to B, therefore the unit measures A the same number of times as A measures B. But the unit measures A according to the units in it; therefore A also measures B according to the units in itself; therefore A by multiplying itself has made B. And A is cube. But, if a cube number by multiplying itself make some number, the product is cube. [] Therefore B is also cube. And, since the four numbers A, B, C, D are in continued proportion, and A is cube, D also is cube. [] For the same reason E is also cube, and similarly all the rest are cube. Q. E. D.

> Greek: 9. ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἑξῆς κατὰ τὸ συνεχὲς ἀριθμοὶ ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα τετράγωνος ᾖ, καὶ οἱ λοιποὶ πάντες τετράγωνοι ἔσονται. καὶ ἐὰν ὁ μετὰ τὴν μονάδα κύβος ᾖ, καὶ οἱ λοιποὶ πάντες κύβοι ἔσονται. ἔστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α τετράγωνος ἔστω· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοι ἔσονται. ὅτι μὲν οὖν ὁ τρίτος ἀπὸ τῆς μονάδος ὁ Β τετράγωνός ἐστι καὶ οἱ ἕνα διαλείποντες πάντες, δέδεικται· λέγω [δή], ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοί εἰσιν. ἐπεὶ γὰρ οἱ Α, Β, Γ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α τετράγωνος, καὶ ὁ Γ [ἄρα] τετράγωνός ἐστιν. πάλιν, ἐπεὶ [καὶ] οἱ β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Β τετράγωνος, καὶ ὁ Δ [ἄρα] τετράγωνός ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοί εἰσιν. ἀλλὰ δὴ ἔστω ὁ Α κύβος· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες κύβοι εἰσίν. ὅτι μὲν οὖν ὁ τέταρτος ἀπὸ τῆς μονάδος ὁ Γ κύβος ἐστὶ καὶ οἱ δύο διαλείποντες πάντες, δέδεικται· λέγω [δή], ὅτι καὶ οἱ λοιποὶ πάντες κύβοι εἰσίν. ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β, ἰσάκις ἄρα ἡ μονὰς τὸν Α μετρεῖ καὶ ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας· καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας· ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν. καί ἐστιν ὁ Α κύβος. ἐὰν δὲ κύβος ἀριθμὸς ἑαυτὸν πολλαπλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἐστίν· καὶ ὁ Β ἄρα κύβος ἐστίν. καὶ ἐπεὶ τέσσαρες ἀριθμοὶ οἱ Α, Β, Γ, Δ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α κύβος, καὶ ὁ Δ ἄρα κύβος ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε κύβος ἐστίν, καὶ ὁμοίως οἱ λοιποὶ πάντες κύβοι εἰσίν· ὅπερ ἔδει δεῖξαι.

[¶9] 10. If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be not square, neither will any other be square except the third from the unit and all those which leave out one. And, if the number after the unit be not cube, neither will any other be cube except the fourth from the unit and all those which leave out two. Let there be as many numbers as we please, A, B, C, D, E, F, beginning from an unit and in continued proportion, and let A, the number after the unit, not be square; I say that neither will any other be square except the third from the unit ltand those which leave out onegt. For, if possible, let C be square. But B is also square; [] [therefore B, C have to one another the ratio which a square number has to a square number]. And, as B is to C, so is A to B; therefore A, B have to one another the ratio which a square number has to a square number; [so that A, B are similar plane numbers]. [, converse] And B is square; therefore A is also square: which is contrary to the hypothesis. Therefore C is not square. Similarly we can prove that neither is any other of the numbers square except the third from the unit and those which leave out one. Next, let A not be cube. I say that neither will any other be cube except the fourth from the unit and those which leave out two. For, if possible, let D be cube. Now C is also cube; for it is fourth from the unit. [] And, as C is to D, so is B to C; therefore B also has to C the ratio which a cube has to a cube. And C is cube; therefore B is also cube. [] And since, as the unit is to A, so is A to B, and the unit measures A according to the units in it, therefore A also measures B according to the units in itself; therefore A by multiplying itself has made the cube number B. But, if a number by multiplying itself make a cube number, it is also itself cube. [] Therefore A is also cube: which is contrary to the hypothesis. Therefore D is not cube. Similarly we can prove that neither is any other of the numbers cube except the fourth from the unit and those which leave out two. Q. E. D.

> Greek: 10. ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ [ἑξῆς] ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα μὴ ᾖ τετράγωνος, οὐδʼ ἄλλος οὐδεὶς τετράγωνος ἔσται χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων πάντων. καὶ ἐὰν ὁ μετὰ τὴν μονάδα κύβος μὴ ᾖ, οὐδὲ ἄλλος οὐδεὶς κύβος ἔσται χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων πάντων. ἔστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α μὴ ἔστω τετράγωνος· λέγω, ὅτι οὐδὲ ἄλλος οὐδεὶς τετράγωνος ἔσται χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος [καὶ τῶν ἕνα διαλειπόντων]. εἰ γὰρ δυνατόν, ἔστω ὁ Γ τετράγωνος. ἔστι δὲ καὶ ὁ Β τετράγωνος· οἱ Β, Γ ἄρα πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β· οἱ Α, Β ἄρα πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ὥστε οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν. καί ἐστι τετράγωνος ὁ Β· τετράγωνος ἄρα ἐστὶ καὶ ὁ Α· ὅπερ οὐχ ὑπέκειτο. οὐκ ἄρα ὁ Γ τετράγωνός ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι οὐδʼ ἄλλος οὐδεὶς τετράγωνός ἐστι χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων. ἀλλὰ δὴ μὴ ἔστω ὁ Α κύβος. λέγω, ὅτι οὐδʼ ἄλλος οὐδεὶς κύβος ἔσται χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων. εἰ γὰρ δυνατόν, ἔστω ὁ Δ κύβος. ἔστι δὲ καὶ ὁ Γ κύβος· τέταρτος γάρ ἐστιν ἀπὸ τῆς μονάδος. καί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, ὁ Β πρὸς τὸν Γ· καὶ ὁ Β ἄρα πρὸς τὸν Γ λόγον ἔχει, ὃν κύβος πρὸς κύβον. καί ἐστιν ὁ Γ κύβος· καὶ ὁ Β ἄρα κύβος ἐστίν. καὶ ἐπεί ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς τὸν Β, ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας, καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας· ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β πεποίηκεν. ἐὰν δὲ ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ αὐτὸς κύβος ἔσται. κύβος ἄρα καὶ ὁ Α· ὅπερ οὐχ ὑπόκειται. οὐκ ἄρα ὁ Δ κύβος ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι οὐδʼ ἄλλος οὐδεὶς κύβος ἐστὶ χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων· ὅπερ ἔδει δεῖξαι.

[¶10] 11. If as many numbers as we please beginning from an unit be in continued proportion, the less measures the greater according to some one of the numbers which have place among the proportional numbers. Let there be as many numbers as we please, B, C, D, E, beginning from the unit A and in continued proportion; I say that B, the least of the numbers B, C, D, E, measures E according to some one of the numbers C, D. For since, as the unit A is to B, so is D to E, therefore the unit A measures the number B the same number of times as D measures E; therefore, alternately, the unit A measures D the same number of times as B measures E. [] But the unit A measures D according to the units in it; therefore B also measures E according to the units in D; so that B the less measures E the greater according to some number of those which have place among the proportional numbers.— Porism. And it is manifest that, whatever place the measuring number has, reckoned from the unit, the same place also has the number according to which it measures, reckoned from the number measured, in the direction of the number before it.— Q. E. D.

> Greek: 11. ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ ἐλάττων τὸν μείζονα μετρεῖ κατά τινα τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον ἀριθμοῖς. ἔστωσαν ἀπὸ μονάδος τῆς Α ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Β, Γ, Δ, Ε· λέγω, ὅτι τῶν Β, Γ, Δ, Ε ὁ ἐλάχιστος ὁ Β τὸν Ε μετρεῖ κατά τινα τῶν Γ, Δ. ἐπεὶ γάρ ἐστιν ὡς ἡ Α μονὰς πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε, ἰσάκις ἄρα ἡ Α μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Δ τὸν Ε· ἐναλλὰξ ἄρα ἰσάκις ἡ Α μονὰς τὸν Δ μετρεῖ καὶ ὁ Β τὸν Ε. ἡ δὲ Α μονὰς τὸν Δ μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας· καὶ ὁ Β ἄρα τὸν Ε μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας· ὥστε ὁ ἐλάσσων ὁ Β τὸν μείζονα τὸν Ε μετρεῖ κατά τινα ἀριθμὸν τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον ἀριθμοῖς. Πόρισμα Καὶ φανερόν, ὅτι ἣν ἔχει τάξιν ὁ μετρῶν ἀπὸ μονάδος, τὴν αὐτὴν ἔχει καὶ ὁ καθʼ ὃν μετρεῖ ἀπὸ τοῦ μετρουμένου ἐπὶ τὸ πρὸ αὐτοῦ. ¯ὅπερ ἔδει δεῖξαι.

[¶11] 12. If as many numbers as we please beginning from an unit be in continued proportion, by however many prime numbers the last is measured, the next to the unit will also be measured by the same. Let there be as many numbers as we please, A, B, C, D, beginning from an unit, and in continued proportion; I say that, by however many prime numbers D is measured, A will also be measured by the same. For let D be measured by any prime number E; I say that E measures A. For suppose it does not; now E is prime, and any prime number is prime to any which it does not measure; [] therefore E, A are prime to one another. And, since E measures D, let it measure it according to F, therefore E by multiplying F has made D. Again, since A measures D according to the units in C, [] therefore A by multiplying C has made D. But, further, E has also by multiplying F made D; therefore the product of A, C is equal to the product of E, F. Therefore, as A is to E, so is F to C. [] But A, E are prime, primes are also least, [] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [] therefore E measures C. Let it measure it according to G; therefore E by multiplying G has made C. But, further, by the theorem before this, A has also by multiplying B made C. [] Therefore the product of A, B is equal to the product of E, G. Therefore, as A is to E, so is G to B. [] But A, E are prime, primes are also least, [] and the least numbers measure those which have the same ratio with them the same number of times, the antecedent the antecedent and the consequent the consequent: [] therefore E measures B. Let it measure it according to H; therefore E by multiplying H has made B. But further A has also by multiplying itself made B; [] therefore the product of E, H is equal to the square on A. Therefore, as E is to A, so is A to H. [] But A, E are prime, primes are also least, [] and the least measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [] therefore E measures A, as antecedent antecedent. But, again, it also does not measure it: which is impossible. Therefore E, A are not prime to one another. Therefore they are composite to one another. But numbers composite to one another are measured by some number. [] And, since E is by hypothesis prime, and the prime is not measured by any number other than itself, therefore E measures A, E, so that E measures A. [But it also measures D; therefore E measures A, D.] Similarly we can prove that, by however many prime numbers D is measured, A will also be measured by the same. Q. E. D.

> Greek: 12. ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὑφʼ ὅσων ἂν ὁ ἔσχατος πρώτων ἀριθμῶν μετρῆται, ὑπὸ τῶν αὐτῶν καὶ ὁ παρὰ τὴν μονάδα μετρηθήσεται. ἔστωσαν ἀπὸ μονάδος ὁποσοιδηποτοῦν ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ· λέγω, ὅτι ὑφʼ ὅσων ἂν ὁ Δ πρώτων ἀριθμῶν μετρῆται, ὑπὸ τῶν αὐτῶν καὶ ὁ Α μετρηθήσεται. μετρείσθω γὰρ ὁ Δ ὑπό τινος πρώτου ἀριθμοῦ τοῦ Ε· λέγω, ὅτι ὁ ε τὸν Α μετρεῖ. μὴ γάρ· καί ἐστιν ὁ Ε πρῶτος, ἅπας δὲ πρῶτος ἀριθμὸς πρὸς ἅπαντα, ὃν μὴ μετρεῖ, πρῶτός ἐστιν· οἱ Ε, Α ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ· ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. πάλιν, ἐπεὶ ὁ Α τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας, ὁ Α ἄρα τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Ε τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν· ὁ ἄρα ἐκ τῶν Α, Γ ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Ζ. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Ε, ὁ Ζ πρὸς τὸν Γ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· μετρεῖ ἄρα ὁ Ε τὸν Γ. μετρείτω αὐτὸν κατὰ τὸν Η· ὁ Ε ἄρα τὸν Η πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν διὰ τὸ πρὸ τούτου καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. ὁ ἄρα ἐκ τῶν Α, Β ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Η. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Ε, ὁ Η πρὸς τὸν Β. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· μετρεῖ ἄρα ὁ Ε τὸν Β. μετρείτω αὐτὸν κατὰ τὸν Θ· ὁ Ε ἄρα τὸν Θ πολλαπλασιάσας τὸν Β πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν· ὁ ἄρα ἐκ τῶν Ε, Θ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Α. ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Α, ὁ Α πρὸς τὸν Θ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· μετρεῖ ἄρα ὁ Ε τὸν Α ὡς ἡγούμενος ἡγούμενον. ἀλλὰ μὴν καὶ οὐ μετρεῖ· ὅπερ ἀδύνατον. οὐκ ἄρα οἱ Ε, Α πρῶτοι πρὸς ἀλλήλους εἰσίν. σύνθετοι ἄρα. οἱ δὲ σύνθετοι ὑπὸ [πρώτου] ἀριθμοῦ τινος μετροῦνται. καὶ ἐπεὶ ὁ Ε πρῶτος ὑπόκειται, ὁ δὲ πρῶτος ὑπὸ ἑτέρου ἀριθμοῦ οὐ μετρεῖται ἢ ὑφʼ ἑαυτοῦ, ὁ Ε ἄρα τοὺς Α, Ε μετρεῖ· ὥστε ὁ Ε τὸν Α μετρεῖ. μετρεῖ δὲ καὶ τὸν Δ· ὁ Ε ἄρα τοὺς Α, Δ μετρεῖ. ὁμοίως δὴ δείξομεν, ὅτι ὑφʼ ὅσων ἂν ὁ Δ πρώτων ἀριθμῶν μετρῆται, ὑπὸ τῶν αὐτῶν καὶ ὁ Α μετρηθήσεται· ὅπερ ἔδει δεῖξαι.

[¶12] 13. If as many numbers as we please beginning from an unit be in continued proportion, and the number after the unit be prime, the greatest will not be measured by any except those which have a place among the proportional numbers. Let there be as many numbers as we please, A, B, C, D, beginning from an unit and in continued proportion, and let A, the number after the unit, be prime; I say that D, the greatest of them, will not be measured by any other number except A, B, C. For, if possible, let it be measured by E, and let E not be the same with any of the numbers A, B, C. It is then manifest that E is not prime. For, if E is prime and measures D, it will also measure A [], which is prime, though it is not the same with it: which is impossible. Therefore E is not prime. Therefore it is composite. But any composite number is measured by some prime number; [] therefore E is measured by some prime number. I say next that it will not be measured by any other prime except A. For, if E is measured by another, and E measures D, that other will also measure D; so that it will also measure A [], which is prime, though it is not the same with it: which is impossible. Therefore A measures E. And, since E measures D, let it measure it according to F. I say that F is not the same with any of the numbers A, B, C. For, if F is the same with one of the numbers A, B, C, and measures D according to E, therefore one of the numbers A, B, C also measures D according to E. But one of the numbers A, B, C measures D according to some one of the numbers A, B, C; [] therefore E is also the same with one of the numbers A, B, C: which is contrary to the hypothesis. Therefore F is not the same as any one of the numbers A, B, C. Similarly we can prove that F is measured by A, by proving again that F is not prime. For, if it is, and measures D, it will also measure A [], which is prime, though it is not the same with it: which is impossible; therefore F is not prime. Therefore it is composite. But any composite number is measured by some prime number; [] therefore F is measured by some prime number. I say next that it will not be measured by any other prime except A. For, if any other prime number measures F, and F measures D, that other will also measure D; so that it will also measure A [], which is prime, though it is not the same with it: which is impossible. Therefore A measures F. And, since E measures D according to F, therefore E by multiplying F has made D. But, further, A has also by multiplying C made D; [] therefore the product of A, C is equal to the product of E, F. Therefore, proportionally, as A is to E, so is F to C. [] But A measures E; therefore F also measures C. Let it measure it according to G. Similarly, then, we can prove that G is not the same with any of the numbers A, B, and that it is measured by A. And, since F measures C according to G therefore F by multiplying G has made C. But, further, A has also by multiplying B made C; [] therefore the product of A, B is equal to the product of F, G. Therefore, proportionally, as A is to F, so is G to B. [] But A measures F; therefore G also measures B. Let it measure it according to H. Similarly then we can prove that H is not the same with A. And, since G measures B according to H, therefore G by multiplying H has made B. But further A has also by multiplying itself made B; [] therefore the product of H, G is equal to the square on A. Therefore, as H is to A, so is A to G. [] But A measures G; lt therefore H also measures A, which is prime, though it is not the same with it: which is absurd. Therefore D the greatest will not be measured by any other number except A, B, C. Q. E. D.

> Greek: 13. ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα πρῶτος ᾖ, ὁ μέγιστος ὑπʼ οὐδενὸς [ἄλλου] μετρηθήσεται παρὲξ τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον ἀριθμοῖς. ἔστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α πρῶτος ἔστω· λέγω, ὅτι ὁ μέγιστος αὐτῶν ὁ Δ ὑπʼ οὐδενὸς ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ. εἰ γὰρ δυνατόν, μετρείσθω ὑπὸ τοῦ Ε, καὶ ὁ Ε μηδενὶ τῶν Α, Β, Γ ἔστω ὁ αὐτός. φανερὸν δή, ὅτι ὁ Ε πρῶτος οὔκ ἐστιν. εἰ γὰρ ὁ Ε πρῶτός ἐστι καὶ μετρεῖ τὸν Δ, καὶ τὸν Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ὁ Ε πρῶτός ἐστιν. σύνθετος ἄρα. πᾶς δὲ σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται· ὁ Ε ἄρα ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. λέγω δή, ὅτι ὑπʼ οὐδενὸς ἄλλου πρώτου μετρηθήσεται πλὴν τοῦ Α. εἰ γὰρ ὑφʼ ἑτέρου μετρεῖται ὁ Ε, ὁ δὲ Ε τὸν Δ μετρεῖ, κἀκεῖνος ἄρα τὸν Δ μετρήσει· ὥστε καὶ τὸν Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον. ὁ Α ἄρα τὸν Ε μετρεῖ. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ. λέγω, ὅτι ὁ Ζ οὐδενὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. εἰ γὰρ ὁ Ζ ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτὸς καὶ μετρεῖ τὸν Δ κατὰ τὸν Ε, καὶ εἷς ἄρα τῶν Α, Β, Γ τὸν Δ μετρεῖ κατὰ τὸν Ε. ἀλλὰ εἷς τῶν Α, Β, Γ τὸν Δ μετρεῖ κατά τινα τῶν Α, Β, Γ· καὶ ὁ Ε ἄρα ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός· ὅπερ οὐχ ὑπόκειται. οὐκ ἄρα ὁ Ζ ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. ὁμοίως δὴ δείξομεν, ὅτι μετρεῖται ὁ Ζ ὑπὸ τοῦ Α, δεικνύντες πάλιν, ὅτι ὁ Ζ οὔκ ἐστι πρῶτος. εἰ γάρ, καὶ μετρεῖ τὸν Δ, καὶ τὸν α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον· οὐκ ἄρα πρῶτός ἐστιν ὁ Ζ· σύνθετος ἄρα. ἅπας δὲ σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται· ὁ Ζ ἄρα ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. λέγω δή, ὅτι ὑφʼ ἑτέρου πρώτου οὐ μετρηθήσεται πλὴν τοῦ α. εἰ γὰρ ἕτερός τις πρῶτος τὸν Ζ μετρεῖ, ὁ δὲ Ζ τὸν Δ μετρεῖ, κἀκεῖνος ἄρα τὸν Δ μετρήσει· ὥστε καὶ τὸν Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον. ὁ Α ἄρα τὸν Ζ μετρεῖ. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ κατὰ τὸν Ζ, ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν· ὁ ἄρα ἐκ τῶν Α, Γ ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Ζ. ἀνάλογον ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Ε, οὕτως ὁ Ζ πρὸς τὸν Γ. ὁ δὲ Α τὸν Ε μετρεῖ· καὶ ὁ Ζ ἄρα τὸν Γ μετρεῖ. μετρείτω αὐτὸν κατὰ τὸν Η. ὁμοίως δὴ δείξομεν, ὅτι ὁ Η οὐδενὶ τῶν Α, Β ἐστιν ὁ αὐτός, καὶ ὅτι μετρεῖται ὑπὸ τοῦ Α. καὶ ἐπεὶ ὁ Ζ τὸν Γ μετρεῖ κατὰ τὸν Η, ὁ Ζ ἄρα τὸν Η πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν· ὁ ἄρα ἐκ τῶν Α, Β ἴσος ἐστὶ τῷ ἐκ τῶν Ζ, Η. ἀνάλογον ἄρα ὡς ὁ Α πρὸς τὸν Ζ, ὁ Η πρὸς τὸν Β. μετρεῖ δὲ ὁ Α τὸν Ζ· μετρεῖ ἄρα καὶ ὁ Η τὸν Β. μετρείτω αὐτὸν κατὰ τὸν Θ. ὁμοίως δὴ δείξομεν, ὅτι ὁ Θ τῷ Α οὐκ ἔστιν ὁ αὐτός. καὶ ἐπεὶ ὁ Η τὸν Β μετρεῖ κατὰ τὸν Θ, ὁ Η ἄρα τὸν Θ πολλαπλασιάσας τὸν Β πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν· ὁ ἄρα ὑπὸ Θ, Η ἴσος ἐστὶ τῷ ἀπὸ τοῦ Α τετραγώνῳ. ἔστιν ἄρα ὡς ὁ Θ πρὸς τὸν Α, ὁ Α πρὸς τὸν Η. μετρεῖ δὲ ὁ Α τὸν Η· μετρεῖ ἄρα καὶ ὁ Θ τὸν Α πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἄτοπον. οὐκ ἄρα ὁ μέγιστος ὁ Δ ὑπὸ ἑτέρου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Α, Β, Γ· ὅπερ ἔδει δεῖξαι.

[¶13] 14. If a number be the least that is measured by prime numbers, it will not be measured by any other prime number except those originally measuring it. For let the number A be the least that is measured by the prime numbers B, C, D; I say that A will not be measured by any other prime number except B, C, D. For, if possible, let it be measured by the prime number E, and let E not be the same with any one of the numbers B, C, D. Now, since E measures A, let it measure it according to F; therefore E by multiplying F has made A. And A is measured by the prime numbers B, C, D. But, if two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers; [] therefore B, C, D will measure one of the numbers E, F. Now they will not measure E; for E is prime and not the same with any one of the numbers B, C, D. Therefore they will measure F, which is less than A: which is impossible, for A is by hypothesis the least number measured by B, C, D. Therefore no prime number will measure A except B, C, D. Q. E. D.

> Greek: 14. ἐὰν ἐλάχιστος ἀριθμὸς ὑπὸ πρώτων ἀριθμῶν μετρῆται, ὑπʼ οὐδενὸς ἄλλου πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν ἐξ ἀρχῆς μετρούντων. ἐλάχιστος γὰρ ἀριθμὸς ὁ Α ὑπὸ πρώτων ἀριθμῶν τῶν Β, Γ, Δ μετρείσθω· λέγω, ὅτι ὁ Α ὑπʼ οὐδενὸς ἄλλου πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Β, Γ, Δ. εἰ γὰρ δυνατόν, μετρείσθω ὑπὸ πρώτου τοῦ Ε, καὶ ὁ Ε μηδενὶ τῶν Β, Γ, Δ ἔστω ὁ αὐτός. καὶ ἐπεὶ ὁ Ε τὸν Α μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ· ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ μετρεῖται ὁ Α ὑπὸ πρώτων ἀριθμῶν τῶν Β, Γ, Δ. ἐὰν δὲ δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, τὸν δὲ γενόμενον ἐξ αὐτῶν μετρῇ τις πρῶτος ἀριθμός, καὶ ἕνα τῶν ἐξ ἀρχῆς μετρήσει· οἱ Β, Γ, Δ ἄρα ἕνα τῶν Ε, Ζ μετρήσουσιν. τὸν μὲν οὖν Ε οὐ μετρήσουσιν· ὁ γὰρ Ε πρῶτός ἐστι καὶ οὐδενὶ τῶν Β, Γ, Δ ὁ αὐτός. τὸν Ζ ἄρα μετροῦσιν ἐλάσσονα ὄντα τοῦ Α· ὅπερ ἀδύνατον. ὁ γὰρ Α ὑπόκειται ἐλάχιστος ὑπὸ τῶν Β, Γ, Δ μετρούμενος. οὐκ ἄρα τὸν Α μετρήσει πρῶτος ἀριθμὸς παρὲξ τῶν Β, Γ, Δ· ὅπερ ἔδει δεῖξαι.

[¶14] 15. If three numbers in continued proportion be the least of those which have the same ratio with them, any two whatever added together will be prime to the remaining number. Let A, B, C, three numbers in continued proportion, be the least of those which have the same ratio with them; I say that any two of the numbers A, B, C whatever added together are prime to the remaining number, namely A, B to C; B, C to A; and further A, C to B. For let two numbers DE, EF, the least of those which have the same ratio with A, B, C, be taken. [] It is then manifest that DE by multiplying itself has made A, and by multiplying EF has made B, and, further, EF by multiplying itself has made C. [] Now, since DE, EF are least, they are prime to one another. [] But, if two numbers be prime to one another, their sum is also prime to each; [] therefore DF is also prime to each of the numbers DE, EF. But further DE is also prime to EF; therefore DF, DE are prime to EF. But, if two numbers be prime to any number, their product is also prime to the other; [] so that the product of FD, DE is prime to EF; hence the product of FD, DE is also prime to the square on EF. [] But the product of FD, DE is the square on DE together with the product of DE, EF; [] therefore the square on DE together with the product of DE, EF is prime to the square on EF. And the square on DE is A, the product of DE, EF is B, and the square on EF is C; therefore A, B added together are prime to C. Similarly we can prove that B, C added together are prime to A. I say next that A, C added together are also prime to B. For, since DF is prime to each of the numbers DE, EF, the square on DF is also prime to the product of DE, EF. [] But the squares on DE, EF together with twice the product of DE, EF are equal to the square on DF; [] therefore the squares on DE, EF together with twice the product of DE, EF are prime to the product of DE, EF. Separando, the squares on DE, EF together with once the product of DE, EF are prime to the product of DE, EF. Therefore, separando again, the squares on DE, EF are prime to the product of DE, EF. And the square on DE is A, the product of DE, EF is B, and the square on EF is C. Therefore A, C added together are prime to B. Q. E. D.

> Greek: 15. ἐὰν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, δύο ὁποιοιοῦν συντεθέντες πρὸς τὸν λοιπὸν πρῶτοί εἰσιν. ἔστωσαν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς οἱ Α, Β, Γ· λέγω, ὅτι τῶν Α, Β, Γ δύο ὁποιοιοῦν συντεθέντες πρὸς τὸν λοιπὸν πρῶτοί εἰσιν, οἱ μὲν Α, Β πρὸς τὸν Γ, οἱ δὲ Β, Γ πρὸς τὸν Α καὶ ἔτι οἱ Α, Γ πρὸς τὸν Β. εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ δύο οἱ ΔΕ, ΕΖ. φανερὸν δή, ὅτι ὁ μὲν ΔΕ ἑαυτὸν πολλαπλασιάσας τὸν Α πεποίηκεν, τὸν δὲ ΕΖ πολλαπλασιάσας τὸν Β πεποίηκεν, καὶ ἔτι ὁ ΕΖ ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν. καὶ ἐπεὶ οἱ ΔΕ, ΕΖ ἐλάχιστοί εἰσιν, πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐὰν δὲ δύο ἀριθμοί πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ συναμφότερος πρὸς ἑκάτερον πρῶτός ἐστιν· καὶ ὁ ΔΖ ἄρα πρὸς ἑκάτερον τῶν ΔΕ, ΕΖ πρῶτός ἐστιν. ἀλλὰ μὴν καὶ ὁ ΔΕ πρὸς τὸν ΕΖ πρῶτός ἐστιν· οἱ ΔΖ, ΔΕ ἄρα πρὸς τὸν ΕΖ πρῶτοί εἰσιν. ἐὰν δὲ δύο ἀριθμοὶ πρός τινα ἀριθμὸν πρῶτοι ὦσιν, καὶ ὁ ἐξ αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτός ἐστιν· ὥστε ὁ ἐκ τῶν ΖΔ, ΔΕ πρὸς τὸν ΕΖ πρῶτός ἐστιν· ὥστε καὶ ὁ ἐκ τῶν ΖΔ, ΔΕ πρὸς τὸν ἀπὸ τοῦ ΕΖ πρῶτός ἐστιν. [ἐὰν γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ ἐκ τοῦ ἑνὸς αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτός ἐστιν]. ἀλλʼ ὁ ἐκ τῶν ΖΔ, ΔΕ ὁ ἀπὸ τοῦ ΔΕ ἐστι μετὰ τοῦ ἐκ τῶν ΔΕ, ΕΖ· ὁ ἄρα ἀπὸ τοῦ ΔΕ μετὰ τοῦ ἐκ τῶν ΔΕ, ΕΖ πρὸς τὸν ἀπὸ τοῦ ΕΖ πρῶτός ἐστιν. καί ἐστιν ὁ μὲν ἀπὸ τοῦ ΔΕ ὁ Α, ὁ δὲ ἐκ τῶν ΔΕ, ΕΖ ὁ Β, ὁ δὲ ἀπὸ τοῦ ΕΖ ὁ Γ· οἱ Α, Β ἄρα συντεθέντες πρὸς τὸν Γ πρῶτοί εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ Β, Γ πρὸς τὸν Α πρῶτοί εἰσιν. λέγω δή, ὅτι καὶ οἱ Α, Γ πρὸς τὸν Β πρῶτοί εἰσιν. ἐπεὶ γὰρ ὁ ΔΖ πρὸς ἑκάτερον τῶν ΔΕ, ΕΖ πρῶτός ἐστιν, καὶ ὁ ἀπὸ τοῦ ΔΖ πρὸς τὸν ἐκ τῶν ΔΕ, ΕΖ πρῶτός ἐστιν. ἀλλὰ τῷ ἀπὸ τοῦ ΔΖ ἴσοι εἰσὶν οἱ ἀπὸ τῶν ΔΕ, ΕΖ μετὰ τοῦ δὶς ἐκ τῶν ΔΕ, ΕΖ· καὶ οἱ ἀπὸ τῶν ΔΕ, ΕΖ ἄρα μετὰ τοῦ δὶς ὑπὸ τῶν ΔΕ, ΕΖ πρὸς τὸν ὑπὸ τῶν ΔΕ, ΕΖ πρῶτοί [εἰσι]. διελόντι οἱ ἀπὸ τῶν ΔΕ, ΕΖ μετὰ τοῦ ἅπαξ ὑπὸ ΔΕ, ΕΖ πρὸς τὸν ὑπὸ ΔΕ, ΕΖ πρῶτοί εἰσιν. ἔτι διελόντι οἱ ἀπὸ τῶν ΔΕ, ΕΖ ἄρα πρὸς τὸν ὑπὸ ΔΕ, ΕΖ πρῶτοί εἰσιν. καί ἐστιν ὁ μὲν ἀπὸ τοῦ ΔΕ ὁ Α, ὁ δὲ ὑπὸ τῶν ΔΕ, ΕΖ ὁ Β, ὁ δὲ ἀπὸ τοῦ ΕΖ ὁ Γ. οἱ Α, Γ ἄρα συντεθέντες πρὸς τὸν Β πρῶτοί εἰσιν· ὅπερ ἔδει δεῖξαι.