ST EW hþ.
τετράγωνον ἀριθμόν. a(nother) square number. ELEMENTS BOOK 8
᾿Επεὶ γὰρ οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν, τῶν Α, Β ἄρα For since A and B are similar plane numbers, one
εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. ἐμπιπτέτω καὶ ἔστω ὁ number thus falls (between) A and B in mean propor-
kzþ
Γ, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον tion [Prop. 8.18]. Let it (so) fall, and let it be C. And
ἐχόντων τοῖς Α, Γ, Β οἱ Δ, Ε, Ζ· οἱ ἄρα ἄκροι αὐτῶν οἱ let the least numbers, D, E, F, having the same ratio
Δ, Ζ τετράγωνοί εἰσιν. καὶ ἐπεί ἐστιν ὡς ὁ Δ πρὸς τὸν Ζ, as A, C, B have been taken [Prop. 8.2]. The outermost
οὕτως ὁ Α πρὸς τὸν Β, καί εἰσιν οἱ Δ, Ζ τετράγωνοι, ὁ Α of them, D and F, are thus square [Prop. 8.2 corr.]. And
ἄρα πρὸς τὸν Β λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς since as D is to F, so A (is) to B, and D and F are square,
τετράγωνον ἀριθμόν· ὅπερ ἔδει δεῖξαι. A thus has to B the ratio which (some) square number
(has) to a(nother) square number. (Which is) the very
thing it was required to show.
Proposition 27
.
Οἱ ὅμοιοι στερεοὶ ἀριθμοὶ πρὸς ἀλλήλους λόγον ἔχου- Similar solid numbers have to one another the ratio
σιν, ὃν κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν. which (some) cube number (has) to a(nother) cube num-
ber.
Α Ε A E
Γ Ζ C F
∆ Η D G
Β Θ B H
῎Εστωσαν ὅμοιοι στερεοὶ ἀριθμοὶ οἱ Α, Β· λέγω, ὅτι ὁ Let A and B be similar solid numbers. I say that A
Α πρὸς τὸν Β λόγον ἔχει, ὃν κύβος ἀριθμὸς πρὸς κύβον has to B the ratio which (some) cube number (has) to
ἀριθμόν. a(nother) cube number.
᾿Επεὶ γὰρ οἱ Α, Β ὅμοιοι στερεοί εἰσιν, τῶν Α, Β ἄρα δύο For since A and B are similar solid (numbers), two
μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. ἐμπιπτέτωσαν οἱ Γ, numbers thus fall (between) A and B in mean proportion
Δ, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον [Prop. 8.19]. Let C and D have (so) fallen. And let the
ἐχόντων τοῖς Α, Γ, Δ, Β ἴσοι αὐτοῖς τὸ πλῆθος οἱ Ε, Ζ, Η, least numbers, E, F, G, H, having the same ratio as A,
Θ· οἱ ἄρα ἄκροι αὐτῶν οἱ Ε, Θ κύβοι εἰσίν. καί ἐστιν ὡς ὁ C, D, B, (and) equal in multitude to them, have been
Ε πρὸς τὸν Θ, οὕτως ὁ Α πρὸς τὸν Β· καὶ ὁ Α ἄρα πρὸς taken [Prop. 8.2]. Thus, the outermost of them, E and
τὸν Β λόγον ἔχει, ὃν κύβος ἀριθμὸς πρὸς κύβον ἀριθμόν· H, are cube [Prop. 8.2 corr.]. And as E is to H, so A (is)
ὅπερ ἔδει δεῖξαι. to B. And thus A has to B the ratio which (some) cube
number (has) to a(nother) cube number. (Which is) the
very thing it was required to show.
251
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ELEMENTS BOOK 9
Applications of Number Theory †
† The propositions contained in Books 7–9 are generally attributed to the school of Pythagoras.
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.
Proposition 1
᾿Εὰν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ πολλαπλασιάσαντες If two similar plane numbers make some (number by)
ἀλλήλους ποιῶσί τινα, ὁ γενόμενος τετράγωνος ἔσται. multiplying one another then the created (number) will
be square.
Α A
Β B
Γ C
∆ D
῎Εστωσαν δύο ὅμοιοι ἐπίπεδοι ἀριθμοὶ οἱ Α, Β, καὶ ὁ Let A and B be two similar plane numbers, and let A
Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ make C (by) multiplying B. I say that C is square.
τετράγωνός ἐστιν. For let A make D (by) multiplying itself. D is thus
῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω. ὁ Δ square. Therefore, since A has made D (by) multiply-
ἄρα τετράγωνός ἐστιν. ἐπεὶ οὖν ὁ Α ἑαυτὸν μὲν πολλα- ing itself, and has made C (by) multiplying B, thus as A
πλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν is to B, so D (is) to C [Prop. 7.17]. And since A and
Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ B are similar plane numbers, one number thus falls (be-
πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί, tween) A and B in mean proportion [Prop. 8.18]. And if
bþ
τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει ἀριθμός. ἐὰν δὲ (some) numbers fall between two numbers in continued
δύο ἀριθμῶν μεταξὺ κατὰ τὸ συνεχὲς ἀνάλογον ἐμπίπτωσιν proportion then, as many (numbers) as fall in (between)
ἀριθμοί, ὅσοι εἰς αὐτοὺς ἐμπίπτουσι, τοσοῦτοι καὶ εἰς τοὺς them (in continued proportion), so many also (fall) in
τὸν αὐτὸν λόγον ἔχοντας· ὥστε καὶ τῶν Δ, Γ εἷς μέσος (between numbers) having the same ratio (as them in
ἀνάλογον ἐμπίπτει ἀριθμός. καί ἐστι τετράγωνος ὁ Δ· continued proportion) [Prop. 8.8]. And hence one num-
τετράγωνος ἄρα καὶ ὁ Γ· ὅπερ ἔδει δεῖξαι. ber falls (between) D and C in mean proportion. And D
is square. Thus, C (is) also square [Prop. 8.22]. (Which
is) the very thing it was required to show.
Proposition 2
.
᾿Εὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσι If two numbers make a square (number by) multiply-
τετράγωνον, ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί. ing one another then they are similar plane numbers.
A
Α
Β B
Γ C
∆ D
῎Εστωσαν δύο ἀριθμοὶ οἱ Α, Β, καὶ ὁ Α τὸν Β πολλα- Let A and B be two numbers, and let A make the
πλασιάσας τετράγωνον τὸν Γ ποιείτω· λέγω, ὅτι οἱ Α, Β square (number) C (by) multiplying B. I say that A and
ὅμοιοι ἐπίπεδοί εἰσιν ἀριθμοί. B are similar plane numbers.
῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· ὁ Δ For let A make D (by) multiplying itself. Thus, D is
ἄρα τετράγωνός ἐστιν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλα- square. And since A has made D (by) multiplying itself,
πλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν and has made C (by) multiplying B, thus as A is to B, so
Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν D (is) to C [Prop. 7.17]. And since D is square, and C
Γ. καὶ ἐπεὶ ὁ Δ τετράγωνός ἐστιν, ἀλλὰ καὶ ὁ Γ, οἱ Δ, Γ (is) also, D and C are thus similar plane numbers. Thus,
ἄρα ὅμοιοι ἐπίπεδοί εἰσιν. τῶν Δ, Γ ἄρα εἷς μέσος ἀνάλογον one (number) falls (between) D and C in mean propor-
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ἐμπίπτει. καί ἐστιν ὡς ὁ Δ πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν tion [Prop. 8.18]. And as D is to C, so A (is) to B. Thus,
Β· καὶ τῶν Α, Β ἄρα εἷς μέσος ἀνάλογον ἐμπίπτει. ἐὰν δὲ one (number) also falls (between) A and B in mean pro-
δύο ἀριθμῶν εἷς μέσος ἀνάλογον ἐμπίπτῃ, ὅμοιοι ἐπίπεδοί portion [Prop. 8.8]. And if one (number) falls (between)
εἰσιν [οἱ] ἀριθμοί· οἱ ἄρα Α, Β ὅμοιοί εἰσιν ἐπίπεδοι· ὅπερ two numbers in mean proportion then [the] numbers are
ἔδει δεῖξαι. similar plane (numbers) [Prop. 8.20]. Thus, A and B are
similar plane (numbers). (Which is) the very thing it was
required to show.
Proposition 3
.
᾿Εὰν κύβος ἀριθμὸς ἑαυτὸν πολλαπλασιάσας ποιῇ τινα, If a cube number makes some (number by) multiply-
ὁ γενόμενος κύβος ἔσται. ing itself then the created (number) will be cube.
Α A
Β B
Γ C
∆ D
Κύβος γὰρ ἀριθμὸς ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β For let the cube number A make B (by) multiplying
ποιείτω· λέγω, ὅτι ὁ Β κύβος ἐστίν. itself. I say that B is cube.
Εἰλήφθω γὰρ τοῦ Α πλευρὰ ὁ Γ, καὶ ὁ Γ ἑαυτὸν πολλα- For let the side C of A have been taken. And let C
πλασιάσας τὸν Δ ποιείτω. φανερὸν δή ἐστιν, ὅτι ὁ Γ τὸν Δ make D by multiplying itself. So it is clear that C has
πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ ἐπεὶ ὁ Γ ἑαυτὸν πολ- made A (by) multiplying D. And since C has made D
λαπλασιάσας τὸν Δ πεποίηκεν, ὁ Γ ἄρα τὸν Δ μετρεῖ κατὰ (by) multiplying itself, C thus measures D according to
τὰς ἐν αὑτῷ μονάδας. ἀλλὰ μὴν καὶ ἡ μονὰς τὸν Γ μετρεῖ the units in it [Def. 7.15]. But, in fact, a unit also mea-
κατὰ τὰς ἐν αὐτῷ μονάδας· ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν sures C according to the units in it [Def. 7.20]. Thus, as
Γ, ὁ Γ πρὸς τὸν Δ. πάλιν, ἐπεὶ ὁ Γ τὸν Δ πολλαπλασιάσας a unit is to C, so C (is) to D. Again, since C has made A
τὸν Α πεποίηκεν, ὁ Δ ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ (by) multiplying D, D thus measures A according to the
μονάδας. μετρεῖ δὲ καὶ ἡ μονὰς τὸν Γ κατὰ τὰς ἐν αὐτῷ units in C. And a unit also measures C according to the
μονάδας· ἔστιν ἄρα ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Δ πρὸς τὸν units in it. Thus, as a unit is to C, so D (is) to A. But,
Α. ἀλλ᾿ ὡς ἡ μονὰς πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Δ· καὶ ὡς as a unit (is) to C, so C (is) to D. And thus as a unit (is)
ἄρα ἡ μονὰς πρὸς τὸν Γ, οὕτως ὁ Γ πρὸς τὸν Δ καὶ ὁ Δ to C, so C (is) to D, and D to A. Thus, two numbers, C
πρὸς τὸν Α. τῆς ἄρα μονάδος καὶ τοῦ Α ἀριθμοῦ δύο μέσοι and D, have fallen (between) a unit and the number A
ἀνάλογον κατὰ τὸ συνεχὲς ἐμπεπτώκασιν ἀριθμοὶ οἱ Γ, Δ. in continued mean proportion. Again, since A has made
πάλιν, ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, B (by) multiplying itself, A thus measures B according
ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας· μετρεῖ δὲ to the units in it. And a unit also measures A according
καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας· ἔστιν ἄρα ὡς to the units in it. Thus, as a unit is to A, so A (is) to B.
dþ
ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς τὸν Β. τῆς δὲ μονάδος καὶ And two numbers have fallen (between) a unit and A in
τοῦ Α δύο μέσοι ἀνάλογον ἐμπεπτώκασιν ἀριθμοί· καὶ τῶν mean proportion. Thus two numbers will also fall (be-
Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπεσοῦνται ἀριθμοί. ἐὰν δὲ tween) A and B in mean proportion [Prop. 8.8]. And if
δύο ἀριθμῶν δύο μέσοι ἀνάλογον ἐμπίπτωσιν, ὁ δὲ πρῶτος two (numbers) fall (between) two numbers in mean pro-
κύβος ᾖ, καὶ ὁ δεύτερος κύβος ἔσται. καί ἐστιν ὁ Α κύβος· portion, and the first (number) is cube, then the second
καὶ ὁ Β ἄρα κύβος ἐστίν· ὅπερ ἔδει δεῖξαι. will also be cube [Prop. 8.23]. And A is cube. Thus, B
is also cube. (Which is) the very thing it was required to
show.
Proposition 4
.
᾿Εὰν κύβος ἀριθμὸς κύβον ἀριθμὸν πολλαπλασιάσας If a cube number makes some (number by) multiply-
ποιῇ τινα, ὁ γενόμενος κύβος ἔσται. ing a(nother) cube number then the created (number)
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will be cube. ELEMENTS BOOK 9
Α A
Β B
Γ C
∆ D
Κύβος γὰρ ἀριθμὸς ὁ Α κύβον ἀριθμὸν τὸν Β πολλα- For let the cube number A make C (by) multiplying
πλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ κύβος ἐστίν. the cube number B. I say that C is cube.
῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· ὁ For let A make D (by) multiplying itself. Thus, D is
Δ ἄρα κύβος ἐστίν. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλα- cube [Prop. 9.3]. And since A has made D (by) multi-
eþ
πλασιάσας τὸν Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν plying itself, and has made C (by) multiplying B, thus
Γ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς as A is to B, so D (is) to C [Prop. 7.17]. And since A
τὸν Γ. καὶ ἐπεὶ οἱ Α, Β κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν οἱ Α, and B are cube, A and B are similar solid (numbers).
Β. τῶν Α, Β ἄρα δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί· Thus, two numbers fall (between) A and B in mean pro-
ὥστε καὶ τῶν Δ, Γ δύο μέσοι ἀνάλογον ἐμπεσοῦνται portion [Prop. 8.19]. Hence, two numbers will also fall
ἀριθμοί. καί ἐστι κύβος ὁ Δ· κύβος ἄρα καὶ ὁ Γ· ὅπερ (between) D and C in mean proportion [Prop. 8.8]. And
ἔδει δεῖξαι. D is cube. Thus, C (is) also cube [Prop. 8.23]. (Which
is) the very thing it was required to show.
Proposition 5
.
᾿Εὰν κύβος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας κύβον If a cube number makes a(nother) cube number (by)
ποιῇ, καὶ ὁ πολλαπλασιασθεὶς κύβος ἔσται. multiplying some (number) then the (number) multi-
plied will also be cube.
Α A
Β B
Γ C
∆ D
Κύβος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλα- For let the cube number A make the cube (number)
πλασιάσας κύβον τὸν Γ ποιείτω· λέγω, ὅτι ὁ Β κύβος ἐστίν. C (by) multiplying some number B. I say that B is cube.
῾Ο γὰρ Α ἑαυτὸν πολλαπλασιάσας τὸν Δ ποιείτω· κύβος For let A make D (by) multiplying itself. D is thus
ἄρα ἐστίν ὁ Δ. καὶ ἐπεὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν cube [Prop. 9.3]. And since A has made D (by) multiply-
Δ πεποίηκεν, τὸν δὲ Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ing itself, and has made C (by) multiplying B, thus as A
þ
ἔστιν ἀρα ὡς ὁ Α πρὸς τὸν Β, ὁ Δ πρὸς τὸν Γ. καὶ ἐπεὶ is to B, so D (is) to C [Prop. 7.17]. And since D and C
οἱ Δ, Γ κύβοι εἰσίν, ὅμοιοι στερεοί εἰσιν. τῶν Δ, Γ ἄρα are (both) cube, they are similar solid (numbers). Thus,
δύο μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. καί ἐστιν ὡς ὁ Δ two numbers fall (between) D and C in mean proportion
πρὸς τὸν Γ, οὕτως ὁ Α πρὸς τὸν Β· καὶ τῶν Α, Β ἄρα δύο [Prop. 8.19]. And as D is to C, so A (is) to B. Thus,
μέσοι ἀνάλογον ἐμπίπτουσιν ἀριθμοί. καί ἐστι κύβος ὁ Α· two numbers also fall (between) A and B in mean pro-
κύβος ἄρα ἐστὶ καὶ ὁ Β· ὅπερ ἔδει δεῖξαι. portion [Prop. 8.8]. And A is cube. Thus, B is also cube
[Prop. 8.23]. (Which is) the very thing it was required to
show.
Proposition 6
.
᾿Εὰν ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ If a number makes a cube (number by) multiplying
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ELEMENTS BOOK 9
αὐτὸς κύβος ἔσται. itself then it itself will also be cube.
Α A
Β B
Γ C
᾿Αριθμὸς γὰρ ὁ Α ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β For let the number A make the cube (number) B (by)
ποιείτω· λέγω, ὅτι καὶ ὁ Α κύβος ἐστίν. multiplying itself. I say that A is also cube.
῾Ο γὰρ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω. ἐπεὶ οὖν For let A make C (by) multiplying B. Therefore, since
ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ A has made B (by) multiplying itself, and has made C
Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Γ ἄρα κύβος ἐστίν. (by) multiplying B, C is thus cube. And since A has made
καὶ ἐπεὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν, ὁ B (by) multiplying itself, A thus measures B according to
Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας. μετρεῖ δὲ the units in (A). And a unit also measures A according
καὶ ἡ μονὰς τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας. ἔστιν ἄρα to the units in it. Thus, as a unit is to A, so A (is) to
ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α πρὸς τὸν Β. καὶ ἐπεὶ B. And since A has made C (by) multiplying B, B thus
ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Β ἄρα τὸν measures C according to the units in A. And a unit also
Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεὶ δὲ καὶ ἡ μονὰς measures A according to the units in it. Thus, as a unit is
τὸν Α κατὰ τὰς ἐν αὐτῷ μονάδας. ἔστιν ἄρα ὡς ἡ μονὰς to A, so B (is) to C. But, as a unit (is) to A, so A (is) to
zþ
πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Γ. ἀλλ᾿ ὡς ἡ μονὰς πρὸς B. And thus as A (is) to B, (so) B (is) to C. And since B
τὸν Α, οὕτως ὁ Α πρὸς τὸν Β· καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, and C are cube, they are similar solid (numbers). Thus,
ὁ Β πρὸς τὸν Γ. καὶ ἐπεὶ οἱ Β, Γ κύβοι εἰσίν, ὅμοιοι στερεοί there exist two numbers in mean proportion (between)
εἰσιν. τῶν Β, Γ ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί. καί B and C [Prop. 8.19]. And as B is to C, (so) A (is) to B.
ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β. καὶ τῶν Α, Β Thus, there also exist two numbers in mean proportion
ἄρα δύο μέσοι ἀνάλογόν εἰσιν ἀριθμοί. καί ἐστιν κύβος ὁ (between) A and B [Prop. 8.8]. And B is cube. Thus, A
Β· κύβος ἄρα ἐστὶ καὶ ὁ Α· ὅπερ ἔδει δεὶξαι. is also cube [Prop. 8.23]. (Which is) the very thing it was
required to show.
Proposition 7
.
᾿Εὰν σύνθετος ἀριθμὸς ἀριθμόν τινα πολλαπλασιάσας If a composite number makes some (number by) mul-
ποιῇ τινα, ὁ γενόμενος στερεὸς ἔσται. tiplying some (other) number then the created (number)
will be solid.
Α A
Β B
Γ C
∆ D
Ε E
Σύνθετος γὰρ ἀριθμὸς ὁ Α ἀριθμόν τινα τὸν Β πολλα- For let the composite number A make C (by) multi-
πλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ στερεός ἐστιν. plying some number B. I say that C is solid.
᾿Επεὶ γὰρ ὁ Α σύνθετός ἐστιν, ὑπὸ ἀριθμοῦ τινος με- For since A is a composite (number), it will be mea-
τρηθήσεται. μετρείσθω ὑπὸ τοῦ Δ, καὶ ὁσάκις ὁ Δ τὸν Α sured by some number. Let it be measured by D. And,
μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν ὁ Δ as many times as D measures A, so many units let there
τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, ὁ Ε ἄρα τὸν Δ be in E. Therefore, since D measures A according to
πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ ἐπεὶ ὁ Α τὸν Β πολ- the units in E, E has thus made A (by) multiplying D
λαπλασιάσας τὸν Γ πεποίηκεν, ὁ δὲ Α ἐστιν ὁ ἐκ τῶν Δ, Ε, [Def. 7.15]. And since A has made C (by) multiplying B,
ὁ ἄρα ἐκ τῶν Δ, Ε τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. and A is the (number created) from (multiplying) D, E,
ὁ Γ ἄρα στερεός ἐστιν, πλευραὶ δὲ αὐτοῦ εἰσιν οἱ Δ, Ε, Β· the (number created) from (multiplying) D, E has thus
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hþ ELEMENTS BOOK 9
ὅπερ ἔδει δεῖξαι. made C (by) multiplying B. Thus, C is solid, and its sides
are D, E, B. (Which is) the very thing it was required to
show.
Proposition 8
.
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu-
ὦσιν, ὁ μὲν τρίτος ἀπὸ τῆς μονάδος τετράγωνος ἔσται ously proportional, (starting) from a unit, then the third
καὶ οἱ ἕνα διαλείποντες, ὁ δὲ τέταρτος κύβος καὶ οἱ from the unit will be square, and (all) those (numbers
δύο διαλείποντες πάντες, ὁ δὲ ἕβδομος κύβος ἅμα καὶ after that) which leave an interval of one (number), and
τετράγωνος καὶ οἱ πέντε διαλείποντες. the fourth (will be) cube, and all those (numbers after
that) which leave an interval of two (numbers), and the
seventh (will be) both cube and square, and (all) those
(numbers after that) which leave an interval of five (num-
bers).
Α A
Β B
Γ C
∆ D
Ε E
Ζ F
῎Εστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογ- Let any multitude whatsoever of numbers, A, B, C,
ον οἱ Α, Β, Γ, Δ, Ε, Ζ· λέγω, ὅτι ὁ μὲν τρίτος ἀπὸ D, E, F, be continuously proportional, (starting) from
τῆς μονάδος ὁ Β τετράγωνός ἐστι καὶ οἱ ἕνα διαλείποντες a unit. I say that the third from the unit, B, is square,
πάντες, ὁ δὲ τέταρτος ὁ Γ κύβος καὶ οἱ δύο διαλείποντες and all those (numbers after that) which leave an inter-
πάντες, ὁ δὲ ἕβδομος ὁ Ζ κύβος ἅμα καὶ τετράγωνος καὶ οἱ val of one (number). And the fourth (from the unit), C,
πέντε διαλείποντες πάντες. (is) cube, and all those (numbers after that) which leave
᾿Επεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Α an interval of two (numbers). And the seventh (from the
πρὸς τὸν Β, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ unit), F, (is) both cube and square, and all those (num-
ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς bers after that) which leave an interval of five (numbers).
ἐν αὐτῷ μονάδας· καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν For since as the unit is to A, so A (is) to B, the unit
τῷ Α μονάδας. ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β thus measures the number A the same number of times
πεποίηκεν· τετράγωνος ἄρα ἐστὶν ὁ Β. καὶ ἐπεὶ οἱ Β, Γ, Δ as A (measures) B [Def. 7.20]. And the unit measures
ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Β τετράγωνός ἐστιν, καὶ ὁ Δ ἄρα the number A according to the units in it. Thus, A also
τετράγωνός ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ζ τετράγωνός measures B according to the units in A. A has thus made
ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ ἕνα διαλείποντες B (by) multiplying itself [Def. 7.15]. Thus, B is square.
πάντες τετράγωνοί εἰσιν. λέγω δή, ὅτι καὶ ὁ τέταρτος ἀπὸ And since B, C, D are continuously proportional, and B
τῆς μονάδος ὁ Γ κύβος ἐστὶ καὶ οἱ δύο διαλείποντες πάντες. is square, D is thus also square [Prop. 8.22]. So, for the
ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν same (reasons), F is also square. So, similarly, we can
Γ, ἰσάκις ἄρα ἡ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Γ. ἡ also show that all those (numbers after that) which leave
δὲ μονὰς τὸν Α ἀριθμὸν μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας· an interval of one (number) are square. So I also say that
καὶ ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας· ὁ Α the fourth (number) from the unit, C, is cube, and all
ἄρα τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν. ἐπεὶ οὖν ὁ those (numbers after that) which leave an interval of two
Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Β πεποίηκεν, τὸν δὲ Β (numbers). For since as the unit is to A, so B (is) to C,
πολλαπλασιάσας τὸν Γ πεποίηκεν, κύβος ἄρα ἐστὶν ὁ Γ. καὶ the unit thus measures the number A the same number
ἐπεὶ οἱ Γ, Δ, Ε, Ζ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ Γ κύβος ἐστίν, of times that B (measures) C. And the unit measures the
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ELEMENTS BOOK 9
καὶ ὁ Ζ ἄρα κύβος ἐστίν. ἐδείχθη δὲ καὶ τετράγωνος· ὁ ἄρα number A according to the units in A. And thus B mea-
ἕβδομος ἀπὸ τῆς μονάδος κύβος τέ ἐστι καὶ τετράγωνος. sures C according to the units in A. A has thus made C
ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ πέντε διαλείποντες πάντες (by) multiplying B. Therefore, since A has made B (by)
κύβοι τέ εἰσι καὶ τετράγωνοι· ὅπερ ἔδει δεῖξαι. multiplying itself, and has made C (by) multiplying B, C
jþ And it was also shown (to be) square. Thus, the seventh
is thus cube. And since C, D, E, F are continuously pro-
portional, and C is cube, F is thus also cube [Prop. 8.23].
(number) from the unit is (both) cube and square. So,
similarly, we can show that all those (numbers after that)
which leave an interval of five (numbers) are (both) cube
and square. (Which is) the very thing it was required to
show.
Proposition 9
.
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἑξῆς κατὰ τὸ συνεχὲς If any multitude whatsoever of numbers is continu-
ἀριθμοὶ ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα τετράγωνος ously proportional, (starting) from a unit, and the (num-
ᾖ, καὶ οἱ λοιποὶ πάντες τετράγωνοι ἔσονται. καὶ ἐὰν ὁ μετὰ ber) after the unit is square, then all the remaining (num-
τὴν μονάδα κύβος ᾖ, καὶ οἱ λοιποὶ πάντες κύβοι ἔσονται. bers) will also be square. And if the (number) after the
unit is cube, then all the remaining (numbers) will also
be cube.
Α A
Β B
Γ C
∆ D
Ε E
Ζ F
῎Εστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν Let any multitude whatsoever of numbers, A, B, C,
ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα D, E, F, be continuously proportional, (starting) from a
ὁ Α τετράγωνος ἔστω· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες unit. And let the (number) after the unit, A, be square. I
τετράγωνοι ἔσονται. say that all the remaining (numbers) will also be square.
῞Οτι μὲν οὖν ὁ τρίτος ἀπὸ τῆς μονάδος ὁ Β τετράγωνός In fact, it has (already) been shown that the third
ἐστι καὶ οἱ ἕνα διαπλείποντες πάντες, δέδεικται· λέγω [δή], (number) from the unit, B, is square, and all those (num-
ὅτι καὶ οἱ λοιποὶ πάντες τετράγωνοί εἰσιν. ἐπεὶ γὰρ οἱ Α, bers after that) which leave an interval of one (number)
Β, Γ ἑξῆς ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α τετράγωνος, καὶ ὁ [Prop. 9.8]. [So] I say that all the remaining (num-
Γ [ἄρα] τετράγωνος ἐστιν. πάλιν, ἐπεὶ [καὶ] οἱ Β, Γ, Δ ἑξῆς bers) are also square. For since A, B, C are continu-
ἀνάλογόν εἰσιν, καί ἐστιν ὁ Β τετράγωνος, καὶ ὁ Δ [ἄρα] ously proportional, and A (is) square, C is [thus] also
τετράγωνός ἐστιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ οἱ λοιποὶ square [Prop. 8.22]. Again, since B, C, D are [also] con-
πάντες τετράγωνοί εἰσιν. tinuously proportional, and B is square, D is [thus] also
᾿Αλλὰ δὴ ἔστω ὁ Α κύβος· λέγω, ὅτι καὶ οἱ λοιποὶ πάντες square [Prop. 8.22]. So, similarly, we can show that all
κύβοι εἰσίν. the remaining (numbers) are also square.
῞Οτι μὲν οὖν ὁ τέταρτος ἀπὸ τῆς μονάδος ὁ Γ κύβος ἐστὶ And so let A be cube. I say that all the remaining
καὶ οἱ δύο διαλείποντες πάντες, δέδεικται· λέγω [δή], ὅτι καὶ (numbers) are also cube.
οἱ λοιποὶ πάντες κύβοι εἰσίν. ἐπεὶ γάρ ἐστιν ὡς ἡ μονὰς πρὸς In fact, it has (already) been shown that the fourth
τὸν Α, οὕτως ὁ Α πρὸς τὸν Β, ἰσάκις ἀρα ἡ μονὰς τὸν Α (number) from the unit, C, is cube, and all those (num-
μετρεῖ καὶ ὁ Α τὸν Β. ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν bers after that) which leave an interval of two (numbers)
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αὐτῷ μονάδας· καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ [Prop. 9.8]. [So] I say that all the remaining (numbers)
μονάδας· ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν. are also cube. For since as the unit is to A, so A (is) to
καί ἐστιν ὁ Α κύβος. ἐὰν δὲ κύβος ἀριθμὸς ἑαυτὸν πολλα- B, the unit thus measures A the same number of times
πλασιάσας ποιῇ τινα, ὁ γενόμενος κύβος ἐστίν· καὶ ὁ Β ἄρα as A (measures) B. And the unit measures A according
κύβος ἐστίν. καὶ ἐπεὶ τέσσαρες ἀριθμοὶ οἱ Α, Β, Γ, Δ ἑξῆς to the units in it. Thus, A also measures B according to
ἀνάλογόν εἰσιν, καί ἐστιν ὁ Α κύβος, καὶ ὁ Δ ἄρα κύβος the units in (A). A has thus made B (by) multiplying it-
iþ four numbers A, B, C, D are continuously proportional,
ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε κύβος ἐστίν, καὶ ὁμοίως οἱ self. And A is cube. And if a cube number makes some
λοιποὶ πάντες κύβοι εἰσίν· ὅπερ ἔδει δεῖξαι. (number by) multiplying itself then the created (number)
is cube [Prop. 9.3]. Thus, B is also cube. And since the
and A is cube, D is thus also cube [Prop. 8.23]. So, for
the same (reasons), E is also cube, and, similarly, all the
remaining (numbers) are cube. (Which is) the very thing
it was required to show.
Proposition 10
.
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ [ἑξῆς] ἀνάλογον If any multitude whatsoever of numbers is [continu-
ὦσιν, ὁ δὲ μετὰ τὴν μονάδα μὴ ᾖ τετράγωνος, οὐδ᾿ ἄλλος ously] proportional, (starting) from a unit, and the (num-
οὐδεὶς τετράγωνος ἔσται χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος ber) after the unit is not square, then no other (number)
καὶ τῶν ἕνα διαλειπόντων πάντων. καὶ ἐὰν ὁ μετὰ τὴν will be square either, apart from the third from the unit,
μονάδα κύβος μὴ ᾖ, οὐδὲ ἄλλος οὐδεὶς κύβος ἔσται χωρὶς and all those (numbers after that) which leave an inter-
τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων val of one (number). And if the (number) after the unit
πάντων. is not cube, then no other (number) will be cube either,
apart from the fourth from the unit, and all those (num-
bers after that) which leave an interval of two (numbers).
Α A
Β B
Γ C
∆ D
Ε E
Ζ F
῎Εστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν Let any multitude whatsoever of numbers, A, B, C,
ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ μετὰ τὴν μονάδα ὁ Α μὴ D, E, F, be continuously proportional, (starting) from
ἔστω τετράγωνος· λέγω, ὅτι οὐδὲ ἄλλος οὐδεὶς τετράγωνος a unit. And let the (number) after the unit, A, not be
ἔσται χωρὶς τοῦ τρίτου ἀπὸ τὴς μονάδος [καὶ τῶν ἕνα δια- square. I say that no other (number) will be square ei-
λειπόντων]. ther, apart from the third from the unit [and (all) those
Εἰ γὰρ δυνατόν, ἔστω ὁ Γ τετράγωνος. ἔστι δὲ καὶ ὁ (numbers after that) which leave an interval of one (num-
Β τετράγωνος· οἱ Β, Γ ἄρα πρὸς ἀλλήλους λόγον ἔχου- ber)].
σιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν. καί For, if possible, let C be square. And B is also
ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β· οἱ Α, Β ἄρα square [Prop. 9.8]. Thus, B and C have to one another
πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς (the) ratio which (some) square number (has) to (some
τετράγωνον ἀριθμόν· ὥστε οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν. other) square number. And as B is to C, (so) A (is)
καί ἐστι τετράγωνος ὁ Β· τετράγωνος ἄρα ἐστὶ καὶ ὁ Α· to B. Thus, A and B have to one another (the) ratio
ὅπερ οὐχ ὑπέκειτο. οὐκ ἄρα ὁ Γ τετράγωνός ἐστιν. ὁμοίως which (some) square number has to (some other) square
δὴ δείξομεν, ὅτι οὐδ᾿ ἄλλος οὐδεὶς τετράγωνός ἐστι χωρὶς number. Hence, A and B are similar plane (numbers)
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τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων. [Prop. 8.26]. And B is square. Thus, A is also square.
᾿Αλλὰ δὴ μὴ ἔστω ὁ Α κύβος. λέγω, ὅτι οὐδ᾿ ἄλλος The very opposite thing was assumed. C is thus not
οὐδεὶς κύβος ἔσται χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος square. So, similarly, we can show that no other (number
καὶ τῶν δύο διαλειπόντων. is) square either, apart from the third from the unit, and
Εἰ γὰρ δυνατόν, ἔστω ὁ Δ κύβος. ἔστι δὲ καὶ ὁ Γ κύβος· (all) those (numbers after that) which leave an interval
τέταρτος γάρ ἐστιν ἀπὸ τῆς μονάδος. καί ἐστιν ὡς ὁ Γ πρὸς of one (number).
τὸν Δ, ὁ Β πρὸς τὸν Γ· καὶ ὁ Β ἄρα πρὸς τὸν Γ λόγον ἔχει, And so let A not be cube. I say that no other (num-
ὃν κύβος πρὸς κύβον. καί ἐστιν ὁ Γ κύβος· καὶ ὁ Β ἄρα ber) will be cube either, apart from the fourth from the
κύβος ἐστίν. καὶ ἐπεί ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς unit, and (all) those (numbers after that) which leave an
τὸν Β, ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας, interval of two (numbers).
καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας· ὁ Α For, if possible, let D be cube. And C is also cube
ἄρα ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β πεποίηκεν. ἐὰν [Prop. 9.8]. For it is the fourth (number) from the unit.
δὲ ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ αὐτὸς And as C is to D, (so) B (is) to C. And B thus has to
κύβος ἔσται. κύβος ἄρα καὶ ὁ Α· ὅπερ οὐχ ὑπόκειται. οὐκ C the ratio which (some) cube (number has) to (some
ἄρα ὁ Δ κύβος ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾿ ἄλλος other) cube (number). And C is cube. Thus, B is also
οὐδεὶς κύβος ἐστὶ χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ cube [Props. 7.13, 8.25]. And since as the unit is to
τῶν δύο διαλειπόντων· ὅπερ ἔδει δεῖξαι. A, (so) A (is) to B, and the unit measures A accord-
ing to the units in it, A thus also measures B according
to the units in (A). Thus, A has made the cube (num-
iaþ cube [Prop. 9.6]. Thus, A (is) also cube. The very oppo-
ber) B (by) multiplying itself. And if a number makes a
cube (number by) multiplying itself then it itself will be
site thing was assumed. Thus, D is not cube. So, simi-
larly, we can show that no other (number) is cube either,
apart from the fourth from the unit, and (all) those (num-
bers after that) which leave an interval of two (numbers).
(Which is) the very thing it was required to show.
.
Proposition 11
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu-
ὦσιν, ὁ ἐλάττων τὸν μείζονα μετρεῖ κατά τινα τῶν ὑπαρχόντ- ously proportional, (starting) from a unit, then a lesser
ων ἐν τοῖς ἀνάλογον ἀριθμοῖς. (number) measures a greater according to some existing
(number) among the proportional numbers.
Α A
Β B
Γ C
∆ D
Ε E
῎Εστωσαν ἀπὸ μονάδος τῆς Α ὁποσοιοῦν ἀριθμοὶ ἑξῆς Let any multitude whatsoever of numbers, B, C, D,
ἀνάλογον οἱ Β, Γ, Δ, Ε· λέγω, ὅτι τῶν Β, Γ, Δ, Ε ὁ E, be continuously proportional, (starting) from the unit
ἐλάχιστος ὁ Β τὸν Ε μετρεῖ κατά τινα τῶν Γ, Δ. A. I say that, for B, C, D, E, the least (number), B,
᾿Επεὶ γάρ ἐστιν ὡς ἡ Α μονὰς πρὸς τὸν Β, οὕτως ὁ Δ measures E according to some (one) of C, D.
πρὸς τὸν Ε, ἰσάκις ἄρα ἡ Α μονὰς τὸν Β ἀριθμὸν μετρεῖ For since as the unit A is to B, so D (is) to E, the
καὶ ὁ Δ τὸν Ε· ἐναλλὰξ ἄρα ἰσάκις ἡ Α μονὰς τὸν Δ μετρεῖ unit A thus measures the number B the same number of
καὶ ὁ Β τὸν Ε. ἡ δὲ Α μονὰς τὸν Δ μετρεῖ κατὰ τὰς ἐν times as D (measures) E. Thus, alternately, the unit A
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ELEMENTS BOOK 9
ìri sma
αὐτῷ μονάδας· καὶ ὁ Β ἄρα τὸν Ε μετρεῖ κατὰ τὰς ἐν τῷ measures D the same number of times as B (measures)
Δ μονάδας· ὥστε ὁ ἐλάσσων ὁ Β τὸν μείζονα τὸν Ε με- E [Prop. 7.15]. And the unit A measures D according to
τρεῖ κατά τινα ἀριθμὸν τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον the units in it. Thus, B also measures E according to the
ἀριθμοῖς. units in D. Hence, the lesser (number) B measures the
greater E according to some existing number among the
proportional numbers (namely, D).
ibþ place from the measured (number), in (the direction of
Corollary
.
Καὶ φανερόν, ὅτι ἣν ἔχει τάξιν ὁ μετρῶν ἀπὸ μονάδος, And (it is) clear that what(ever relative) place the
τὴν αὐτὴν ἔχει καὶ ὁ καθ᾿ ὃν μετρεῖ ἀπὸ τοῦ μετρουμένου measuring (number) has from the unit, the (number)
ἐπὶ τὸ πρὸ αὐτοῦ. ὅπερ ἔδει δεῖξαι. according to which it measures has the same (relative)
the number) before it. (Which is) the very thing it was
required to show.
.
Proposition 12
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu-
ὦσιν, ὑφ᾿ ὅσων ἂν ὁ ἔσχατος πρώτων ἀριθμῶν μετρῆται, ously proportional, (starting) from a unit, then however
ὑπὸ τῶν αὐτῶν καὶ ὁ παρὰ τὴν μονάδα μετρηθήσεται. many prime numbers the last (number) is measured by,
the (number) next to the unit will also be measured by
the same (prime numbers).
Α Ε A E
Β Ζ B F
Γ Η C G
∆ Θ D H
῎Εστωσαν ἀπὸ μονάδος ὁποσοιδηποτοῦν ἀριθμοὶ ἀνάλογ- Let any multitude whatsoever of numbers, A, B, C,
ον οἱ Α, Β, Γ, Δ· λέγω, ὅτι ὑφ᾿ ὅσων ἂν ὁ Δ πρώτων D, be (continuously) proportional, (starting) from a unit.
ἀριθμῶν μετρῆται, ὑπὸ τῶν αὐτῶν καὶ ὁ Α μετρηθήσεται. I say that however many prime numbers D is measured
Μετρείσθω γὰρ ὁ Δ ὑπό τινος πρώτου ἀριθμοῦ τοῦ Ε· by, A will also be measured by the same (prime numbers).
λέγω, ὅτι ὁ Ε τὸν Α μετρεῖ. μὴ γάρ· καί ἐστιν ὁ Ε πρῶτος, For let D be measured by some prime number E. I
ἅπας δὲ πρῶτος ἀριθμὸς πρὸς ἅπαντα, ὃν μὴ μετρεῖ, πρῶτός say that E measures A. For (suppose it does) not. E is
ἐστιν· οἱ Ε, Α ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ἐπεὶ prime, and every prime number is prime to every num-
ὁ Ε τὸν Δ μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ· ὁ Ε ἄρα ber which it does not measure [Prop. 7.29]. Thus, E
τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. πάλιν, ἐπεὶ ὁ Α and A are prime to one another. And since E measures
τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας, ὁ Α ἄρα τὸν Γ D, let it measure it according to F. Thus, E has made
πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Ε τὸν Ζ D (by) multiplying F. Again, since A measures D ac-
πολλαπλασιάσας τὸν Δ πεποίηκεν· ὁ ἄρα ἐκ τῶν Α, Γ ἴσος cording to the units in C [Prop. 9.11 corr.], A has thus
ἐστὶ τῷ ἐκ τῶν Ε, Ζ. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Ε, ὁ Ζ made D (by) multiplying C. But, in fact, E has also
πρὸς τὸν Γ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, made D (by) multiplying F. Thus, the (number cre-
οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ated) from (multiplying) A, C is equal to the (number
ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν created) from (multiplying) E, F. Thus, as A is to E,
ἑπόμενον· μετρεῖ ἄρα ὁ Ε τὸν Γ. μετρείτω αὐτὸν κατὰ τὸν (so) F (is) to C [Prop. 7.19]. And A and E (are) prime
Η· ὁ Ε ἄρα τὸν Η πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ (to one another), and (numbers) prime (to one another
μὴν διὰ τὸ πρὸ τούτου καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν are) also the least (of those numbers having the same ra-
Γ πεποίηκεν. ὁ ἄρα ἐκ τῶν Α, Β ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Η. tio as them) [Prop. 7.21], and the least (numbers) mea-
ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Ε, ὁ Η πρὸς τὸν Β. οἱ δὲ Α, Ε sure those (numbers) having the same ratio as them an
πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ equal number of times, the leading (measuring) the lead-
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ELEMENTS BOOK 9
μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας αὐτοῖς ἰσάκις ὅ τε ing, and the following the following [Prop. 7.20]. Thus,
ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· E measures C. Let it measure it according to G. Thus,
μετρεῖ ἄρα ὁ Ε τὸν Β. μετρείτω αὐτὸν κατὰ τὸν Θ· ὁ Ε ἄρα E has made C (by) multiplying G. But, in fact, via the
τὸν Θ πολλαπλασιάσας τὸν Β πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α (proposition) before this, A has also made C (by) multi-
ἑαυτὸν πολλαπλασιάσας τὸν Β πεποίηκεν· ὁ ἄρα ἐκ τῶν Ε, plying B [Prop. 9.11 corr.]. Thus, the (number created)
Θ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Α. ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Α, ὁ Α from (multiplying) A, B is equal to the (number created)
πρὸς τὸν Θ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ from (multiplying) E, G. Thus, as A is to E, (so) G
δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις (is) to B [Prop. 7.19]. And A and E (are) prime (to
ὅ ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· one another), and (numbers) prime (to one another are)
μετρεῖ ἄρα ὁ Ε τὸν Α ὡς ἡγούμενος ἡγούμενον. ἀλλὰ μὴν also the least (of those numbers having the same ratio
καὶ οὐ μετρεῖ· ὅπερ ἀδύνατον. οὐκ ἄρα οἱ Ε, Α πρῶτοι πρὸς as them) [Prop. 7.21], and the least (numbers) measure
ἀλλήλους εἰσίν. σύνθετοι ἄρα. οἱ δὲ σύνθετοι ὑπὸ [πρώτου] those (numbers) having the same ratio as them an equal
ἀριθμοῦ τινος μετροῦνται. καὶ ἐπεὶ ὁ Ε πρῶτος ὑπόκειται, ὁ number of times, the leading (measuring) the leading,
δὲ πρῶτος ὑπὸ ἑτέρου ἀριθμοῦ οὐ μετρεῖται ἢ ὑφ᾿ ἑαυτοῦ, ὁ and the following the following [Prop. 7.20]. Thus, E
Ε ἄρα τοὺς Α, Ε μετρεῖ· ὥστε ὁ Ε τὸν Α μετρεῖ. μετρεῖ δὲ measures B. Let it measure it according to H. Thus,
καὶ τὸν Δ· ὁ Ε ἄρα τοὺς Α, Δ μετρεῖ. ὁμοίως δὴ δείξομεν, E has made B (by) multiplying H. But, in fact, A has
ὅτι ὑφ᾿ ὅσων ἂν ὁ Δ πρώτων ἀριθμῶν μετρῆται, ὑπὸ τῶν also made B (by) multiplying itself [Prop. 9.8]. Thus,
αὐτῶν καὶ ὁ Α μετρηθήσεται· ὅπερ ἔδει δεῖξαι. the (number created) from (multiplying) E, H is equal
to the (square) on A. Thus, as E is to A, (so) A (is)
to H [Prop. 7.19]. And A and E are prime (to one an-
other), and (numbers) prime (to one another are) also
the least (of those numbers having the same ratio as
them) [Prop. 7.21], and the least (numbers) measure
those (numbers) having the same ratio as them an equal
number of times, the leading (measuring) the leading,
and the following the following [Prop. 7.20]. Thus, E
measures A, as the leading (measuring the) leading. But,
in fact, (E) also does not measure (A). The very thing
(is) impossible. Thus, E and A are not prime to one
another. Thus, (they are) composite (to one another).
And (numbers) composite (to one another) are (both)
measured by some [prime] number [Def. 7.14]. And
igþ [Def. 7.11], E thus measures (both) A and E. Hence, E
since E is assumed (to be) prime, and a prime (number)
is not measured by another number (other) than itself
measures A. And it also measures D. Thus, E measures
(both) A and D. So, similarly, we can show that however
many prime numbers D is measured by, A will also be
measured by the same (prime numbers). (Which is) the
very thing it was required to show.
Proposition 13
.
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον If any multitude whatsoever of numbers is continu-
ὦσιν, ὁ δὲ μετὰ τὴν μονάδα πρῶτος ᾖ, ὁ μέγιστος ὑπ᾿ ously proportional, (starting) from a unit, and the (num-
οὐδενὸς [ἄλλου] μετρηθήσεται παρὲξ τῶν ὑπαρχόντων ἐν ber) after the unit is prime, then the greatest (number)
τοῖς ἀνάλογον ἀριθμοῖς. will be measured by no [other] (numbers) except (num-
῎Εστωσαν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον bers) existing among the proportional numbers.
οἱ Α, Β, Γ, Δ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α πρῶτος ἔστω· Let any multitude whatsoever of numbers, A, B, C,
λέγω, ὅτι ὁ μέγιστος αὐτῶν ὁ Δ ὑπ᾿ οὐδενὸς ἄλλου με- D, be continuously proportional, (starting) from a unit.
τρηθήσεται παρὲξ τῶν Α, Β, Γ. And let the (number) after the unit, A, be prime. I say
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ELEMENTS BOOK 9
that the greatest of them, D, will be measured by no other
(numbers) except A, B, C.
Α Ε A E
Β Ζ B F
Γ Η C G
∆ Θ D H
Εἰ γὰρ δυνατόν, μετρείσθω ὑπὸ τοῦ Ε, καὶ ὁ Ε μηδενὶ For, if possible, let it be measured by E, and let E not
τῶν Α, Β, Γ ἔστω ὁ αὐτός. φανερὸν δή, ὅτι ὁ Ε πρῶτος be the same as one of A, B, C. So it is clear that E is
οὔκ ἐστιν. εἰ γὰρ ὁ Ε πρῶτός ἐστι καὶ μετρεῖ τὸν Δ, καὶ τὸν not prime. For if E is prime, and measures D, then it will
Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν also measure A, (despite A) being prime (and) not being
ἀδύνατον. οὐκ ἄρα ὁ Ε πρῶτός ἐστιν. σύνθετος ἄρα. πᾶς the same as it [Prop. 9.12]. The very thing is impossible.
δὲ σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται· Thus, E is not prime. Thus, (it is) composite. And every
ὁ Ε ἄρα ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. λέγω δή, ὅτι composite number is measured by some prime number
ὑπ᾿ οὐδενὸς ἄλλου πρώτου μετρηθήσεται πλὴν τοῦ Α. εἰ γὰρ [Prop. 7.31]. Thus, E is measured by some prime num-
ὑφ᾿ ἑτέρου μετρεῖται ὁ Ε, ὁ δὲ Ε τὸν Δ μετρεῖ, κἀκεῖνος ἄρα ber. So I say that it will be measured by no other prime
τὸν Δ μετρήσει· ὥστε καὶ τὸν Α μετρήσει πρῶτον ὄντα μὴ number than A. For if E is measured by another (prime
ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον. ὁ Α ἄρα τὸν Ε number), and E measures D, then this (prime number)
μετρεῖ. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν will thus also measure D. Hence, it will also measure A,
Ζ. λέγω, ὅτι ὁ Ζ οὐδενὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. εἰ γὰρ ὁ (despite A) being prime (and) not being the same as it
Ζ ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτὸς καὶ μετρεῖ τὸν Δ κατὰ τὸν [Prop. 9.12]. The very thing is impossible. Thus, A mea-
Ε, καὶ εἷς ἄρα τῶν Α, Β, Γ τὸν Δ μετρεῖ κατά τὸν Ε. ἀλλὰ sures E. And since E measures D, let it measure it ac-
εἷς τῶν Α, Β, Γ τὸν Δ μετρεῖ κατά τινα τῶν Α, Β, Γ· καὶ ὁ cording to F. I say that F is not the same as one of A, B,
Ε ἄρα ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός· ὅπερ οὐχ ὑπόκειται. C. For if F is the same as one of A, B, C, and measures
οὐκ ἄρα ὁ Ζ ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. ὁμοίως δὴ D according to E, then one of A, B, C thus also measures
δείξομεν, ὅτι μετρεῖται ὁ Ζ ὑπὸ τοῦ Α, δεικνύντες πάλιν, D according to E. But one of A, B, C (only) measures
ὅτι ὁ Ζ οὔκ ἐστι πρῶτος. εἰ γὰρ, καὶ μετρεῖ τὸν Δ, καὶ τὸν D according to some (one) of A, B, C [Prop. 9.11]. And
Α μετρήσει πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν thus E is the same as one of A, B, C. The very oppo-
ἀδύνατον· οὐκ ἄρα πρῶτός ἐστιν ὁ Ζ· σύνθετος ἄρα. ἅπας site thing was assumed. Thus, F is not the same as one
δὲ σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται· ὁ of A, B, C. Similarly, we can show that F is measured
Ζ ἄρα ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. λέγω δή, ὅτι ὑφ᾿ by A, (by) again showing that F is not prime. For if (F
ἑτέρου πρώτου οὐ μετρηθήσεται πλὴν τοῦ Α. εἰ γὰρ ἕτερός is prime), and measures D, then it will also measure A,
τις πρῶτος τὸν Ζ μετρεῖ, ὁ δὲ Ζ τὸν Δ μετρεῖ, κἀκεῖνος (despite A) being prime (and) not being the same as it
ἄρα τὸν Δ μετρήσει· ὥστε καὶ τὸν Α μετρήσει πρῶτον ὄντα [Prop. 9.12]. The very thing is impossible. Thus, F is
μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἐστὶν ἀδύνατον. ὁ Α ἄρα τὸν Ζ not prime. Thus, (it is) composite. And every composite
μετρεῖ. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ κατὰ τὸν Ζ, ὁ Ε ἄρα τὸν number is measured by some prime number [Prop. 7.31].
Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν Thus, F is measured by some prime number. So I say
Γ πολλαπλασιάσας τὸν Δ πεποίηκεν· ὁ ἄρα ἐκ τῶν Α, Γ that it will be measured by no other prime number than
ἴσος ἐστὶ τῷ ἐκ τῶν Ε, Ζ. ἀνάλογον ἄρα ἐστὶν ὡς ὁ Α πρὸς A. For if some other prime (number) measures F, and
τὸν Ε, οὕτως ὁ Ζ πρὸς τὸν Γ. ὁ δὲ Α τὸν Ε μετρεῖ· καὶ ὁ Ζ F measures D, then this (prime number) will thus also
ἄρα τὸν Γ μετρεῖ. μετρείτω αὐτὸν κατὰ τὸν Η. ὁμοίως δὴ measure D. Hence, it will also measure A, (despite A)
δείξομεν, ὅτι ὁ Η οὐδενὶ τῶν Α, Β ἐστιν ὁ αὐτός, καὶ ὅτι being prime (and) not being the same as it [Prop. 9.12].
μετρεῖται ὑπὸ τοῦ Α. καὶ ἐπεὶ ὁ Ζ τὸν Γ μετρεῖ κατὰ τὸν Η, The very thing is impossible. Thus, A measures F. And
ὁ Ζ ἄρα τὸν Η πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν since E measures D according to F, E has thus made
καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν· ὁ ἄρα ἐκ D (by) multiplying F. But, in fact, A has also made D
τῶν Α, Β ἴσος ἐστὶ τῷ ἐκ τῶν Ζ, Η. ἀνάλογον ἄρα ὡς ὁ Α (by) multiplying C [Prop. 9.11 corr.]. Thus, the (number
πρὸς τὸν Ζ, ὁ Η πρὸς τὸν Β. μετρεῖ δὲ ὁ Α τὸν Ζ· μετρεῖ created) from (multiplying) A, C is equal to the (number
ἄρα καὶ ὁ Η τὸν Β. μετρείτω αὐτὸν κατὰ τὸν Θ. ὁμοίως δὴ created) from (multiplying) E, F. Thus, proportionally,
δείξομεν, ὅτι ὁ Θ τῷ Α οὐκ ἔστιν ὁ αὐτός. καὶ ἐπεὶ ὁ Η τὸν as A is to E, so F (is) to C [Prop. 7.19]. And A measures
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ELEMENTS BOOK 9
Β μετρεῖ κατὰ τὸν Θ, ὁ Η ἄρα τὸν Θ πολλαπλασιάσας τὸν E. Thus, F also measures C. Let it measure it according
Β πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας to G. So, similarly, we can show that G is not the same
τὸν Β πεποίηκεν· ὁ ἄρα ὑπὸ Θ, Η ἴσος ἐστὶ τῷ ἀπὸ τοῦ Α as one of A, B, and that it is measured by A. And since
τετραγώνῳ· ἔστιν ἄρα ὡς ὁ Θ πρὸς τὸν Α, ὁ Α πρὸς τὸν Η. F measures C according to G, F has thus made C (by)
μετρεῖ δὲ ὁ Α τὸν Η· μετρεῖ ἄρα καὶ ὁ Θ τὸν Α πρῶτον ὄντα multiplying G. But, in fact, A has also made C (by) mul-
μὴ ὢν αὐτῷ ὁ αὐτός· ὅπερ ἄτοπον. οὐκ ἄρα ὁ μέγιστος ὁ tiplying B [Prop. 9.11 corr.]. Thus, the (number created)
Δ ὑπὸ ἑτέρου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Α, Β, Γ· from (multiplying) A, B is equal to the (number created)
ὅπερ ἔδει δεῖξαι. from (multiplying) F, G. Thus, proportionally, as A (is)
to F, so G (is) to B [Prop. 7.19]. And A measures F.
Thus, G also measures B. Let it measure it according to
H. So, similarly, we can show that H is not the same as
A. And since G measures B according to H, G has thus
made B (by) multiplying H. But, in fact, A has also made
idþ on A. Thus, as H is to A, (so) A (is) to G [Prop. 7.19].
B (by) multiplying itself [Prop. 9.8]. Thus, the (number
created) from (multiplying) H, G is equal to the square
And A measures G. Thus, H also measures A, (despite
A) being prime (and) not being the same as it. The very
thing (is) absurd. Thus, the greatest (number) D cannot
be measured by another (number) except (one of) A, B,
C. (Which is) the very thing it was required to show.
Proposition 14
.
᾿Εὰν ἐλάχιστος ἀριθμὸς ὑπὸ πρώτων ἀριθμῶν μετρῆται, If a least number is measured by (some) prime num-
ὑπ᾿ οὑδενὸς ἄλλου πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ bers then it will not be measured by any other prime
τῶν ἐξ ἀρχῆς μετρούντων. number except (one of) the original measuring (num-
bers).
A B
Α Β
Ε Γ E C
Ζ ∆ F D
᾿Ελάχιστος γὰρ ἀριθμὸς ὁ Α ὑπὸ πρώτων ἀριθμῶν τῶν For let A be the least number measured by the prime
Β, Γ, Δ μετρείσθω· λέγω, ὅτι ὁ Α ὑπ᾿ οὐδενὸς ἄλλου numbers B, C, D. I say that A will not be measured by
πρώτου ἀριθμοῦ μετρηθήσεται παρὲξ τῶν Β, Γ, Δ. any other prime number except (one of) B, C, D.
Εἰ γὰρ δυνατόν, μετρείσθω ὑπὸ πρώτου τοῦ Ε, καὶ ὁ For, if possible, let it be measured by the prime (num-
Ε μηδενὶ τῶν Β, Γ, Δ ἔστω ὁ αὐτός. καὶ ἐπεὶ ὁ Ε τὸν ber) E. And let E not be the same as one of B, C, D.
Α μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν Ζ· ὁ Ε ἄρα τὸν Ζ And since E measures A, let it measure it according to F.
πολλαπλασιάσας τὸν Α πεποίηκεν. καὶ μετρεῖται ὁ Α ὑπὸ Thus, E has made A (by) multiplying F. And A is mea-
πρώτων ἀριθμῶν τῶν Β, Γ, Δ. ἐὰν δὲ δύο ἀριθμοὶ πολ- sured by the prime numbers B, C, D. And if two num-
λαπλασιάσαντες ἀλλήλους ποιῶσί τινα, τὸν δὲ γενόμενον bers make some (number by) multiplying one another,
ἐξ αὐτῶν μετρῇ τις πρῶτος ἀριθμός, καὶ ἕνα τῶν ἐξ ἀρχῆς and some prime number measures the number created
μετρήσει· οἱ Β, Γ, Δ ἄρα ἕνα τῶν Ε, Ζ μετρήσουσιν. τὸν from them, then (the prime number) will also measure
μὲν οὖν Ε οὐ μετρήσουσιν· ὁ γὰρ Ε πρῶτός ἐστι καὶ οὐδενὶ one of the original (numbers) [Prop. 7.30]. Thus, B, C,
τῶν Β, Γ, Δ ὁ αὐτός. τὸν Ζ ἄρα μετροῦσιν ἐλάσσονα ὄντα D will measure one of E, F. In fact, they do not measure
τοῦ Α· ὅπερ ἀδύνατον. ὁ γὰρ Α ὑπόκειται ἐλάχιστος ὑπὸ E. For E is prime, and not the same as one of B, C, D.
τῶν Β, Γ, Δ μετρούμενος. οὐκ ἄρα τὸν Α μετρήσει πρῶτος Thus, they (all) measure F, which is less than A. The
ἀριθμὸς παρὲξ τῶν Β, Γ, Δ· ὅπερ ἔδει δεῖξαι. very thing (is) impossible. For A was assumed (to be) the
least (number) measured by B, C, D. Thus, no prime
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ieþ number can measure A except (one of) B, C, D. (Which
ELEMENTS BOOK 9
is) the very thing it was required to show.
.
Proposition 15
᾿Εὰν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν ἐλάχιστοι τῶν If three continuously proportional numbers are the
τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς, δύο ὁποιοιοῦν συν- least of those (numbers) having the same ratio as them
τεθέντες πρὸς τὸν λοιπὸν πρῶτοί εἰσιν. then two (of them) added together in any way are prime
to the remaining (one).
∆ Ε Ζ D E F
Α A
Β B
Γ C
῎Εστωσαν τρεῖς ἀριθμοὶ ἑξῆς ἀνάλογον ἐλάχιστοι τῶν Let A, B, C be three continuously proportional num-
τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς οἱ Α, Β, Γ· λέγω, ὅτι τῶν bers (which are) the least of those (numbers) having the
Α, Β, Γ δύο ὁποιοιοῦν συντεθέντες πρὸς τὸν λοιπὸν πρῶτοι same ratio as them. I say that two of A, B, C added to-
εἰσιν, οἱ μὲν Α, Β πρὸς τὸν Γ, οἱ δὲ Β, Γ πρὸς τὸν Α καὶ gether in any way are prime to the remaining (one), (that
ἔτι οἱ Α, Γ πρὸς τὸν Β. is) A and B (prime) to C, B and C to A, and, further, A
Εἰλήφθωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν and C to B.
λόγον ἐχόντων τοῖς Α, Β, Γ δύο οἱ ΔΕ, ΕΖ. φανερὸν Let the two least numbers, DE and EF, having the
δή, ὅτι ὁ μὲν ΔΕ ἑαυτὸν πολλαπλασιάσας τὸν Α πεποίηκεν, same ratio as A, B, C, have been taken [Prop. 8.2].
τὸν δὲ ΕΖ πολλαπλασιάσας τὸν Β πεποίηκεν, καὶ ἔτι ὁ ΕΖ So it is clear that DE has made A (by) multiplying it-
ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν. καὶ ἐπεὶ οἱ ΔΕ, self, and has made B (by) multiplying EF, and, fur-
ΕΖ ἐλάχιστοί εἰσιν, πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐὰν δὲ ther, EF has made C (by) multiplying itself [Prop. 8.2].
δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ συναμφότερος And since DE, EF are the least (of those numbers hav-
πρὸς ἑκάτερον πρῶτός ἐστιν· καὶ ὁ ΔΖ ἄρα πρὸς ἑκάτερον ing the same ratio as them), they are prime to one an-
τῶν ΔΕ, ΕΖ πρῶτός ἐστιν. ἀλλὰ μὴν καὶ ὁ ΔΕ πρὸς τὸν other [Prop. 7.22]. And if two numbers are prime to
ΕΖ πρῶτός ἐστιν· οἱ ΔΖ, ΔΕ ἄρα πρὸς τὸν ΕΖ πρῶτοί εἰσιν. one another then the sum (of them) is also prime to each
ἐὰν δὲ δύο ἀριθμοὶ πρός τινα ἀριθμὸν πρῶτοι ὦσιν, καὶ ὁ [Prop. 7.28]. Thus, DF is also prime to each of DE, EF.
ἐξ αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτός ἐστιν· ὥστε But, in fact, DE is also prime to EF. Thus, DF, DE
ὁ ἐκ τῶν ΖΔ, ΔΕ πρὸς τὸν ΕΖ πρῶτός ἐστιν· ὥστε καὶ ὁ are (both) prime to EF. And if two numbers are (both)
ἐκ τῶν ΖΔ, ΔΕ πρὸς τὸν ἀπὸ τοῦ ΕΖ πρῶτός ἐστιν. [ἐὰν prime to some number then the (number) created from
γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ ἐκ τοῦ ἑνὸς (multiplying) them is also prime to the remaining (num-
αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτός ἐστιν]. ἀλλ᾿ ὁ ber) [Prop. 7.24]. Hence, the (number created) from
ἐκ τῶν ΖΔ, ΔΕ ὁ ἀπὸ τοῦ ΔΕ ἐστι μετὰ τοῦ ἐκ τῶν ΔΕ, (multiplying) FD, DE is prime to EF. Hence, the (num-
ΕΖ· ὁ ἄρα ἀπὸ τοῦ ΔΕ μετὰ τοῦ ἐκ τῶν ΔΕ, ΕΖ πρὸς τὸν ber created) from (multiplying) FD, DE is also prime
ἀπὸ τοῦ ΕΖ πρῶτός ἐστιν. καί ἐστιν ὁ μὲν ἀπὸ τοῦ ΔΕ to the (square) on EF [Prop. 7.25]. [For if two num-
ὁ Α, ὁ δὲ ἐκ τῶν ΔΕ, ΕΖ ὁ Β, ὁ δὲ ἀπὸ τοῦ ΕΖ ὁ Γ· οἱ bers are prime to one another then the (number) created
Α, Β ἄρα συντεθέντες πρὸς τὸν Γ πρῶτοί εἰσιν. ὁμοίως δὴ from (squaring) one of them is prime to the remaining
δείξομεν, ὅτι καὶ οἱ Β, Γ πρὸς τὸν Α πρῶτοί εἰσιν. λέγω (number).] But the (number created) from (multiplying)
δή, ὅτι καὶ οἱ Α, Γ πρὸς τὸν Β πρῶτοί εἰσιν. ἐπεὶ γὰρ ὁ ΔΖ FD, DE is the (square) on DE plus the (number cre-
πρὸς ἑκάτερον τῶν ΔΕ, ΕΖ πρῶτός ἐστιν, καὶ ὁ ἀπὸ τοῦ ated) from (multiplying) DE, EF [Prop. 2.3]. Thus, the
ΔΖ πρὸς τὸν ἐκ τῶν ΔΕ, ΕΖ πρῶτός ἐστιν. ἀλλὰ τῷ ἀπὸ (square) on DE plus the (number created) from (multi-
τοῦ ΔΖ ἴσοι εἰσὶν οἱ ἀπὸ τῶν ΔΕ, ΕΖ μετὰ τοῦ δὶς ἐκ τῶν plying) DE, EF is prime to the (square) on EF. And
ΔΕ, ΕΖ· καὶ οἱ ἀπὸ τῶν ΔΕ, ΕΖ ἄρα μετὰ τοῦ δὶς ὑπὸ τῶν the (square) on DE is A, and the (number created) from
ΔΕ, ΕΖ πρὸς τὸν ὑπὸ τῶν ΔΕ, ΕΖ πρῶτοί [εἰσι]. διελόντι (multiplying) DE, EF (is) B, and the (square) on EF
οἱ ἀπὸ τῶν ΔΕ, ΕΖ μετὰ τοῦ ἅπαξ ὑπὸ ΔΕ, ΕΖ πρὸς τὸν (is) C. Thus, A, B summed is prime to C. So, similarly,
ὑπὸ ΔΕ, ΕΖ πρῶτοί εἰσιν. ἔτι διελόντι οἱ ἀπὸ τῶν ΔΕ, ΕΖ we can show that B, C (summed) is also prime to A. So
ἄρα πρὸς τὸν ὑπὸ ΔΕ, ΕΖ πρῶτοί εἰσιν. καί ἐστιν ὁ μὲν I say that A, C (summed) is also prime to B. For since
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ἀπὸ τοῦ ΔΕ ὁ Α, ὁ δὲ ὑπὸ τῶν ΔΕ, ΕΖ ὁ Β, ὁ δὲ ἀπὸ τοῦ DF is prime to each of DE, EF then the (square) on DF
ΕΖ ὁ Γ. οἱ Α, Γ ἄρα συντεθέντες πρὸς τὸν Β πρῶτοί εἰσιν· is also prime to the (number created) from (multiplying)
ὅπερ ἔδει δεῖξαι. DE, EF [Prop. 7.25]. But, the (sum of the squares) on
DE, EF plus twice the (number created) from (multiply-
ing) DE, EF is equal to the (square) on DF [Prop. 2.4].
And thus the (sum of the squares) on DE, EF plus twice
the (rectangle contained) by DE, EF [is] prime to the
(rectangle contained) by DE, EF. By separation, the
(sum of the squares) on DE, EF plus once the (rect-
angle contained) by DE, EF is prime to the (rectangle
iþ contained) by DE, EF. And the (square) on DE is A,
†
contained) by DE, EF. Again, by separation, the (sum
of the squares) on DE, EF is prime to the (rectangle
and the (rectangle contained) by DE, EF (is) B, and
the (square) on EF (is) C. Thus, A, C summed is prime
to B. (Which is) the very thing it was required to show.
2
2
2
2
† Since if α β measures α + β + 2 α β then it also measures α + β + α β, and vice versa.
Proposition 16
.
᾿Εὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, οὐκ ἔσται If two numbers are prime to one another then as the
ὡς ὁ πρῶτος πρὸς τὸν δεύτερον, οὕτως ὁ δεύτερος πρὸς first is to the second, so the second (will) not (be) to some
ἄλλον τινά. other (number).
Α A
Β B
Γ C
Δύο γὰρ ἀριθμοὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους ἔστω- For let the two numbers A and B be prime to one
σαν· λέγω, ὅτι οὐκ ἔστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β another. I say that as A is to B, so B is not to some other
πρὸς ἄλλον τινά. (number).
Εἰ γὰρ δυνατόν, ἔστω ὡς ὁ Α πρὸς τὸν Β, ὁ Β πρὸς For, if possible, let it be that as A (is) to B, (so)
τὸν Γ. οἱ δὲ Α, Β πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ B (is) to C. And A and B (are) prime (to one an-
ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας other). And (numbers) prime (to one another are) also
ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν the least (of those numbers having the same ratio as
ἑπόμενον· μετρεῖ ἄρα ὁ Α τὸν Β ὡς ἡγούμενος ἡγούμενον. them) [Prop. 7.21]. And the least numbers measure
izþ And (A) also measures itself. Thus, A measures A and B,
μετρεῖ δὲ καὶ ἑαυτόν· ὁ Α ἄρα τοὺς Α, Β μετρεῖ πρώτους those (numbers) having the same ratio (as them) an
ὄντας πρὸς ἀλλήλους· ὅπερ ἄτοπον. οὐκ ἄρα ἔσται ὡς ὁ Α equal number of times, the leading (measuring) the lead-
πρὸς τὸν Β, οὕτως ὁ Β πρὸς τὸν Γ· ὅπερ ἔδει δεῖξαι. ing, and the following the following [Prop. 7.20]. Thus,
A measures B, as the leading (measuring) the leading.
which are prime to one another. The very thing (is) ab-
surd. Thus, as A (is) to B, so B cannot be to C. (Which
is) the very thing it was required to show.
Proposition 17
.
᾿Εὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ δὲ If any multitude whatsoever of numbers is continu-
ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, οὐκ ἔσται ὡς ὁ ously proportional, and the outermost of them are prime
πρῶτος πρὸς τὸν δεύτερον, οὕτως ὁ ἔσχατος πρὸς ἄλλον to one another, then as the first (is) to the second, so the
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τινά. last will not be to some other (number).
῎Εστωσαν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Let A, B, C, D be any multitude whatsoever of con-
Β, Γ, Δ, οἱ δὲ ἄκροι αὐτῶν οἱ Α, Δ πρῶτοι πρὸς ἀλλήλους tinuously proportional numbers. And let the outermost
ἔστωσαν· λέγω, ὅτι οὐκ ἔστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ of them, A and D, be prime to one another. I say that as
Δ πρὸς ἄλλον τινά. A is to B, so D (is) not to some other (number).
Α A
Β B
Γ C
∆ D
Ε E
Εἰ γὰρ δυνατόν, ἔστω ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ For, if possible, let it be that as A (is) to B, so D
πρὸς τὸν Ε· ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, ὁ Β πρὸς (is) to E. Thus, alternately, as A is to D, (so) B (is)
τὸν Ε. οἱ δὲ Α, Δ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ to E [Prop. 7.13]. And A and D are prime (to one
ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας another). And (numbers) prime (to one another are)
ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν also the least (of those numbers having the same ra-
ἑπόμενον. μετρεῖ ἄρα ὁ Α τὸν Β. καί ἐστιν ὡς ὁ Α πρὸς tio as them) [Prop. 7.21]. And the least numbers mea-
τὸν Β, ὁ Β πρὸς τὸν Γ. καὶ ὁ Β ἄρα τὸν Γ μετρεῖ΄· ὥστε sure those (numbers) having the same ratio (as them) an
καὶ ὁ Α τὸν Γ μετρεῖ. καὶ ἐπεί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, equal number of times, the leading (measuring) the lead-
ὁ Γ πρὸς τὸν Δ, μετρεῖ δὲ ὁ Β τὸν Γ, μετρεῖ ἄρα καὶ ὁ Γ ing, and the following the following [Prop. 7.20]. Thus,
τὸν Δ. ἀλλ᾿ ὁ Α τὸν Γ ἐμέτρει· ὥστε ὁ Α καὶ τὸν Δ μετρεῖ. A measures B. And as A is to B, (so) B (is) to C. Thus, B
μετρεῖ δὲ καὶ ἑαυτόν. ὁ Α ἄρα τοὺς Α, Δ μετρεῖ πρώτους also measures C. And hence A measures C [Def. 7.20].
ihþ and D, which are prime to one another. The very thing is
ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσται And since as B is to C, (so) C (is) to D, and B mea-
ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς ἄλλον τινά· ὅπερ ἔδει sures C, C thus also measures D [Def. 7.20]. But, A was
δεῖξαι. (found to be) measuring C. And hence A also measures
D. And (A) also measures itself. Thus, A measures A
impossible. Thus, as A (is) to B, so D cannot be to some
other (number). (Which is) the very thing it was required
to show.
Proposition 18
.
Δύο ἀριθμῶν δοθέντων ἐπισκέψασθαι, εἰ δυνατόν ἐστιν For two given numbers, to investigate whether it is
αὐτοῖς τρίτον ἀνάλογον προσευρεῖν. possible to find a third (number) proportional to them.
Α Γ A C
Β ∆ B D
῎Εστωσαν οἱ δοθέντες δύο ἀριθμοὶ οἱ Α, Β, καὶ Let A and B be the two given numbers. And let it
δέον ἔστω ἐπισκέψασθαι, εἰ δυνατόν ἐστιν αὐτοῖς τρίτον be required to investigate whether it is possible to find a
ἀνάλογον προσευρεῖν. third (number) proportional to them.
Οἱ δὴ Α, Β ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ. καὶ εἰ So A and B are either prime to one another, or not.
πρῶτοι πρὸς ἀλλήλους εἰσίν, δέδεικται, ὅτι ἀδύνατόν ἐστιν And if they are prime to one another then it has (already)
αὐτοῖς τρίτον ἀνάλογον προσευρεῖν. been show that it is impossible to find a third (number)
᾿Αλλὰ δὴ μὴ ἔστωσαν οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, proportional to them [Prop. 9.16].
καὶ ὁ Β ἑαυτον πολλαπλασιάσας τὸν Γ ποιείτω. ὁ Α δὴ τὸν And so let A and B not be prime to one another. And
Γ ἤτοι μετρεῖ ἢ οὐ μετρεῖ. μετρείτω πρότερον κατὰ τὸν Δ· let B make C (by) multiplying itself. So A either mea-
ὁ Α ἄρα τὸν Δ πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλα μὴν sures, or does not measure, C. Let it first of all measure
καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν· ὁ ἄρα (C) according to D. Thus, A has made C (by) multiply-
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ἐκ τῶν Α, Δ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Β. ἔστιν ἄρα ὡς ὁ Α ing D. But, in fact, B has also made C (by) multiplying
πρὸς τὸν Β, ὁ Β πρὸς τὸν Δ· τοῖς Α, Β ἄρα τρίτος ἀριθμὸς itself. Thus, the (number created) from (multiplying) A,
ἀνάλογον προσηύρηται ὁ Δ. D is equal to the (square) on B. Thus, as A is to B, (so)
᾿Αλλὰ δὴ μὴ μετρείτω ὁ Α τὸν Γ· λέγω, ὅτι τοῖς Α, Β B (is) to D [Prop. 7.19]. Thus, a third number has been
ἀδύνατόν ἐστι τρίτον ἀνάλογον προσευρεῖν ἀριθμόν. εἰ γὰρ found proportional to A, B, (namely) D.
δυνατόν, προσηυρήσθω ὁ Δ. ὁ ἄρα ἐκ τῶν Α, Δ ἴσος ἐστὶ And so let A not measure C. I say that it is impossi-
τῷ ἀπὸ τοῦ Β. ὁ δὲ ἀπὸ τοῦ Β ἐστιν ὁ Γ· ὁ ἄρα ἐκ τῶν Α, ble to find a third number proportional to A, B. For, if
Δ ἴσος ἐστὶ τῷ Γ. ὥστε ὁ Α τὸν Δ πολλαπλασιάσας τὸν possible, let it have been found, (and let it be) D. Thus,
Γ πεποίηκεν· ὁ Α ἄρα τὸν Γ μετρεῖ κατὰ τὸν Δ. ἀλλα μὴν the (number created) from (multiplying) A, D is equal to
ijþ fact, also assumed (to be) not measuring (C). The very
ὑπόκειται καὶ μὴ μετρῶν· ὅπερ ἄτοπον. οὐκ ἄρα δυνατόν the (square) on B [Prop. 7.19]. And the (square) on B
ἐστι τοῖς Α, Β τρίτον ἀνάλογον προσευρεῖν ἀριθμὸν, ὅταν is C. Thus, the (number created) from (multiplying) A,
ὁ Α τὸν Γ μὴ μετρῇ· ὅπερ ἔδει δεῖξαι. D is equal to C. Hence, A has made C (by) multiplying
D. Thus, A measures C according to D. But (A) was, in
thing (is) absurd. Thus, it is not possible to find a third
number proportional to A, B when A does not measure
C. (Which is) the very thing it was required to show.
.
†
Proposition 19
Τριῶν ἀριθμῶν δοθέντων ἐπισκέψασθαι, πότε δυνατόν For three given numbers, to investigate when it is pos-
ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν. sible to find a fourth (number) proportional to them.
Α A
Β B
Γ C
∆ D
Ε E
῎Εστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ οἱ Α, Β, Γ, καὶ δέον Let A, B, C be the three given numbers. And let it be
ἔστω επισκέψασθαι, πότε δυνατόν ἐστιν αὐτοῖς τέταρτον required to investigate when it is possible to find a fourth
ἀνάλογον προσευρεῖν. (number) proportional to them.
῎Ητοι οὖν οὔκ εἰσιν ἑξῆς ἀνάλογον, καὶ οἱ ἄκροι αὐτῶν In fact, (A, B, C) are either not continuously pro-
πρῶτοι πρὸς ἀλλήλους εἰσίν, ἢ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ portional and the outermost of them are prime to one
ἄκροι αὐτῶν οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους, ἢ οὕτε ἑξῆς another, or are continuously proportional and the outer-
εἰσιν ἀνάλογον, οὔτε οἱ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους most of them are not prime to one another, or are neither
εἰσίν, ἢ καὶ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ ἄκροι αὐτῶν πρῶτοι continuously proportional nor are the outermost of them
πρὸς ἀλλήλους εἰσίν. prime to one another, or are continuously proportional
Εἰ μὲν οὖν οἱ Α, Β, Γ ἑξῆς εἰσιν ἀνάλογον, καὶ οἱ and the outermost of them are prime to one another.
ἄκροι αὐτῶν οἱ Α, Γ πρῶτοι πρὸς ἀλλήλους εἰσίν, δέδεικται, In fact, if A, B, C are continuously proportional, and
ὅτι ἀδύνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν the outermost of them, A and C, are prime to one an-
ἀριθμόν. μὴ ἔστωσαν δὴ οἱ Α, Β, Γ ἑξῆς ἀνάλογον τῶν other, (then) it has (already) been shown that it is im-
ἀκρῶν πάλιν ὄντων πρώτων πρὸς ἀλλήλους. λέγω, ὅτι possible to find a fourth number proportional to them
καὶ οὕτως ἀδύνατόν ἐστιν αὐτοῖς τέταρτον ἀνάλογον προ- [Prop. 9.17]. So let A, B, C not be continuously propor-
σευρεῖν. εἰ γὰρ δυνατόν, προσευρήσθω ὁ Δ, ὥστε εἶναι ὡς tional, (with) the outermost of them again being prime to
τὸν Α πρὸς τὸν Β, τὸν Γ πρὸς τὸν Δ, καὶ γεγονέτω ὡς ὁ Β one another. I say that, in this case, it is also impossible
πρὸς τὸν Γ, ὁ Δ πρὸς τὸν Ε. καὶ ἐπεί ἐστιν ὡς μὲν ὁ Α πρὸς to find a fourth (number) proportional to them. For, if
τὸν Β, ὁ Γ πρὸς τὸν Δ, ὡς δὲ ὁ Β πρὸς τὸν Γ, ὁ Δ πρὸς possible, let it have been found, (and let it be) D. Hence,
τὸν Ε, δι᾿ ἴσου ἄρα ὡς ὁ Α πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Ε. οἱ it will be that as A (is) to B, (so) C (is) to D. And let it be
δὲ Α, Γ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι contrived that as B (is) to C, (so) D (is) to E. And since
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μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ὅ τε ἡγούμενος as A is to B, (so) C (is) to D, and as B (is) to C, (so) D
τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον. μετρεῖ ἄρα ὁ (is) to E, thus, via equality, as A (is) to C, (so) C (is) to E
Α τὸν Γ ὡς ἡγούμενος ἡγούμενον. μετρεῖ δὲ καὶ ἑαυτόν· [Prop. 7.14]. And A and C (are) prime (to one another).
ὁ Α ἄρα τοὺς Α, Γ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· And (numbers) prime (to one another are) also the least
ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοῖς Α, Β, Γ δυνατόν ἐστι (numbers having the same ratio as them) [Prop. 7.21].
τέταρτον ἀνάλογον προσευρεῖν. And the least (numbers) measure those numbers having
᾿Αλλά δὴ πάλιν ἔστωσαν οἱ Α, Β, Γ ἑξῆς ἀνάλογον, οἱ δὲ the same ratio as them (the same number of times), the
Α, Γ μὴ ἔστωσαν πρῶτοι πρὸς ἀλλήλους. λέγω, ὅτι δυνατόν leading (measuring) the leading, and the following the
ἐστιν αὐτοῖς τέταρτον ἀνάλογον προσευρεῖν. ὁ γὰρ Β τὸν Γ following [Prop. 7.20]. Thus, A measures C, (as) the
πολλαπλασιάσας τὸν Δ ποιείτω· ὁ Α ἄρα τὸν Δ ἤτοι μετρεῖ leading (measuring) the leading. And it also measures
ἢ οὐ μετρεῖ. μετρείτω αὐτὸν πρότερον κατὰ τὸν Ε· ὁ Α ἄρα itself. Thus, A measures A and C, which are prime to
τὸν Ε πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ one another. The very thing is impossible. Thus, it is not
Β τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν· ὁ ἄρα ἐκ τῶν possible to find a fourth (number) proportional to A, B,
Α, Ε ἴσος ἐστὶ τῷ ἐκ τῶν Β, Γ. ἀνάλογον ἄρα [ἐστὶν] ὡς ὁ C.
Α πρὸς τὸν Β, ὁ Γ πρὸς τὸν Ε· τοὶς Α, Β, Γ ἄρα τέταρτος And so let A, B, C again be continuously propor-
ἀνάλογον προσηύρηται ὁ Ε. tional, and let A and C not be prime to one another. I
᾿Αλλὰ δὴ μὴ μετρείτω ὁ Α τὸν Δ· λέγω, ὅτι ἀδύνατόν say that it is possible to find a fourth (number) propor-
ἐστι τοῖς Α, Β, Γ τέταρτον ἀνάλογον προσευρεῖν ἀριθμόν. tional to them. For let B make D (by) multiplying C.
εἰ γὰρ δυνατόν, προσευρήσθω ὁ Ε· ὁ ἄρα ἐκ τῶν Α, Ε ἴσος Thus, A either measures or does not measure D. Let it,
ἐστὶ τῷ ἐκ τῶν Β, Γ. ἀλλὰ ὁ ἑκ τῶν Β, Γ ἐστιν ὁ Δ· καὶ first of all, measure (D) according to E. Thus, A has
ὁ ἐκ τῶν Α, Ε ἄρα ἴσος ἐστὶ τῷ Δ. ὁ Α ἄρα τὸν Ε πολλα- made D (by) multiplying E. But, in fact, B has also made
πλασιάσας τὸν Δ πεποίηκεν· ὁ Α ἄρα τὸν Δ μετρεῖ κατὰ τὸν D (by) multiplying C. Thus, the (number created) from
Ε· ὥστε μετρεῖ ὁ Α τὸν Δ. ἀλλὰ καὶ οὐ μετρεῖ· ὅπερ ἄτοπον. (multiplying) A, E is equal to the (number created) from
οὐκ ἄρα δυνάτον ἐστι τοῖς Α, Β, Γ τέταρτον ἀνάλογον προ- (multiplying) B, C. Thus, proportionally, as A [is] to B,
σευρεῖν ἀριθμόν, ὅταν ὁ Α τὸν Δ μὴ μετρῇ. ἀλλὰ δὴ οἱ Α, Β, (so) C (is) to E [Prop. 7.19]. Thus, a fourth (number)
Γ μήτε ἑξῆς ἔστωσαν ἀνάλογον μήτε οἱ ἄκροι πρῶτοι πρὸς proportional to A, B, C has been found, (namely) E.
ἀλλήλους. καὶ ὁ Β τὸν Γ πολλαπλασιάσας τὸν Δ ποιείτω. And so let A not measure D. I say that it is impossible
ὁμοίως δὴ δειχθήσεται, ὅτι εἰ μὲν μετρεῖ ὁ Α τὸν Δ, δυ- to find a fourth number proportional to A, B, C. For, if
νατόν ἐστιν αὐτοῖς ἀνάλογον προσευρεῖν, εἰ δὲ οὐ μετρεῖ, possible, let it have been found, (and let it be) E. Thus,
ἀδύνατον· ὅπερ ἔδει δεῖξαι. the (number created) from (multiplying) A, E is equal to
the (number created) from (multiplying) B, C. But, the
(number created) from (multiplying) B, C is D. And thus
the (number created) from (multiplying) A, E is equal to
D. Thus, A has made D (by) multiplying E. Thus, A
measures D according to E. Hence, A measures D. But,
it also does not measure (D). The very thing (is) absurd.
Thus, it is not possible to find a fourth number propor-
tional to A, B, C when A does not measure D. And so
(let) A, B, C (be) neither continuously proportional, nor
(let) the outermost of them (be) prime to one another.
And let B make D (by) multiplying C. So, similarly, it
can be show that if A measures D then it is possible to
find a fourth (number) proportional to (A, B, C), and
impossible if (A) does not measure (D). (Which is) the
very thing it was required to show.
† The proof of this proposition is incorrect. There are, in fact, only two cases. Either A, B, C are continuously proportional, with A and C prime
to one another, or not. In the first case, it is impossible to find a fourth proportional number. In the second case, it is possible to find a fourth
proportional number provided that A measures B times C. Of the four cases considered by Euclid, the proof given in the second case is incorrect,
since it only demonstrates that if A : B :: C : D then a number E cannot be found such that B : C :: D : E. The proofs given in the other three
270
ST EW jþ.
cases are correct. kþ ELEMENTS BOOK 9
.
Proposition 20
Οἱ πρῶτοι ἀριθμοὶ πλείους εἰσὶ παντὸς τοῦ προτεθέντος The (set of all) prime numbers is more numerous than
πλήθους πρώτων ἀριθμῶν. any assigned multitude of prime numbers.
Α Η A G
Β B
Γ C
Ε ∆ Ζ E D F
῎Εστωσαν οἱ προτεθέντες πρῶτοι ἀριθμοὶ οἱ Α, Β, Γ· Let A, B, C be the assigned prime numbers. I say that
λέγω, ὅτι τῶν Α, Β, Γ πλείους εἰσὶ πρῶτοι ἀριθμοί. the (set of all) primes numbers is more numerous than A,
Εἰλήφθω γὰρ ὁ ὑπὸ τῶν Α, Β, Γ ἐλάχιστος μετρούμενος B, C.
καὶ ἔστω ΔΕ, καὶ προσκείσθω τῷ ΔΕ μονὰς ἡ ΔΖ. ὁ δὴ ΕΖ For let the least number measured by A, B, C have
ἤτοι πρῶτός ἐστιν ἢ οὔ. ἔστω πρότερον πρῶτος· εὐρημένοι been taken, and let it be DE [Prop. 7.36]. And let the
ἄρα εἰσὶ πρῶτοι ἀριθμοὶ οἱ Α, Β, Γ, ΕΖ πλείους τῶν Α, Β, unit DF have been added to DE. So EF is either prime,
Γ. or not. Let it, first of all, be prime. Thus, the (set of)
᾿Αλλὰ δὴ μὴ ἔστω ὁ ΕΖ πρῶτος· ὑπὸ πρώτου ἄρα τινὸς prime numbers A, B, C, EF, (which is) more numerous
ἀριθμοῦ μετρεῖται. μετρείσθω ὑπὸ πρώτου τοῦ Η· λέγω, than A, B, C, has been found.
ὅτι ὁ Η οὐδενὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. εἰ γὰρ δυνατόν, And so let EF not be prime. Thus, it is measured by
ἔστω. οἱ δὲ Α, Β, Γ τὸν ΔΕ μετροῦσιν· καὶ ὁ Η ἄρα τὸν some prime number [Prop. 7.31]. Let it be measured by
ΔΕ μετρήσει. μετρεῖ δὲ καὶ τὸν ΕΖ· καὶ λοιπὴν τὴν ΔΖ the prime (number) G. I say that G is not the same as
μονάδα μετρήσει ὁ Η ἀριθμὸς ὤν· ὄπερ ἄτοπον. οὐκ ἄρα ὁ any of A, B, C. For, if possible, let it be (the same). And
Η ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. καὶ ὑπόκειται πρῶτος. A, B, C (all) measure DE. Thus, G will also measure
kaþ the same as one of A, B, C. And it was assumed (to be)
εὑρημένοι ἄρα εἰσὶ πρῶτοι ἀριθμοὶ πλείους τοῦ προτεθέντος DE. And it also measures EF. (So) G will also mea-
πλήθους τῶν Α, Β, Γ οἱ Α, Β, Γ, Η· ὅπερ ἔδει δεῖξαι. sure the remainder, unit DF, (despite) being a number
[Prop. 7.28]. The very thing (is) absurd. Thus, G is not
prime. Thus, the (set of) prime numbers A, B, C, G,
(which is) more numerous than the assigned multitude
(of prime numbers), A, B, C, has been found. (Which is)
the very thing it was required to show.
.
Proposition 21
᾿Εὰν ἄρτιοι ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, ὁ ὅλος If any multitude whatsoever of even numbers is added
ἄρτιός ἐστιν. together then the whole is even.
Α Β Γ ∆ Ε A B C D E
Συγκείσθωσαν γὰρ ἄρτιοι ἀριθμοὶ ὁποσοιοῦν οἱ ΑΒ, For let any multitude whatsoever of even numbers,
ΒΓ, ΓΔ, ΔΕ· λέγω, ὅτι ὅλος ὁ ΑΕ ἄρτιός ἐστιν. AB, BC, CD, DE, lie together. I say that the whole,
᾿Επεὶ γὰρ ἕκαστος τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ ἄρτιός ἐστιν, AE, is even.
ἔχει μέρος ἥμισυ· ὥστε καὶ ὅλος ὁ ΑΕ ἔχει μέρος ἥμισυ. For since everyone of AB, BC, CD, DE is even, it
ἄρτιος δὲ ἀριθμός ἐστιν ὁ δίχα διαιρούμενος· ἄρτιος ἄρα has a half part [Def. 7.6]. And hence the whole AE has
ἐστὶν ὁ ΑΕ· ὅπερ ἔδει δεῖξαι. a half part. And an even number is one (which can be)
divided in half [Def. 7.6]. Thus, AE is even. (Which is)
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ST EW jþ. kbþ ELEMENTS BOOK 9
Proposition 22
.
᾿Εὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ the very thing it was required to show.
If any multitude whatsoever of odd numbers is added
πλῆθος αὐτῶν ἄρτιον ᾖ, ὁ ὅλος ἄρτιος ἔσται. together, and the multitude of them is even, then the
whole will be even.
Α Β Γ ∆ Ε A B C D E
Συγκείσθωσαν γὰρ περισσοὶ ἀριθμοὶ ὁσοιδηποτοῦν For let any even multitude whatsoever of odd num-
ἄρτιοι τὸ πλῆθος οἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ· λέγω, ὅτι ὅλος bers, AB, BC, CD, DE, lie together. I say that the
kgþ
ὁ ΑΕ ἄρτιός ἐστιν. whole, AE, is even.
᾿Επεὶ γὰρ ἕκαστος τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ περιττός ἐστιν, For since everyone of AB, BC, CD, DE is odd then, a
ἀφαιρεθείσης μονάδος ἀφ᾿ ἑκάστου ἕκαστος τῶν λοιπῶν unit being subtracted from each, everyone of the remain-
ἄρτιος ἔσται· ὥστε καὶ ὁ συγκείμενος ἐξ αὐτῶν ἄρτιος ders will be (made) even [Def. 7.7]. And hence the sum
ἔσται. ἔστι δὲ καὶ τὸ πλῆθος τῶν μονάδων ἄρτιον. καὶ of them will be even [Prop. 9.21]. And the multitude
ὅλος ἄρα ὁ ΑΕ ἄρτιός ἐστιν· ὅπερ ἔδει δεῖξαι. of the units is even. Thus, the whole AE is also even
[Prop. 9.21]. (Which is) the very thing it was required to
show.
Proposition 23
.
᾿Εὰν περισσοὶ ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, τὸ δὲ If any multitude whatsoever of odd numbers is added
πλῆθος αὐτῶν περισσὸν ᾖ, καὶ ὁ ὅλος περισσὸς ἔσται. together, and the multitude of them is odd, then the
whole will also be odd.
Α Β Γ Ε ∆ A B C E D
Συγκείσθωσαν γὰρ ὁποσοιοῦν περισσοὶ ἀριθμοί, ὧν τὸ For let any multitude whatsoever of odd numbers,
kdþ
πλῆθος περισσὸν ἔστω, οἱ ΑΒ, ΒΓ, ΓΔ· λέγω, ὅτι καὶ ὅλος AB, BC, CD, lie together, and let the multitude of them
ὁ ΑΔ περισσός ἐστιν. be odd. I say that the whole, AD, is also odd.
᾿Αφῃρήσθω ἀπὸ τοῦ ΓΔ μονὰς ἡ ΔΕ· λοιπὸς ἄρα ὁ ΓΕ For let the unit DE have been subtracted from CD.
ἄρτιός ἐστιν. ἔστι δὲ καὶ ὁ ΓΑ ἄρτιος· καὶ ὅλος ἄρα ὁ ΑΕ The remainder CE is thus even [Def. 7.7]. And CA
ἄρτιός ἐστιν. καί ἐστι μονὰς ἡ ΔΕ. περισσὸς ἄρα ἐστὶν ὁ is also even [Prop. 9.22]. Thus, the whole AE is also
ΑΔ· ὅπερ ἔδει δεῖξαι. even [Prop. 9.21]. And DE is a unit. Thus, AD is odd
[Def. 7.7]. (Which is) the very thing it was required to
show.
Proposition 24
.
᾿Εὰν ἀπὸ ἀρτίου ἀριθμοῦ ἄρτιος ἀφαιρεθῇ, ὁ λοιπὸς If an even (number) is subtracted from an(other) even
ἄρτιος ἔσται. number then the remainder will be even.
Α Γ Β A C B
᾿Απὸ γὰρ ἀρτίου τοῦ ΑΒ ἄρτιος ἀφῃρήσθω ὁ ΒΓ· λέγω, For let the even (number) BC have been subtracted
ὅτι ὁ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν. from the even number AB. I say that the remainder CA
᾿Επεὶ γὰρ ὁ ΑΒ ἄρτιός ἐστιν, ἔχει μέρος ἥμισυ. διὰ τὰ is even.
αὐτὰ δὴ καὶ ὁ ΒΓ ἔχει μέρος ἥμισυ· ὥστε καὶ λοιπὸς [ὁ ΓΑ For since AB is even, it has a half part [Def. 7.6]. So,
ἔχει μέρος ἥμισυ] ἄρτιος [ἄρα] ἐστὶν ὁ ΑΓ· ὅπερ ἔδει δεῖξαι. for the same (reasons), BC also has a half part. And
hence the remainder [CA has a half part]. [Thus,] AC is
even. (Which is) the very thing it was required to show.
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ST EW jþ. keþ ELEMENTS BOOK 9
Proposition 25
.
᾿Εὰν ἀπὸ ἀρτίου ἀριθμοῦ περισσὸς ἀφαιρεθῇ, ὁ λοιπὸς If an odd (number) is subtracted from an even num-
περισσὸς ἔσται. ber then the remainder will be odd.
Α Γ ∆ Β A C D B
kþ
᾿Απὸ γὰρ ἀρτίου τοῦ ΑΒ περισσὸς ἀφῃρήσθω ὁ ΒΓ· For let the odd (number) BC have been subtracted
λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ περισσός ἐστιν. from the even number AB. I say that the remainder CA
᾿Αφῃρήσθω γὰρ ἀπὸ τοῦ ΒΓ μονὰς ἡ ΓΔ· ὁ ΔΒ ἄρα is odd.
ἄρτιός ἐστιν. ἔστι δὲ καὶ ὁ ΑΒ ἄρτιος· καὶ λοιπὸς ἄρα ὁ For let the unit CD have been subtracted from BC.
ΑΔ ἄρτιός ἐστιν. καί ἐστι μονὰς ἡ ΓΔ· ὁ ΓΑ ἄρα περισσός DB is thus even [Def. 7.7]. And AB is also even. And
ἐστιν· ὅπερ ἔδει δεῖξαι. thus the remainder AD is even [Prop. 9.24]. And CD is
a unit. Thus, CA is odd [Def. 7.7]. (Which is) the very
thing it was required to show.
.
Proposition 26
᾿Εὰν ἀπὸ περισσοῦ ἀριθμοῦ περισσὸς ἀφαιρεθῇ, ὁ λοιπὸς If an odd (number) is subtracted from an odd number
ἄρτιος ἔσται. then the remainder will be even.
Α Γ ∆ Β A C D B
kzþ
᾿Απὸ γὰρ περισσοῦ τοῦ ΑΒ περισσὸς ἀφῃρήσθω ὁ ΒΓ· For let the odd (number) BC have been subtracted
λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν. from the odd (number) AB. I say that the remainder CA
᾿Επεὶ γὰρ ὁ ΑΒ περισσός ἐστιν, ἀφῃρήσθω μονὰς ἡ ΒΔ· is even.
λοιπὸς ἄρα ὁ ΑΔ ἄρτιός ἐστιν. διὰ τὰ αὐτὰ δὴ καὶ ὁ ΓΔ For since AB is odd, let the unit BD have been
ἄρτιός ἐστιν· ὥστε καὶ λοιπὸς ὁ ΓΑ ἄρτιός ἐστιν· ὅπερ ἔδει subtracted (from it). Thus, the remainder AD is even
δεῖξαι. [Def. 7.7]. So, for the same (reasons), CD is also
even. And hence the remainder CA is even [Prop. 9.24].
(Which is) the very thing it was required to show.
.
Proposition 27
᾿Εὰν ἀπὸ περισσοῦ ἀριθμοῦ ἄρτιος ἀφαιρεθῇ, ὁ λοιπὸς If an even (number) is subtracted from an odd num-
περισσὸς ἔσται. ber then the remainder will be odd.
Α ∆ Γ Β A D C B
khþ the remainder CD is also even [Prop. 9.24]. CA (is) thus
᾿Απὸ γὰρ περισσοῦ τοῦ ΑΒ ἄρτιος ἀφῃρήσθω ὁ ΒΓ· For let the even (number) BC have been subtracted
λέγω, ὅτι ὁ λοιπὸς ὁ ΓΑ περισσός ἐστιν. from the odd (number) AB. I say that the remainder CA
᾿Αφῃρήσθω [γὰρ] μονὰς ἡ ΑΔ· ὁ ΔΒ ἄρα ἄρτιός ἐστιν. is odd.
ἔστι δὲ καὶ ὁ ΒΓ ἄρτιος· καὶ λοιπὸς ἄρα ὁ ΓΔ ἄρτιός ἐστιν. [For] let the unit AD have been subtracted (from AB).
περισσὸς ἄρα ὁ ΓΑ· ὅπερ ἔδει δεῖξαι. DB is thus even [Def. 7.7]. And BC is also even. Thus,
odd [Def. 7.7]. (Which is) the very thing it was required
to show.
.
Proposition 28
᾿Εὰν περισσὸς ἀριθμὸς ἄρτιον πολλαπλασιάσας ποιῇ If an odd number makes some (number by) multiply-
τινα, ὁ γενόμενος ἄρτιος ἔσται. ing an even (number) then the created (number) will be
even.
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ST EW jþ.
Α A ELEMENTS BOOK 9
Β B
Γ C
Περισσὸς γὰρ ἀριθμὸς ὁ Α ἄρτιον τὸν Β πολλα- For let the odd number A make C (by) multiplying
kjþ
πλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ ἄρτιός ἐστιν. the even (number) B. I say that C is even.
᾿Επεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, For since A has made C (by) multiplying B, C is thus
ὁ Γ ἄρα σύγκειται ἐκ τοσούτων ἴσων τῷ Β, ὅσαι εἰσὶν ἐν composed out of so many (magnitudes) equal to B, as
τῷ Α μονάδες. καί ἐστιν ὁ Β ἄρτιος· ὁ Γ ἄρα σύγκειται many as (there) are units in A [Def. 7.15]. And B is
ἐξ ἀρτίων. ἐὰν δὲ ἄρτιοι ἀριθμοὶ ὁποσοιοῦν συντεθῶσιν, ὁ even. Thus, C is composed out of even (numbers). And
ὅλος ἄρτιός ἐστιν. ἄρτιος ἄρα ἐστὶν ὁ Γ· ὅπερ ἔδει δεῖξαι. if any multitude whatsoever of even numbers is added
together then the whole is even [Prop. 9.21]. Thus, C is
even. (Which is) the very thing it was required to show.
.
Proposition 29
᾿Εὰν περισσὸς ἀριθμὸς περισσὸν ἀριθμὸν πολλαπλασιάς- If an odd number makes some (number by) multiply-
ας ποιῇ τινα, ὁ γενόμενος περισσὸς ἔσται. ing an odd (number) then the created (number) will be
odd.
Α A
Β B
Γ C
Περισσὸς γὰρ ἀριθμὸς ὁ Α περισσὸν τὸν Β πολλα- For let the odd number A make C (by) multiplying
lþ
πλασιάσας τὸν Γ ποιείτω· λέγω, ὅτι ὁ Γ περισσός ἐστιν. the odd (number) B. I say that C is odd.
᾿Επεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, For since A has made C (by) multiplying B, C is thus
ὁ Γ ἄρα σύγκειται ἐκ τοσούτων ἴσων τῷ Β, ὅσαι εἰσὶν ἐν τῷ composed out of so many (magnitudes) equal to B, as
Α μονάδες. καί ἐστιν ἑκάτερος τῶν Α, Β περισσός· ὁ Γ ἄρα many as (there) are units in A [Def. 7.15]. And each
σύγκειται ἐκ περισσῶν ἀριθμῶν, ὧν τὸ πλῆθος περισσόν of A, B is odd. Thus, C is composed out of odd (num-
ἐστιν. ὥστε ὁ Γ περισσός ἐστιν· ὅπερ ἔδει δεῖξαι. bers), (and) the multitude of them is odd. Hence C is odd
[Prop. 9.23]. (Which is) the very thing it was required to
show.
Proposition 30
.
᾿Εὰν περισσὸς ἀριθμὸς ἄρτιον ἀριθμὸν μετρῇ, καὶ τὸν If an odd number measures an even number then it
ἥμισυν αὐτοῦ μετρήσει. will also measure (one) half of it.
Α A
Β B
Γ C
Περισσὸς γὰρ ἀριθμὸς ὁ Α ἄρτιον τὸν Β μετρείτω· λέγω, For let the odd number A measure the even (number)
ὅτι καὶ τὸν ἥμισυν αὐτοῦ μετρήσει. B. I say that (A) will also measure (one) half of (B).
᾿Επεὶ γὰρ ὁ Α τὸν Β μετρεῖ, μετρείτω αὐτὸν κατὰ τὸν For since A measures B, let it measure it according to
Γ· λέγω, ὅτι ὁ Γ οὐκ ἔστι περισσός. εἰ γὰρ δυνατόν, ἔστω. C. I say that C is not odd. For, if possible, let it be (odd).
καὶ ἐπεὶ ὁ Α τὸν Β μετρεῖ κατὰ τὸν Γ, ὁ Α ἄρα τὸν Γ And since A measures B according to C, A has thus made
πολλαπλασιάσας τὸν Β πεποίηκεν. ὁ Β ἄρα σύγκειται ἐκ B (by) multiplying C. Thus, B is composed out of odd
περισσῶν ἀριθμῶν, ὧν τὸ πλῆθος περισσόν ἐστιν. ὁ Β ἄρα numbers, (and) the multitude of them is odd. B is thus
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laþ
ELEMENTS BOOK 9
περισσός ἐστιν· ὅπερ ἄτοπον· ὑπόκειται γὰρ ἄρτιος. οὐκ odd [Prop. 9.23]. The very thing (is) absurd. For (B)
ἄρα ὁ Γ περισσός ἐστιν· ἄρτιος ἄρα ἐστὶν ὁ Γ. ὥστε ὁ Α was assumed (to be) even. Thus, C is not odd. Thus, C
τὸν Β μετρεῖ ἀρτιάκις. διὰ δὴ τοῦτο καὶ τὸν ἥμισυν αὐτοῦ is even. Hence, A measures B an even number of times.
μετρήσει· ὅπερ ἔδει δεῖξαι. So, on account of this, (A) will also measure (one) half
of (B). (Which is) the very thing it was required to show.
.
Proposition 31
᾿Εὰν περισσὸς ἀριθμὸς πρός τινα ἀριθμὸν πρῶτος ᾖ, καὶ If an odd number is prime to some number then it will
πρὸς τὸν διπλασίονα αὐτοῦ πρῶτος ἔσται. also be prime to its double.
Α A
Β B
Γ C
∆ D
Περισσὸς γὰρ ἀριθμὸς ὁ Α πρός τινα ἀριθμὸν τὸν Β For let the odd number A be prime to some number
πρῶτος ἔστω, τοῦ δὲ Β διπλασίων ἔστω ὁ Γ· λέγω, ὅτι ὁ Α B. And let C be double B. I say that A is [also] prime to
[καὶ] πρὸς τὸν Γ πρῶτός ἐστιν. C.
Εἰ γὰρ μή εἰσιν [οἱ Α, Γ] πρῶτοι, μετρήσει τις αὐτοὺς For if [A and C] are not prime (to one another) then
ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. καί ἐστιν ὁ Α περισσός· some number will measure them. Let it measure (them),
περισσὸς ἄρα καὶ ὁ Δ. καὶ ἐπεὶ ὁ Δ περισσὸς ὢν τὸν Γ and let it be D. And A is odd. Thus, D (is) also odd.
lbþ
μετρεῖ, καί ἐστιν ὁ Γ ἄρτιος, καὶ τὸν ἥμισυν ἄρα τοῦ Γ And since D, which is odd, measures C, and C is even,
μετρήσει [ὁ Δ]. τοῦ δὲ Γ ἥμισύ ἐστιν ὁ Β· ὁ Δ ἄρα τὸν [D] will thus also measure half of C [Prop. 9.30]. And B
Β μετρεῖ. μετρεῖ δὲ καὶ τὸν Α. ὁ Δ ἄρα τοὺς Α, Β μετρεῖ is half of C. Thus, D measures B. And it also measures
πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον. οὐκ A. Thus, D measures (both) A and B, (despite) them
ἄρα ὁ Α πρὸς τὸν Γ πρῶτος οὔκ ἐστιν. οἱ Α, Γ ἄρα πρῶτοι being prime to one another. The very thing is impossible.
πρὸς ἀλλήλους εἰσίν· ὅπερ ἔδει δεῖξαι. Thus, A is not unprime to C. Thus, A and C are prime to
one another. (Which is) the very thing it was required to
show.
.
Proposition 32
Τῶν ἀπὸ δύαδος διπλασιαζομένων ἀριθμων ἕκαστος Each of the numbers (which is continually) doubled,
ἀρτιάκις ἄρτιός ἐστι μόνον. (starting) from a dyad, is an even-times-even (number)
only.
Α A
Β B
Γ C
∆ D
᾿Απὸ γὰρ δύαδος τῆς Α δεδιπλασιάσθωσαν ὁσοιδη- For let any multitude of numbers whatsoever, B, C,
ποτοῦν ἀριθμοὶ οἱ Β, Γ, Δ· λέγω, ὅτι οἱ Β, Γ, Δ ἀρτιάκις D, have been (continually) doubled, (starting) from the
ἄρτιοί εἰσι μόνον. dyad A. I say that B, C, D are even-times-even (num-
῞Οτι μὲν οὖν ἕκαστος [τῶν Β, Γ, Δ] ἀρτιάκις ἄρτιός bers) only.
ἐστιν, φανερόν· ἀπὸ γὰρ δυάδος ἐστὶ διπλασιασθείς. λέγω, In fact, (it is) clear that each [of B, C, D] is an
ὅτι καὶ μόνον. ἐκκείσθω γὰρ μονάς. ἐπεὶ οὖν ἀπὸ μονάδος even-times-even (number). For it is doubled from a dyad
ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογόν εἰσιν, ὁ δὲ μετὰ τὴν [Def. 7.8]. I also say that (they are even-times-even num-
μονάδα ὁ Α πρῶτός ἐστιν, ὁ μέγιστος τῶν Α, Β, Γ, Δ ὁ bers) only. For let a unit be laid down. Therefore, since
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ELEMENTS BOOK 9
Δ ὑπ᾿ οὐδενὸς ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ. καί any multitude of numbers whatsoever are continuously
lgþ D is an even-time-even (number) only [Def. 7.8]. So,
ἐστιν ἕκαστος τῶν Α, Β, Γ ἄρτιος· ὁ Δ ἄρα ἀρτιάκις ἄρτιός proportional, starting from a unit, and the (number) A af-
ἐστι μόνον. ὁμοίως δὴ δείξομεν, ὅτι [καὶ] ἑκάτερος τῶν Β, ter the unit is prime, the greatest of A, B, C, D, (namely)
Γ ἀρτιάκις ἄρτιός ἐστι μόνον· ὅπερ ἔδει δεῖξαι. D, will not be measured by any other (numbers) except
A, B, C [Prop. 9.13]. And each of A, B, C is even. Thus,
similarly, we can show that each of B, C is [also] an even-
time-even (number) only. (Which is) the very thing it was
required to show.
Proposition 33
.
᾿Εὰν ἀριθμὸς τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις πε- If a number has an odd half then it is an even-time-
ρισσός ἐστι μόνον. odd (number) only.
Α A
᾿Αριθμὸς γὰρ ὁ Α τὸν ἥμισυν ἐχέτω περισσόν· λέγω, ὅτι For let the number A have an odd half. I say that A is
ὁ Α ἀρτιάκις περισσός ἐστι μόνον. an even-times-odd (number) only.
῞Οτι μὲν οὖν ἀρτιάκις περισσός ἐστιν, φανερόν· ὁ γὰρ In fact, (it is) clear that (A) is an even-times-odd
ἥμισυς αὐτοῦ περισσὸς ὢν μετρεῖ αὐτὸν ἀρτιάκις, λέγω δή, (number). For its half, being odd, measures it an even
ldþ (despite) being odd. The very thing is absurd. Thus, A
ὅτι καὶ μόνον. εἰ γὰρ ἔσται ὁ Α καὶ ἀρτιάκις ἄρτιος, με- number of times [Def. 7.9]. So I also say that (it is
τρηθήσεται ὑπὸ ἀρτίου κατὰ ἄρτιον ἀριθμόν· ὥστε καὶ ὁ an even-times-odd number) only. For if A is also an
ἥμισυς αὐτοῦ μετρηθήσεται ὑπὸ ἀρτίου ἀριθμοῦ περισσὸς even-times-even (number) then it will be measured by an
ὤν· ὅπερ ἐστὶν ἄτοπον. ὁ Α ἄρα ἀρτιάκις περισσός ἐστι even (number) according to an even number [Def. 7.8].
μόνον· ὅπερ ἔδει δεῖξαι. Hence, its half will also be measured by an even number,
is an even-times-odd (number) only. (Which is) the very
thing it was required to show.
.
Proposition 34
᾿Εὰν ἀριθμὸς μήτε τῶν ἀπὸ δυάδος διπλασιαζομένων ᾖ, If a number is neither (one) of the (numbers) doubled
μήτε τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις τε ἄρτιός ἐστι καὶ from a dyad, nor has an odd half, then it is (both) an
ἀρτιάκις περισσός. even-times-even and an even-times-odd (number).
Α A
᾿Αριθμὸς γὰρ ὁ Α μήτε τῶν ἀπὸ δυάδος διπλασιαζομένων For let the number A neither be (one) of the (num-
ἔστω μήτε τὸν ἥμισυν ἐχέτω περισσόν· λέγω, ὅτι ὁ Α bers) doubled from a dyad, nor let it have an odd half.
ἀρτιάκις τέ ἐστιν ἄρτιος καὶ ἀρτιάκις περισσός. I say that A is (both) an even-times-even and an even-
῞Οτι μὲν οὖν ὁ Α ἀρτιάκις ἐστὶν ἄρτιος, φανερόν· τὸν times-odd (number).
γὰρ ἥμισυν οὐκ ἔχει περισσόν. λέγω δή, ὅτι καὶ ἀρτιάκις πε- In fact, (it is) clear that A is an even-times-even (num-
ρισσός ἐστιν. ἐὰν γὰρ τὸν Α τέμνωμεν δίχα καὶ τὸν ἥμισυν ber) [Def. 7.8]. For it does not have an odd half. So I
αὐτοῦ δίχα καὶ τοῦτο ἀεὶ ποιῶμεν, καταντήσομεν εἴς τινα say that it is also an even-times-odd (number). For if we
ἀριθμὸν περισσόν, ὃς μετρήσει τὸν Α κατὰ ἄρτιον ἀριθμόν. cut A in half, and (then cut) its half in half, and we do
εἰ γὰρ οὔ, καταντήσομεν εἰς δυάδα, καὶ ἔσται ὁ Α τῶν ἀπὸ this continually, then we will arrive at some odd num-
δυάδος διπλασιαζομένων· ὅπερ οὐχ ὑπόκειται. ὥστε ὁ Α ber which will measure A according to an even number.
ἀρτιάκις περισσόν ἐστιν. ἐδείχθη δὲ καὶ ἀρτιάκις ἄρτιος. ὁ For if not, we will arrive at a dyad, and A will be (one)
Α ἄρα ἀρτιάκις τε ἄρτιός ἐστι καὶ ἀρτιάκις περισσός· ὅπερ of the (numbers) doubled from a dyad. The very oppo-
ἔδει δεῖξαι. site thing (was) assumed. Hence, A is an even-times-odd
(number) [Def. 7.9]. And it was also shown (to be) an
even-times-even (number). Thus, A is (both) an even-
times-even and an even-times-odd (number). (Which is)
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.
†
Proposition 35
᾿Εὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, ἀφαι- the very thing it was required to show.
If there is any multitude whatsoever of continually
ρεθῶσι δὲ ἀπό τε τοῦ δευτέρου καὶ τοῦ ἐσχάτου ἴσοι τῷ proportional numbers, and (numbers) equal to the first
πρώτῳ, ἔσται ὡς ἡ τοῦ δευτέρου ὑπεροχὴ πρὸς τὸν πρῶτον, are subtracted from (both) the second and the last, then
οὕτως ἡ τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ as the excess of the second (number is) to the first, so the
πάντας. excess of the last will be to (the sum of) all those (num-
bers) before it.
Α A
Β Η Γ B G C
∆ D
Ε Λ Κ Θ Ζ E L K H F
῎Εστωσαν ὁποσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Let A, BC, D, EF be any multitude whatsoever of
ΒΓ, Δ, ΕΖ ἀφχόμενοι ἀπὸ ἐλαχίστου τοῦ Α, καὶ ἀφῃρήσθω continuously proportional numbers, beginning from the
ἀπὸ τοῦ ΒΓ καὶ τοῦ ΕΖ τῲ Α ἴσος ἑκάτερος τῶν ΒΗ, ΖΘ· least A. And let BG and FH, each equal to A, have been
λέγω, ὅτι ἐστὶν ὡς ὁ ΗΓ πρὸς τὸν Α, οὕτως ὁ ΕΘ πρὸς subtracted from BC and EF (respectively). I say that as
τοὺς Α, ΒΓ, Δ. GC is to A, so EH is to A, BC, D.
Κείσθω γὰρ τῷ μὲν ΒΓ ἴσος ὁ ΖΚ, τῷ δὲ Δ ἴσος ὁ ΖΛ. For let FK be made equal to BC, and FL to D. And
καὶ ἐπεὶ ὁ ΖΚ τῷ ΒΓ ἴσος ἐστίν, ὧν ὁ ΖΘ τῷ ΒΗ ἴσος ἐστίν, since FK is equal to BC, of which FH is equal to BG,
λοιπὸς ἄρα ὁ ΘΚ λοιπῷ τῷ ΗΓ ἐστιν ἴσος. καὶ ἐπεί ἐστιν ὡς the remainder HK is thus equal to the remainder GC.
ὁ ΕΖ πρὸς τὸν Δ, οὕτως ὁ Δ πρὸς τὸν ΒΓ καὶ ὁ ΒΓ πρὸς And since as EF is to D, so D (is) to BC, and BC to
τὸν Α, ἴσος δὲ ὁ μὲν Δ τῷ ΖΛ, ὁ δὲ ΒΓ τῷ ΖΚ, ὁ δὲ Α τῷ A [Prop. 7.13], and D (is) equal to FL, and BC to FK,
ΖΘ, ἔστιν ἄρα ὡς ὁ ΕΖ πρὸς τὸν ΖΛ, οὕτως ὁ ΛΖ πρὸς τὸν and A to FH, thus as EF is to FL, so LF (is) to FK, and
ΖΚ καὶ ὁ ΖΚ πρὸς τὸν ΖΘ. διελόντι, ὡς ὁ ΕΛ πρὸς τὸν ΛΖ, FK to FH. By separation, as EL (is) to LF, so LK (is)
οὕτως ὁ ΛΚ πρὸς τὸν ΖΚ καὶ ὁ ΚΘ πρὸς τὸν ΖΘ. ἔστιν ἄρα to FK, and KH to FH [Props. 7.11, 7.13]. And thus as
καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως one of the leading (numbers) is to one of the following,
ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους· ἔστιν so (the sum of) all of the leading (numbers is) to (the
ἄρα ὡς ὁ ΚΘ πρὸς τὸν ΖΘ, οὕτως οἱ ΕΛ, ΛΚ, ΚΘ πρὸς sum of) all of the following [Prop. 7.12]. Thus, as KH
τοὺς ΛΖ, ΖΚ, ΘΖ. ἴσος δὲ ὁ μὲν ΚΘ τῷ ΓΗ, ὁ δὲ ΖΘ τῷ is to FH, so EL, LK, KH (are) to LF, FK, HF. And
Α, οἱ δὲ ΛΖ, ΖΚ, ΘΖ τοὶς Δ, ΒΓ, Α· ἔστιν ἄρα ὡς ὁ ΓΗ KH (is) equal to CG, and FH to A, and LF, FK, HF
lþ was required to show.
πρὸς τὸν Α, οὕτως ὁ ΕΘ πρὸς τοὺς Δ, ΒΓ, Α. ἔστιν ἄρα to D, BC, A. Thus, as CG is to A, so EH (is) to D,
ὡς ἡ τοῦ δευτέρου ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ τοῦ BC, A. Thus, as the excess of the second (number) is to
ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας· ὅπερ ἔδει the first, so the excess of the last (is) to (the sum of) all
δεῖξαι. those (numbers) before it. (Which is) the very thing it
† This proposition allows us to sum a geometric series of the form a, a r, a r , a r , · · · a r n−1 . According to Euclid, the sum S n satisfies
3
2
n
n
(a r − a)/a = (a r − a)/S n. Hence, S n = a (r − 1)/(r − 1). †
Proposition 36
.
᾿Εὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἐκτεθῶσιν ἐν If any multitude whatsoever of numbers is set out con-
τῇ διπλασίονι ἀναλογίᾳ, ἕως οὗ ὁ σύμπας συντεθεὶς πρῶτος tinuously in a double proportion, (starting) from a unit,
γένηται, καὶ ὁ σύμπας ἐπὶ τὸν ἔσχατον πολλαπλασιασθεὶς until the whole sum added together becomes prime, and
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ποιῇ τινα, ὁ γενόμενος τέλειος ἔσται. the sum multiplied into the last (number) makes some
᾿Απὸ γὰρ μονάδος ἐκκείσθωσαν ὁσοιδηποτοῦν ἀριθμ- (number), then the (number so) created will be perfect.
οὶ ἐν τῇ διπλασίονι ἀναλογίᾳ, ἕως οὗ ὁ σύμπας συντεθεὶς For let any multitude of numbers, A, B, C, D, be set
πρῶτος γένηται, οἱ Α, Β, Γ, Δ, καὶ τῷ σύμπαντι ἴσος ἔστω out (continuouly) in a double proportion, until the whole
ὁ Ε, καὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν ΖΗ ποιείτω. λέγω, sum added together is made prime. And let E be equal to
ὅτι ὁ ΖΗ τέλειός ἐστιν. the sum. And let E make FG (by) multiplying D. I say
that FG is a perfect (number).
A
Α
Β B
Γ C
∆ D
Ε E
Θ Ν Κ H N K
Λ L
Μ M
Ζ Ξ Η F O G
Ο P
Π Q
῞Οσοι γάρ εἰσιν οἱ Α, Β, Γ, Δ τῷ πλήθει, τοσοῦτοι ἀπὸ For as many as is the multitude of A, B, C, D, let so
τοῦ Ε εἰλήφθωσαν ἐν τῇ διπλασίονι ἀναλογίᾳ οἱ Ε, ΘΚ, Λ, many (numbers), E, HK, L, M, have been taken in a
Μ· δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, οὕτως ὁ Ε πρὸς double proportion, (starting) from E. Thus, via equal-
τὸν Μ. ὁ ἄρα ἐκ τῶν Ε, Δ ἴσος ἐστὶ τῷ ἐκ τῶν Α, Μ. καί ity, as A is to D, so E (is) to M [Prop. 7.14]. Thus, the
ἐστιν ὁ ἐκ τῶν Ε, Δ ὁ ΖΗ· καὶ ὁ ἐκ τῶν Α, Μ ἄρα ἐστὶν ὁ (number created) from (multiplying) E, D is equal to the
ΖΗ. ὁ Α ἄρα τὸν Μ πολλαπλασιάσας τὸν ΖΗ πεποίηκεν· ὁ (number created) from (multiplying) A, M. And FG is
Μ ἄρα τὸν ΖΗ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. καί ἐστι the (number created) from (multiplying) E, D. Thus,
δυὰς ὁ Α· διπλάσιος ἄρα ἐστὶν ὁ ΖΗ τοῦ Μ. εἰσὶ δὲ καὶ οἱ Μ, FG is also the (number created) from (multiplying) A,
Λ, ΘΚ, Ε ἑξῆς διπλάσιοι ἀλλήλων· οἱ Ε, ΘΚ, Λ, Μ, ΖΗ ἄρα M [Prop. 7.19]. Thus, A has made FG (by) multiplying
ἑξῆς ἀνάλογόν εἰσιν ἐν τῇ διπλασίονι ἀναλογίᾳ. ἀφῃρήσθω M. Thus, M measures FG according to the units in A.
δὴ ἀπὸ τοῦ δευτέρου τοῦ ΘΚ καὶ τοῦ ἐσχάτου τοῦ ΖΗ τῷ And A is a dyad. Thus, FG is double M. And M, L,
πρώτῳ τῷ Ε ἴσος ἑκάτερος τῶν ΘΝ, ΖΞ· ἔστιν ἄρα ὡς ἡ HK, E are also continuously double one another. Thus,
τοῦ δευτέρου ἀριθμοῦ ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ E, HK, L, M, FG are continuously proportional in a
τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας. ἔστιν double proportion. So let HN and FO, each equal to the
ἄρα ὡς ὁ ΝΚ πρὸς τὸν Ε, οὕτως ὁ ΞΗ πρὸς τοὺς Μ, Λ, first (number) E, have been subtracted from the second
ΚΘ, Ε. καί ἐστιν ὁ ΝΚ ἴσος τῷ Ε· καὶ ὁ ΞΗ ἄρα ἴσος ἐστὶ (number) HK and the last FG (respectively). Thus, as
τοῖς Μ, Λ, ΘΚ, Ε. ἔστι δὲ καὶ ὁ ΖΞ τῷ Ε ἴσος, ὁ δὲ Ε the excess of the second number is to the first, so the ex-
τοῖς Α, Β, Γ, Δ καὶ τῇ μονάδι. ὅλος ἄρα ὁ ΖΗ ἴσος ἐστὶ cess of the last (is) to (the sum of) all those (numbers)
τοῖς τε Ε, ΘΚ, Λ, Μ καὶ τοῖς Α, Β, Γ, Δ καὶ τῇ μονάδι· before it [Prop. 9.35]. Thus, as NK is to E, so OG (is)
καὶ μετρεῖται ὑπ᾿ αὐτῶν. λέγω, ὅτι καὶ ὁ ΖΗ ὐπ᾿ οὐδενὸς to M, L, KH, E. And NK is equal to E. And thus OG
ἄλλου μετρηθήσεται παρὲξ τῶν Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ is equal to M, L, HK, E. And FO is also equal to E,
καὶ τῆς μονάδος. εἰ γὰρ δυνατόν, μετρείτω τις τὸν ΖΗ ὁ and E to A, B, C, D, and a unit. Thus, the whole of FG
Ο, καὶ ὁ Ο μηδενὶ τῶν Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ ἔστω ὁ is equal to E, HK, L, M, and A, B, C, D, and a unit.
αὐτός. καὶ ὁσάκις ὁ Ο τὸν ΖΗ μετρεῖ, τοσαῦται μονάδες And it is measured by them. I also say that FG will be
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ἔστωσαν ἐν τῷ Π· ὁ Π ἄρα τὸν Ο πολλαπλασιάσας τὸν ΖΗ measured by no other (numbers) except A, B, C, D, E,
πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν HK, L, M, and a unit. For, if possible, let some (num-
ΖΗ πεποίηκεν· ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Π, ὁ Ο πρὸς τὸν ber) P measure FG, and let P not be the same as any
Δ. καὶ ἐπεὶ ἀπὸ μονάδος ἑξῆς ἀνάλογόν εἰσιν οἱ Α, Β, Γ, of A, B, C, D, E, HK, L, M. And as many times as P
Δ, ὁ Δ ἄρα ὑπ᾿ οὐδενὸς ἄλλου ἀριθμοῦ μετρηθήσεται παρὲξ measures FG, so many units let there be in Q. Thus, Q
τῶν Α, Β, Γ. καὶ ὑπόκειται ὁ Ο οὐδενὶ τῶν Α, Β, Γ ὁ αὐτός· has made FG (by) multiplying P. But, in fact, E has also
οὐκ ἄρα μετρήσει ὁ Ο τὸν Δ. ἀλλ᾿ ὡς ὁ Ο πρὸς τὸν Δ, ὁ made FG (by) multiplying D. Thus, as E is to Q, so P
Ε πρὸς τὸν Π· οὐδὲ ὁ Ε ἄρα τὸν Π μετρεῖ. καί ἐστιν ὁ Ε (is) to D [Prop. 7.19]. And since A, B, C, D are con-
πρῶτος· πᾶς δὲ πρῶτος ἀριθμὸς πρὸς ἅπαντα, ὃν μὴ μετρεῖ, tinually proportional, (starting) from a unit, D will thus
πρῶτός [ἐστιν]. οἱ Ε, Π ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. οἱ not be measured by any other numbers except A, B, C
δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν [Prop. 9.13]. And P was assumed not (to be) the same
αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον as any of A, B, C. Thus, P does not measure D. But,
καὶ ὁ ἑπόμενος τὸν ἑπόμενον· καί ἐστιν ὡς ὁ Ε πρὸς τὸν Π, as P (is) to D, so E (is) to Q. Thus, E does not mea-
ὁ Ο πρὸς τὸν Δ. ἰσάκις ἄρα ὁ Ε τὸν Ο μετρεῖ καὶ ὁ Π τὸν sure Q either [Def. 7.20]. And E is a prime (number).
Δ. ὁ δὲ Δ ὑπ᾿ οὐδενὸς ἄλλου μετρεῖται παρὲξ τῶν Α, Β, Γ· And every prime number [is] prime to every (number)
ὁ Π ἄρα ἑνὶ τῶν Α, Β, Γ ἐστιν ὁ αὐτός. ἔστω τῷ Β ὁ αὐτός. which it does not measure [Prop. 7.29]. Thus, E and Q
καὶ ὅσοι εἰσὶν οἱ Β, Γ, Δ τῷ πλήθει τοσοῦτοι εἰλήφθωσαν are prime to one another. And (numbers) prime (to one
ἀπὸ τοῦ Ε οἱ Ε, ΘΚ, Λ. καί εἰσιν οἱ Ε, ΘΚ, Λ τοῖς Β, Γ, Δ another are) also the least (of those numbers having the
ἐν τῷ αὐτῷ λόγω· δι᾿ ἴσου ἄρα ἐστὶν ὡς ὁ Β πρὸς τὸν Δ, ὁ same ratio as them) [Prop. 7.21], and the least (num-
Ε πρὸς τὸν Λ. ὁ ἄρα ἐκ τῶν Β, Λ ἴσος ἐστὶ τῷ ἐκ τῶν Δ, bers) measure those (numbers) having the same ratio as
Ε· ἀλλ᾿ ὁ ἐκ τῶν Δ, Ε ἴσος ἐστὶ τῷ ἐκ τῶν Π, Ο· καὶ ὁ ἐκ them an equal number of times, the leading (measuring)
τῶν Π, Ο ἄρα ἴσος ἐστὶ τῷ ἐκ τῶν Β, Λ. ἔστιν ἄρα ὡς ὁ Π the leading, and the following the following [Prop. 7.20].
πρὸς τὸν Β, ὁ Λ πρὸς τὸν Ο. καί ἐστιν ὁ Π τῷ Β ὁ αὐτός· And as E is to Q, (so) P (is) to D. Thus, E measures P
καὶ ὁ Λ ἄρα τῳ Ο ἐστιν ὁ αὐτός· ὅπερ ἀδύνατον· ὁ γὰρ Ο the same number of times as Q (measures) D. And D
ὑπόκειται μηδενὶ τῶν ἐκκειμένων ὁ αὐτός· οὐκ ἄρα τὸν ΖΗ is not measured by any other (numbers) except A, B, C.
μετρήσει τις ἀριθμὸς παρὲξ τῶν Α, Β, Γ, Δ, Ε, ΘΚ, Λ, Μ Thus, Q is the same as one of A, B, C. Let it be the same
καὶ τῆς μονάδος. καὶ ἐδείχη ὁ ΖΗ τοῖς Α, Β, Γ, Δ, Ε, ΘΚ, as B. And as many as is the multitude of B, C, D, let so
Λ, Μ καὶ τῇ μονάδι ἴσος. τέλειος δὲ ἀριθμός ἐστιν ὁ τοῖς many (of the set out numbers) have been taken, (start-
ἑαυτοῦ μέρεσιν ἴσος ὤν· τέλειος ἄρα ἐστὶν ὁ ΖΗ· ὅπερ ἔδει ing) from E, (namely) E, HK, L. And E, HK, L are in
δεῖξαι. the same ratio as B, C, D. Thus, via equality, as B (is)
to D, (so) E (is) to L [Prop. 7.14]. Thus, the (number
created) from (multiplying) B, L is equal to the (num-
ber created) from multiplying D, E [Prop. 7.19]. But,
the (number created) from (multiplying) D, E is equal
to the (number created) from (multiplying) Q, P. Thus,
the (number created) from (multiplying) Q, P is equal
to the (number created) from (multiplying) B, L. Thus,
as Q is to B, (so) L (is) to P [Prop. 7.19]. And Q is the
same as B. Thus, L is also the same as P. The very thing
(is) impossible. For P was assumed not (to be) the same
as any of the (numbers) set out. Thus, FG cannot be
measured by any number except A, B, C, D, E, HK, L,
M, and a unit. And FG was shown (to be) equal to (the
sum of) A, B, C, D, E, HK, L, M, and a unit. And a
perfect number is one which is equal to (the sum of) its
own parts [Def. 7.22]. Thus, FG is a perfect (number).
(Which is) the very thing it was required to show.
n
† This proposition demonstrates that perfect numbers take the form 2 n−1 (2 − 1) provided that 2 − 1 is a prime number. The ancient Greeks
n
knew of four perfect numbers: 6, 28, 496, and 8128, which correspond to n = 2, 3, 5, and 7, respectively.
279
280
ELEMENTS BOOK 10
Incommensurable Magnitudes †
† The theory of incommensurable magntidues set out in this book is generally attributed to Theaetetus of Athens. In the footnotes throughout
′
this book, k, k , etc. stand for distinct ratios of positive integers.
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ST EW iþ. Vroi ELEMENTS BOOK 10
Definitions
.
αʹ. Σύμμετρα μεγέθη λέγεται τὰ τῷ αὐτῷ μετρῳ με- 1. Those magnitudes measured by the same measure
τρούμενα, ἀσύμμετρα δέ, ὧν μηδὲν ἐνδέχεται κοινὸν μέτρον are said (to be) commensurable, but (those) of which no
γενέσθαι. (magnitude) admits to be a common measure (are said
βʹ. Εὐθεῖαι δυνάμει σύμμετροί εἰσιν, ὅταν τὰ ἀπ᾿ αὐτῶν to be) incommensurable. †
τετράγωνα τῷ αὐτῷ χωρίῳ μετρῆται, ἀσύμμετροι δέ, ὅταν 2. (Two) straight-lines are commensurable in square ‡
τοῖς ἀπ᾿ αὐτῶν τετραγώνοις μηδὲν ἐνδέχηται χωρίον κοινὸν when the squares on them are measured by the same
μέτρον γενέσθαι. area, but (are) incommensurable (in square) when no
γʹ. Τούτων ὑποκειμένων δείκνυται, ὅτι τῇ προτεθείσῃ area admits to be a common measure of the squares on
εὐθείᾳ ὑπάρχουσιν εὐθεῖαι πλήθει ἄπειροι σύμμετροί τε καὶ them. §
ἀσύμμετροι αἱ μὲν μήκει μόνον, αἱ δὲ καὶ δυνάμει. καλείσθω 3. These things being assumed, it is proved that there
οὖν ἡ μὲν προτεθεῖσα εὐθεῖα ῥητή, καὶ αἱ ταύτῃ σύμμετροι exist an infinite multitude of straight-lines commensu-
εἴτε μήκει καὶ δυνάμει εἴτε δυνάμει μόνον ῥηταί, αἱ δὲ ταύτῃ rable and incommensurable with an assigned straight-
ἀσύμμετροι ἄλογοι καλείσθωσαν. line—those (incommensurable) in length only, and those
δʹ. Καὶ τὸ μὲν ἀπὸ τῆς προτεθείσης εὐθείας τετράγω- also (commensurable or incommensurable) in square. ¶
νον ῥητόν, καὶ τὰ τούτῳ σύμμετρα ῥητά, τὰ δὲ τούτῳ Therefore, let the assigned straight-line be called ratio-
ἀσύμμετρα ἄλογα καλείσθω, καὶ αἱ δυνάμεναι αὐτὰ ἄλογοι, nal. And (let) the (straight-lines) commensurable with it,
εἰ μὲν τετράγωνα εἴη, αὐταὶ αἱ πλευραί, εἰ δὲ ἕτερά τινα either in length and square, or in square only, (also be
εὐθύγραμμα, αἱ ἴσα αὐτοῖς τετράγωνα ἀναγράφουσαι. called) rational. But let the (straight-lines) incommensu-
rable with it be called irrational. ∗
4. And let the square on the assigned straight-line be
called rational. And (let areas) commensurable with it
(also be called) rational. But (let areas) incommensu-
rable with it (be called) irrational, and (let) their square-
$
roots (also be called) irrational—the sides themselves, if
the (areas) are squares, and the (straight-lines) describ-
ing squares equal to them, if the (areas) are some other
rectilinear (figure). k
† In other words, two magnitudes α and β are commensurable if α : β :: 1 : k, and incommensurable otherwise.
‡ Literally, “in power”.
§ In other words, two straight-lines of length α and β are commensurable in square if α : β :: 1 : k 1/2 , and incommensurable in square otherwise.
Likewise, the straight-lines are commensurable in length if α : β :: 1 : k, and incommensurable in length otherwise.
aþ
¶ To be more exact, straight-lines can either be commensurable in square only, incommensurable in length only, or commenusrable/incommensurable
in both length and square, with an assigned straight-line.
∗ Let the length of the assigned straight-line be unity. Then rational straight-lines have lengths expressible as k or k 1/2 , depending on whether
the lengths are commensurable in length, or in square only, respectively, with unity. All other straight-lines are irrational.
$ The square-root of an area is the length of the side of an equal area square.
k The area of the square on the assigned straight-line is unity. Rational areas are expressible as k. All other areas are irrational. Thus, squares
whose sides are of rational length have rational areas, and vice versa. †
Proposition 1
.
Δύο μεγεθῶν ἀνίσων ἐκκειμένων, ἐὰν ἀπὸ τοῦ μείζονος If, from the greater of two unequal magnitudes
ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ καὶ τοῦ καταλειπομένου μεῖζον (which are) laid out, (a part) greater than half is sub-
ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ γίγνηται, λειφθήσεταί τι μέγεθος, tracted, and (if from) the remainder (a part) greater than
ὃ ἔσται ἔλασσον τοῦ ἐκκειμένου ἐλάσσονος μεγέθους. half (is subtracted), and (if) this happens continually,
῎Εστω δύο μεγέθη ἄνισα τὰ ΑΒ, Γ, ὧν μεῖζον τὸ ΑΒ· then some magnitude will (eventually) be left which will
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λέγω, ὅτι, ἐαν ἀπὸ τοῦ ΑΒ ἀφαιρεθῇ μεῖζον ἢ τὸ ἥμισυ be less than the lesser laid out magnitude.
καὶ τοῦ καταλειπομένου μεῖζον ἢ τὸ ἥμισυ, καὶ τοῦτο ἀεὶ Let AB and C be two unequal magnitudes, of which
γίγνηται, λειφθήσεταί τι μέγεθος, ὃ ἔσται ἔλασσον τοῦ Γ (let) AB (be) the greater. I say that if (a part) greater
μεγέθους. than half is subtracted from AB, and (if a part) greater
than half (is subtracted) from the remainder, and (if) this
happens continually, then some magnitude will (eventu-
ally) be left which will be less than the magnitude C.
Α Κ Θ Β A K H B
Γ C
∆ Ζ Η Ε D F G E
Τὸ Γ γὰρ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ ΑΒ For C, when multiplied (by some number), will some-
μεῖζον. πεπολλαπλασιάσθω, καὶ ἔστω τὸ ΔΕ τοῦ μὲν Γ times be greater than AB [Def. 5.4]. Let it have been
πολλαπλάσιον, τοῦ δὲ ΑΒ μεῖζον, καὶ διῃρήσθω τὸ ΔΕ εἰς (so) multiplied. And let DE be (both) a multiple of C,
τὰ τῷ Γ ἴσα τὰ ΔΖ, ΖΗ, ΗΕ, καὶ ἀφῃρήσθω ἀπὸ μὲν τοῦ and greater than AB. And let DE have been divided into
ΑΒ μεῖζον ἢ τὸ ἥμισυ τὸ ΒΘ, ἀπὸ δὲ τοῦ ΑΘ μεῖζον ἢ τὸ the (divisions) DF, FG, GE, equal to C. And let BH,
ἥμισυ τὸ ΘΚ, καὶ τοῦτο ἀεὶ γιγνέσθω, ἕως ἂν αἱ ἐν τῷ ΑΒ (which is) greater than half, have been subtracted from
διαιρέσεις ἰσοπληθεῖς γένωνται ταῖς ἐν τῷ ΔΕ διαιρέσεσιν. AB. And (let) HK, (which is) greater than half, (have
῎Εστωσαν οὖν αἱ ΑΚ, ΚΘ, ΘΒ διαιρέσεις ἰσοπληθεῖς been subtracted) from AH. And let this happen continu-
οὖσαι ταῖς ΔΖ, ΖΗ, ΗΕ· καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΔΕ τοῦ ally, until the divisions in AB become equal in number to
ΑΒ, καὶ ἀφῄρηται ἀπὸ μὲν τοῦ ΔΕ ἔλασσον τοῦ ἡμίσεως τὸ the divisions in DE.
ΕΗ, ἀπὸ δὲ τοῦ ΑΒ μεῖζον ἢ τὸ ἥμισυ τὸ ΒΘ, λοιπὸν ἄρα Therefore, let the divisions (in AB) be AK, KH, HB,
τὸ ΗΔ λοιποῦ τοῦ ΘΑ μεῖζόν ἐστιν. καὶ ἐπεὶ μεῖζόν ἐστι τὸ being equal in number to DF, FG, GE. And since DE is
ΗΔ τοῦ ΘΑ, καὶ ἀφῄρηται τοῦ μὲν ΗΔ ἥμισυ τὸ ΗΖ, τοῦ greater than AB, and EG, (which is) less than half, has
δὲ ΘΑ μεῖζον ἢ τὸ ἥμισυ τὸ ΘΚ, λοιπὸν ἄρα τὸ ΔΖ λοιποῦ been subtracted from DE, and BH, (which is) greater
τοῦ ΑΚ μεῖζόν ἐστιν. ἴσον δὲ τὸ ΔΖ τῷ Γ· καὶ τὸ Γ ἄρα than half, from AB, the remainder GD is thus greater
τοῦ ΑΚ μεῖζόν ἐστιν. ἔλασσον ἄρα τὸ ΑΚ τοῦ Γ. than the remainder HA. And since GD is greater than
Καταλείπεται ἄρα ἀπὸ τοῦ ΑΒ μεγέθους τὸ ΑΚ μέγεθος HA, and the half GF has been subtracted from GD, and
ἔλασσον ὂν τοῦ ἐκκειμένου ἐλάσσονος μεγέθους τοῦ Γ· HK, (which is) greater than half, from HA, the remain-
ὅπερ ἔδει δεῖξαι. — ὁμοίως δὲ δειχθήσεται, κἂν ἡμίση ᾖ τὰ der DF is thus greater than the remainder AK. And DF
ἀφαιρούμενα. (is) equal to C. C is thus also greater than AK. Thus,
bþ laid out magnitude C, is left over from the magnitude
AK (is) less than C.
Thus, the magnitude AK, which is less than the lesser
AB. (Which is) the very thing it was required to show. —
(The theorem) can similarly be proved even if the (parts)
subtracted are halves.
† This theorem is the basis of the so-called method of exhaustion, and is generally attributed to Eudoxus of Cnidus.
Proposition 2
.
᾿Εὰν δύο μεγεθῶν [ἐκκειμένων] ἀνίσων ἀνθυφαιρουμένου If the remainder of two unequal magnitudes (which
ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος τὸ καταλειπόμενον are) [laid out] never measures the (magnitude) before it,
μηδέποτε καταμετρῇ τὸ πρὸ ἑαυτοῦ, ἀσύμμετρα ἔσται τὰ (when) the lesser (magnitude is) continually subtracted
μεγέθη. in turn from the greater, then the (original) magnitudes
Δύο γὰρ μεγεθῶν ὄντων ἀνίσων τῶν ΑΒ, ΓΔ καὶ will be incommensurable.
ἐλάσσονος τοῦ ΑΒ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος For, AB and CD being two unequal magnitudes, and
ἀπὸ τοῦ μείζονος τὸ περιλειπόμενον μηδέποτε καταμε- AB (being) the lesser, let the remainder never measure
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τρείτω τὸ πρὸ ἑαυτοῦ· λέγω, ὅτι ἀσύμμετρά ἐστι τὰ ΑΒ, the (magnitude) before it, (when) the lesser (magnitude
ΓΔ μεγέθη. is) continually subtracted in turn from the greater. I say
that the magnitudes AB and CD are incommensurable.
Α Η Β A G B
Ε E
Γ Ζ ∆ C F D
Εἰ γάρ ἐστι σύμμετρα, μετρήσει τι αὐτὰ μέγεθος. με- For if they are commensurable then some magnitude
τρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Ε· καὶ τὸ μὲν ΑΒ τὸ ΖΔ will measure them (both). If possible, let it (so) measure
καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΓΖ, τὸ δὲ ΓΖ τὸ (them), and let it be E. And let AB leave CF less than
ΒΗ καταμετροῦν λειπέτω ἑαυτοῦ ἔλασσον τὸ ΑΗ, καὶ τοῦτο itself (in) measuring FD, and let CF leave AG less than
ἀεὶ γινέσθω, ἕως οὗ λειφθῇ τι μέγεθος, ὅ ἐστιν ἔλασσον τοῦ itself (in) measuring BG, and let this happen continually,
Ε. γεγονέτω, καὶ λελείφθω τὸ ΑΗ ἔλασσον τοῦ Ε. ἐπεὶ οὖν until some magnitude which is less than E is left. Let
†
τὸ Ε τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΔΖ μετρεῖ, καὶ τὸ Ε ἄρα (this) have occurred, and let AG, (which is) less than
τὸ ΖΔ μετρήσει. μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ· καὶ λοιπὸν ἄρα E, have been left. Therefore, since E measures AB, but
τὸ ΓΖ μετρήσει. ἀλλὰ τὸ ΓΖ τὸ ΒΗ μετρεῖ· καὶ τὸ Ε ἄρα AB measures DF, E will thus also measure FD. And it
τὸ ΒΗ μετρεῖ. μετρεῖ δὲ καὶ ὅλον τὸ ΑΒ· καὶ λοιπὸν ἄρα τὸ also measures the whole (of) CD. Thus, it will also mea-
ΑΗ μετρήσει, τὸ μεῖζον τὸ ἔλασσον· ὅπερ ἐστὶν ἀδύνατον. sure the remainder CF. But, CF measures BG. Thus, E
οὐκ ἄρα τὰ ΑΒ, ΓΔ μεγέθη μετρήσει τι μέγεθος· ἀσύμμετρα also measures BG. And it also measures the whole (of)
gþ magnitudes AB and CD. Thus, the magnitudes AB and
ἄρα ἐστὶ τὰ ΑΒ, ΓΔ μεγέθη. AB. Thus, it will also measure the remainder AG, the
᾿Εὰν ἄρα δύο μεγεθῶν ἀνίσων, καὶ τὰ ἑξῆς. greater (measuring) the lesser. The very thing is impos-
sible. Thus, some magnitude cannot measure (both) the
CD are incommensurable [Def. 10.1].
Thus, if . . . of two unequal magnitudes, and so on . . . .
† The fact that this will eventually occur is guaranteed by Prop. 10.1.
.
Proposition 3
Δύο μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον αὐτῶν To find the greatest common measure of two given
κοινὸν μέτρον εὑρεῖν. commensurable magnitudes.
Α Ζ Β A F B
Γ Ε ∆ C E D
Η G
῎Εστω τὰ δοθέντα δύο μεγέθη σύμμετρα τὰ ΑΒ, ΓΔ, Let AB and CD be the two given magnitudes, of
ὧν ἔλασσον τὸ ΑΒ· δεῖ δὴ τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν which (let) AB (be) the lesser. So, it is required to find
μέτρον εὑρεῖν. the greatest common measure of AB and CD.
Τὸ ΑΒ γὰρ μέγεθος ἤτοι μετρεῖ τὸ ΓΔ ἢ οὔ. εἰ μὲν For the magnitude AB either measures, or (does) not
οὖν μετρεῖ, μετρεῖ δὲ καὶ ἑαυτό, τὸ ΑΒ ἄρα τῶν ΑΒ, ΓΔ (measure), CD. Therefore, if it measures (CD), and
κοινὸν μέτρον ἐστίν· καὶ φανερόν, ὅτι καὶ μέγιστον. μεῖζον (since) it also measures itself, AB is thus a common mea-
γὰρ τοῦ ΑΒ μεγέθους τὸ ΑΒ οὐ μετρήσει. sure of AB and CD. And (it is) clear that (it is) also (the)
Μὴ μετρείτω δὴ τὸ ΑΒ τὸ ΓΔ. καὶ ἀνθυφαιρουμένου greatest. For a (magnitude) greater than magnitude AB
ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, τὸ περιλειπόμενον cannot measure AB.
μετρήσει ποτὲ τὸ πρὸ ἑαυτοῦ διὰ τὸ μὴ εἶναι ἀσύμμετρα τὰ So let AB not measure CD. And continually subtract-
ΑΒ, ΓΔ· καὶ τὸ μὲν ΑΒ τὸ ΕΔ καταμετροῦν λειπέτω ἑαυτοῦ ing in turn the lesser (magnitude) from the greater, the
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ἔλασσον τὸ ΕΓ, τὸ δὲ ΕΓ τὸ ΖΒ καταμετροῦν λειπέτω remaining (magnitude) will (at) some time measure the
ἑαυτοῦ ἔλασσον τὸ ΑΖ, τὸ δὲ ΑΖ τὸ ΓΕ μετρείτω. (magnitude) before it, on account of AB and CD not be-
᾿Επεὶ οὖν τὸ ΑΖ τὸ ΓΕ μετρεῖ, ἀλλὰ τὸ ΓΕ τὸ ΖΒ μετρεῖ, ing incommensurable [Prop. 10.2]. And let AB leave EC
καὶ τὸ ΑΖ ἄρα τὸ ΖΒ μετρήσει. μετρεῖ δὲ καὶ ἑαυτό· καὶ less than itself (in) measuring ED, and let EC leave AF
ὅλον ἄρα τὸ ΑΒ μετρήσει τὸ ΑΖ. ἀλλὰ τὸ ΑΒ τὸ ΔΕ μετρεῖ· less than itself (in) measuring FB, and let AF measure
καὶ τὸ ΑΖ ἄρα τὸ ΕΔ μετρήσει. μετρεῖ δὲ καὶ τὸ ΓΕ· καί CE.
ὅλον ἄρα τὸ ΓΔ μετρεῖ· τὸ ΑΖ ἄρα τῶν ΑΒ, ΓΔ κοινὸν Therefore, since AF measures CE, but CE measures
μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή, ἔσται FB, AF will thus also measure FB. And it also mea-
τι μέγεθος μεῖζον τοῦ ΑΖ, ὃ μετρήσει τὰ ΑΒ, ΓΔ. ἔστω τὸ sures itself. Thus, AF will also measure the whole (of)
Η. ἐπεὶ οὖν τὸ Η τὸ ΑΒ μετρεῖ, ἀλλὰ τὸ ΑΒ τὸ ΕΔ μετρεῖ, AB. But, AB measures DE. Thus, AF will also mea-
καὶ τὸ Η ἄρα τὸ ΕΔ μετρήσει. μετρεῖ δὲ καὶ ὅλον τὸ ΓΔ· sure ED. And it also measures CE. Thus, it also mea-
καὶ λοιπὸν ἄρα τὸ ΓΕ μετρήσει τὸ Η. ἀλλὰ τὸ ΓΕ τὸ ΖΒ sures the whole of CD. Thus, AF is a common measure
μετρεῖ· καὶ τὸ Η ἄρα τὸ ΖΒ μετρήσει. μετρεῖ δὲ καὶ ὅλον of AB and CD. So I say that (it is) also (the) greatest
τὸ ΑΒ, καὶ λοιπὸν τὸ ΑΖ μετρήσει, τὸ μεῖζον τὸ ἔλασσον· (common measure). For, if not, there will be some mag-
ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα μεῖζόν τι μέγεθος τοῦ ΑΖ nitude, greater than AF, which will measure (both) AB
τὰ ΑΒ, ΓΔ μετρήσει· τὸ ΑΖ ἄρα τῶν ΑΒ, ΓΔ τὸ μέγιστον and CD. Let it be G. Therefore, since G measures AB,
κοινὸν μέτρον ἐστίν. but AB measures ED, G will thus also measure ED. And
Δύο ἄρα μεγεθῶν συμμέτρων δοθέντων τῶν ΑΒ, ΓΔ it also measures the whole of CD. Thus, G will also mea-
τὸ μέγιστον κοινὸν μέτρον ηὕρηται· ὅπερ ἔδει δεῖξαι. sure the remainder CE. But CE measures FB. Thus, G
will also measure FB. And it also measures the whole
ìri sma possible. Thus, some magnitude greater than AF cannot
(of) AB. And (so) it will measure the remainder AF,
the greater (measuring) the lesser. The very thing is im-
measure (both) AB and CD. Thus, AF is the greatest
common measure of AB and CD.
Thus, the greatest common measure of two given
dþ
commensurable magnitudes, AB and CD, has been
found. (Which is) the very thing it was required to show.
.
Corollary
᾿Εκ δὴ τούτου φανερόν, ὅτι, ἐὰν μέγεθος δύο μεγέθη So (it is) clear, from this, that if a magnitude measures
μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει. two magnitudes then it will also measure their greatest
common measure.
Proposition 4
.
Τριῶν μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον To find the greatest common measure of three given
αὐτῶν κοινὸν μέτρον εὑρεῖν. commensurable magnitudes.
Α A
Β B
Γ C
∆ Ε Ζ D E F
῎Εστω τὰ δοθέντα τρία μεγέθη σύμμετρα τὰ Α, Β, Γ· Let A, B, C be the three given commensurable mag-
δεῖ δὴ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν. nitudes. So it is required to find the greatest common
Εἰλήφθω γὰρ δύο τῶν Α, Β τὸ μέγιστον κοινὸν μέτρον, measure of A, B, C.
καὶ ἔστω τὸ Δ· τὸ δὴ Δ τὸ Γ ἤτοι μετρεῖ ἢ οὔ [μετρεῖ]. For let the greatest common measure of the two (mag-
μετρείτω πρότερον. ἐπεὶ οὖν τὸ Δ τὸ Γ μετρεῖ, μετρεῖ δὲ nitudes) A and B have been taken [Prop. 10.3], and let it
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καὶ τὰ Α, Β, τὸ Δ ἄρα τὰ Α, Β, Γ μετρεῖ· τὸ Δ ἄρα τῶν Α, be D. So D either measures, or [does] not [measure], C.
Β, Γ κοινὸν μέτρον ἐστίν. καὶ φανερόν, ὅτι καὶ μέγιστον· Let it, first of all, measure (C). Therefore, since D mea-
μεῖζον γὰρ τοῦ Δ μεγέθους τὰ Α, Β οὐ μετρεῖ. sures C, and it also measures A and B, D thus measures
Μὴ μετρείτω δὴ τὸ Δ τὸ Γ. λέγω πρῶτον, ὅτι σύμμετρά A, B, C. Thus, D is a common measure of A, B, C. And
ἐστι τὰ Γ, Δ. ἐπεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, Γ, μετρήσει (it is) clear that (it is) also (the) greatest (common mea-
τι αὐτὰ μέγεθος, ὃ δηλαδὴ καὶ τὰ Α, Β μετρήσει· ὥστε sure). For no magnitude larger than D measures (both)
καὶ τὸ τῶν Α, Β μέγιστον κοινὸν μέτρον τὸ Δ μετρήσει. A and B.
μετρεῖ δὲ καὶ τὸ Γ· ὥστε τὸ εἰρημένον μέγεθος μετρήσει τὰ So let D not measure C. I say, first, that C and D are
Γ, Δ· σύμμετρα ἄρα ἐστὶ τὰ Γ, Δ. εἰλήφθω οὖν αὐτῶν τὸ commensurable. For if A, B, C are commensurable then
μέγιστον κοινὸν μέτρον, καὶ ἔστω τὸ Ε. ἐπεὶ οὖν τὸ Ε τὸ some magnitude will measure them which will clearly
Δ μετρεῖ, ἀλλὰ τὸ Δ τὰ Α, Β μετρεῖ, καὶ τὸ Ε ἄρα τὰ Α, Β also measure A and B. Hence, it will also measure D, the
μετρήσει. μετρεῖ δὲ καὶ τὸ Γ. τὸ Ε ἄρα τὰ Α, Β, Γ μετρεῖ· greatest common measure of A and B [Prop. 10.3 corr.].
τὸ Ε ἄρα τῶν Α, Β, Γ κοινόν ἐστι μέτρον. λέγω δή, ὅτι καὶ And it also measures C. Hence, the aforementioned mag-
μέγιστον. εἰ γὰρ δυνατόν, ἔστω τι τοῦ Ε μεῖζον μέγεθος nitude will measure (both) C and D. Thus, C and D are
τὸ Ζ, καὶ μετρείτω τὰ Α, Β, Γ. καὶ ἐπεὶ τὸ Ζ τὰ Α, Β, Γ commensurable [Def. 10.1]. Therefore, let their greatest
μετρεῖ, καὶ τὰ Α, Β ἄρα μετρήσει καὶ τὸ τῶν Α, Β μέγιστον common measure have been taken [Prop. 10.3], and let
κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον κοινὸν it be E. Therefore, since E measures D, but D measures
μέτρον ἐστὶ τὸ Δ· τὸ Ζ ἄρα τὸ Δ μετρεῖ. μετρεῖ δὲ καὶ τὸ (both) A and B, E will thus also measure A and B. And
Γ· τὸ Ζ ἄρα τὰ Γ, Δ μετρεῖ· καὶ τὸ τῶν Γ, Δ ἄρα μέγιστον it also measures C. Thus, E measures A, B, C. Thus, E
κοινὸν μέτρον μετρήσει τὸ Ζ. ἔστι δὲ τὸ Ε· τὸ Ζ ἄρα τὸ Ε is a common measure of A, B, C. So I say that (it is) also
μετρήσει, τὸ μεῖζον τὸ ἔλασσον· ὅπερ ἐστὶν ἀδύνατον. οὐκ (the) greatest (common measure). For, if possible, let F
ἄρα μεῖζόν τι τοῦ Ε μεγέθους [μέγεθος] τὰ Α, Β, Γ μετρεῖ· be some magnitude greater than E, and let it measure A,
τὸ Ε ἄρα τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον ἐστίν, ἐὰν B, C. And since F measures A, B, C, it will thus also
μὴ μετρῇ τὸ Δ τὸ Γ, ἐὰν δὲ μετρῇ, αὐτὸ τὸ Δ. measure A and B, and will (thus) measure the greatest
Τριῶν ἄρα μεγεθῶν συμμέτρων δοθέντων τὸ μέγιστον common measure of A and B [Prop. 10.3 corr.]. And D
κοινὸν μέτρον ηὕρηται [ὅπερ ἔδει δεῖξαι]. is the greatest common measure of A and B. Thus, F
measures D. And it also measures C. Thus, F measures
(both) C and D. Thus, F will also measure the greatest
common measure of C and D [Prop. 10.3 corr.]. And it is
E. Thus, F will measure E, the greater (measuring) the
ìri sma B, C. Thus, if D does not measure C then E is the great-
lesser. The very thing is impossible. Thus, some [magni-
tude] greater than the magnitude E cannot measure A,
est common measure of A, B, C. And if it does measure
(C) then D itself (is the greatest common measure).
Thus, the greatest common measure of three given
commensurable magnitudes has been found. [(Which is)
the very thing it was required to show.]
Corollary
.
᾿Εκ δὴ τούτου φανερόν, ὅτι, ἐὰν μέγεθος τρία μεγέθη So (it is) clear, from this, that if a magnitude measures
μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει. three magnitudes then it will also measure their greatest
῾Ομοίως δὴ καὶ ἐπὶ πλειόνων τὸ μέγιστον κοινὸν μέτρον common measure.
ληφθήσεται, καὶ τὸ πόρισμα προχωρήσει. ὅπερ ἔδει δεῖξαι. So, similarly, the greatest common measure of more
(magnitudes) can also be taken, and the (above) corol-
lary will go forward. (Which is) the very thing it was
required to show.
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.
Proposition 5
Τὰ σύμμετρα μεγέθη πρὸς ἄλληλα λόγον ἔχει, ὃν Commensurable magnitudes have to one another the
ἀριθμὸς πρὸς ἀριθμόν. ratio which (some) number (has) to (some) number.
Α Β Γ A B C
∆ Ε D E
῎Εστω σύμμετρα μεγέθη τὰ Α, Β· λέγω, ὅτι τὸ Α πρὸς Let A and B be commensurable magnitudes. I say
τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. that A has to B the ratio which (some) number (has) to
᾿Επεὶ γὰρ σύμμετρά ἐστι τὰ Α, Β, μετρήσει τι αὐτὰ (some) number.
μέγεθος. μετρείτω, καὶ ἔστω τὸ Γ. καὶ ὁσάκις τὸ Γ τὸ For if A and B are commensurable (magnitudes) then
Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Δ, ὁσάκις δὲ some magnitude will measure them. Let it (so) measure
τὸ Γ τὸ Β μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. (them), and let it be C. And as many times as C measures
᾿Επεὶ οὖν τὸ Γ τὸ Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας, A, so many units let there be in D. And as many times as
μετρεῖ δὲ καὶ ἡ μονὰς τὸν Δ κατὰ τὰς ἐν αὐτῷ μονάδας, C measures B, so many units let there be in E.
ἰσάκις ἄρα ἡ μονὰς τὸν Δ μετρεῖ ἀριθμὸν καὶ τὸ Γ μέγεθος Therefore, since C measures A according to the units
τὸ Α· ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς in D, and a unit also measures D according to the units
τὸν Δ· ἀνάπαλιν ἄρα, ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς in it, a unit thus measures the number D as many times
τὴν μονάδα. πάλιν ἐπεὶ τὸ Γ τὸ Β μετρεῖ κατὰ τὰς ἐν τῷ as the magnitude C (measures) A. Thus, as C is to A,
†
Ε μονάδας, μετρεῖ δὲ καὶ ἡ μονὰς τὸν Ε κατὰ τὰς ἐν αὐτῷ so a unit (is) to D [Def. 7.20]. Thus, inversely, as A (is)
μονάδας, ἰσάκις ἄρα ἡ μονὰς τὸν Ε μετρεῖ καὶ τὸ Γ τὸ Β· to C, so D (is) to a unit [Prop. 5.7 corr.]. Again, since
ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Β, οὕτως ἡ μονὰς πρὸς τὸν Ε. C measures B according to the units in E, and a unit
ἐδείχθη δὲ καὶ ὡς τὸ Α πρὸς τὸ Γ, ὁ Δ πρὸς τὴν μονάδα· also measures E according to the units in it, a unit thus
δι᾿ ἴσου ἄρα ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Δ ἀριθμὸς measures E the same number of times that C (measures)
πρὸς τὸν Ε. B. Thus, as C is to B, so a unit (is) to E [Def. 7.20]. And
Τὰ ἄρα σύμμετρα μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον it was also shown that as A (is) to C, so D (is) to a unit.
þ to one another the ratio which the number D (has) to the
ἔχει, ὃν ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε· ὅπερ ἔδει δεῖξαι. Thus, via equality, as A is to B, so the number D (is) to
the (number) E [Prop. 5.22].
Thus, the commensurable magnitudes A and B have
number E. (Which is) the very thing it was required to
show.
† There is a slight logical gap here, since Def. 7.20 applies to four numbers, rather than two number and two magnitudes.
Proposition 6
.
᾿Εὰν δύο μεγέθη πρὸς ἄλληλα λόγον ἔχῃ, ὃν ἀριθμὸς If two magnitudes have to one another the ratio which
πρὸς ἀριθμόν, σύμμετρα ἔσται τὰ μεγέθη. (some) number (has) to (some) number then the magni-
tudes will be commensurable.
Α Β A B
∆ Ε D E
Γ Ζ C F
Δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον ἐχέτω, ὃν For let the two magnitudes A and B have to one an-
ἀριθμὸς ὁ Δ πρὸς ἀριθμὸν τὸν Ε· λέγω, ὅτι σύμμετρά ἐστι other the ratio which the number D (has) to the number
τὰ Α, Β μεγέθη. E. I say that the magnitudes A and B are commensu-
῞Οσαι γάρ εἰσιν ἐν τῷ Δ μονάδες, εἰς τοσαῦτα ἴσα rable.
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διῃρήσθω τὸ Α, καὶ ἑνὶ αὐτῶν ἴσον ἔστω τὸ Γ· ὅσαι δέ For, as many units as there are in D, let A have been
εἰσιν ἐν τῷ Ε μονάδες, ἐκ τοσούτων μεγεθῶν ἴσων τῷ Γ divided into so many equal (divisions). And let C be
συγκείσθω τὸ Ζ. equal to one of them. And as many units as there are
᾿Επεὶ οὖν, ὅσαι εἰσὶν ἐν τῷ Δ μονάδες, τοσαῦτά εἰσι καὶ in E, let F be the sum of so many magnitudes equal to
ἐν τῷ Α μεγέθη ἴσα τῷ Γ, ὃ ἄρα μέρος ἐστὶν ἡ μονὰς τοῦ C.
Δ, τὸ αὐτὸ μέρος ἐστὶ καὶ τὸ Γ τοῦ Α· ἔστιν ἄρα ὡς τὸ Γ Therefore, since as many units as there are in D, so
πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ. μετρεῖ δὲ ἡ μονὰς many magnitudes equal to C are also in A, therefore
τὸν Δ ἀριθμόν· μετρεῖ ἄρα καὶ τὸ Γ τὸ Α. καὶ ἐπεί ἐστιν whichever part a unit is of D, C is also the same part of
ὡς τὸ Γ πρὸς τὸ Α, οὕτως ἡ μονὰς πρὸς τὸν Δ [ἀριθμόν], A. Thus, as C is to A, so a unit (is) to D [Def. 7.20]. And
ἀνάπαλιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ ἀριθμὸς πρὸς a unit measures the number D. Thus, C also measures
τὴν μονάδα. πάλιν ἐπεί, ὅσαι εἰσὶν ἐν τῷ Ε μονάδες, τοσαῦτά A. And since as C is to A, so a unit (is) to the [number]
εἰσι καὶ ἐν τῷ Ζ ἴσα τῷ Γ, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ, D, thus, inversely, as A (is) to C, so the number D (is)
οὕτως ἡ μονὰς πρὸς τὸν Ε [ἀριθμόν]. ἐδείχθη δὲ καὶ ὡς to a unit [Prop. 5.7 corr.]. Again, since as many units as
τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς τὴν μονάδα· δι᾿ ἴσου ἄρα there are in E, so many (magnitudes) equal to C are also
ἐστὶν ὡς τὸ Α πρὸς τὸ Ζ, οὕτως ὁ Δ πρὸς τὸν Ε. ἀλλ᾿ ὡς ὁ in F, thus as C is to F, so a unit (is) to the [number] E
Δ πρὸς τὸν Ε, οὕτως ἐστὶ τὸ Α πρὸς τὸ Β· καὶ ὡς ἄρα τὸ Α [Def. 7.20]. And it was also shown that as A (is) to C,
πρὸς τὸ Β, οὕτως καὶ πρὸς τὸ Ζ. τὸ Α ἄρα πρὸς ἑκάτερον so D (is) to a unit. Thus, via equality, as A is to F, so D
τῶν Β, Ζ τὸν αὐτὸν ἔχει λόγον· ἴσον ἄρα ἐστὶ τὸ Β τῷ Ζ. (is) to E [Prop. 5.22]. But, as D (is) to E, so A is to B.
ìri sma C measures (both) A and B. Thus, A is commensurable
μετρεῖ δὲ τὸ Γ τὸ Ζ· μετρεῖ ἄρα καὶ τὸ Β. ἀλλὰ μὴν καὶ τὸ And thus as A (is) to B, so (it) also is to F [Prop. 5.11].
Α· τὸ Γ ἄρα τὰ Α, Β μετρεῖ. σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Thus, A has the same ratio to each of B and F. Thus, B is
Β. equal to F [Prop. 5.9]. And C measures F. Thus, it also
᾿Εὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς. measures B. But, in fact, (it) also (measures) A. Thus,
with B [Def. 10.1].
Thus, if two magnitudes . . . to one another, and so on
. . . .
.
Corollary
᾿Εκ δὴ τούτου φανερόν, ὅτι, ἐὰν ὦσι δύο ἀριθμοί, ὡς So it is clear, from this, that if there are two numbers,
οἱ Δ, Ε, καὶ εὐθεῖα, ὡς ἡ Α, δύνατόν ἐστι ποιῆσαι ὡς ὁ like D and E, and a straight-line, like A, then it is possible
Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν, οὕτως τὴν εὐθεῖαν πρὸς to contrive that as the number D (is) to the number E,
εὐθεῖαν. ἐὰν δὲ καὶ τῶν Α, Ζ μέση ἀνάλογον ληφθῇ, ὡς ἡ so the straight-line (is) to (another) straight-line (i.e., F).
Β, ἔσται ὡς ἡ Α πρὸς τὴν Ζ, οὕτως τὸ ἀπὸ τῆς Α πρὸς τὸ And if the mean proportion, (say) B, is taken of A and
ἀπὸ τῆς Β, τουτέστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως F, then as A is to F, so the (square) on A (will be) to the
zþ the number E, so the (figure) on the straight-line A (is)
τὸ ἀπὸ τῆς πρώτης πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ (square) on B. That is to say, as the first (is) to the third,
ὁμοίως ἀναγραφόμενον. ἀλλ᾿ ὡς ἡ Α πρὸς τὴν Ζ, οὕτως so the (figure) on the first (is) to the similar, and similarly
ἐστὶν ὁ Δ ἀριθμος πρὸς τὸν Ε ἀριθμόν· γέγονεν ἄρα καὶ described, (figure) on the second [Prop. 6.19 corr.]. But,
ὡς ὁ Δ ἀριθμὸς πρὸς τὸν Ε ἀριθμόν, οὕτως τὸ ἀπὸ τῆς Α as A (is) to F, so the number D is to the number E. Thus,
εὐθείας πρὸς τὸ ἀπὸ τῆς Β εὐθείας· ὅπερ ἔδει δεῖξαι. it has also been contrived that as the number D (is) to
to the (similar figure) on the straight-line B. (Which is)
the very thing it was required to show.
Proposition 7
.
Τὰ ἀσύμμετρα μεγέθη πρὸς ἄλληλα λόγον οὐκ ἔχει, ὃν Incommensurable magnitudes do not have to one an-
ἀριθμὸς πρὸς ἀριθμόν. other the ratio which (some) number (has) to (some)
῎Εστω ἀσύμμετρα μεγέθη τὰ Α, Β· λέγω, ὅτι τὸ Α πρὸς number.
τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. Let A and B be incommensurable magnitudes. I say
that A does not have to B the ratio which (some) number
(has) to (some) number.
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hþ
Β B
Εἰ γὰρ ἔχει τὸ Α πρὸς τὸ Β λόγον, ὃν ἀριθμὸς πρὸς For if A has to B the ratio which (some) number (has)
ἀριθμόν, σύμμετρον ἔσται τὸ Α τῷ Β. οὐκ ἔστι δέ· οὐκ ἄρα to (some) number then A will be commensurable with B
τὸ Α πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. [Prop. 10.6]. But it is not. Thus, A does not have to B
Τὰ ἄρα ἀσύμμετρα μεγέθη πρὸς ἄλληλα λόγον οὐκ ἔχει, the ratio which (some) number (has) to (some) number.
Thus, incommensurable numbers do not have to one
καὶ τὰ ἑξῆς. another, and so on . . . .
.
Proposition 8
᾿Εὰν δύο μεγέθη πρὸς ἄλληλα λόγον μὴ ἔχῃ, ὃν ἀριθμὸς If two magnitudes do not have to one another the ra-
πρὸς ἀριθμόν, ἀσύμμετρα ἔσται τὰ μεγέθη. tio which (some) number (has) to (some) number then
the magnitudes will be incommensurable.
Α A
Β B
Δύο γὰρ μεγέθη τὰ Α, Β πρὸς ἄλληλα λόγον μὴ ἐχέτω, For let the two magnitudes A and B not have to one
ὃν ἀριθμὸς πρὸς ἀριθμόν· λέγω, ὅτι ἀσύμμετρά ἐστι τὰ Α, another the ratio which (some) number (has) to (some)
jþ the magnitudes A and B are incommensurable.
Β μεγέθη. number. I say that the magnitudes A and B are incom-
Εἰ γὰρ ἔσται σύμμετρα, τὸ Α πρὸς τὸ Β λόγον ἕξει, ὃν mensurable.
ἀριθμὸς πρὸς ἀριθμόν. οὐκ ἔχει δέ. ἀσύμμετρα ἄρα ἐστὶ τὰ For if they are commensurable, A will have to B
Α, Β μεγέθη. the ratio which (some) number (has) to (some) number
᾿Εὰν ἄρα δύο μεγέθη πρὸς ἄλληλα, καὶ τὰ ἑξῆς. [Prop. 10.5]. But it does not have (such a ratio). Thus,
Thus, if two magnitudes . . . to one another, and so on
. . . .
Proposition 9
.
Τὰ ἀπὸ τῶν μήκει συμμέτρων εὐθειῶν τετράγωνα Squares on straight-lines (which are) commensurable
πρὸς ἄλληλα λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς in length have to one another the ratio which (some)
τετράγωνον ἀριθμόν· καὶ τὰ τετράγωνα τὰ πρὸς ἄλληλα square number (has) to (some) square number. And
λόγον ἔχοντα, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον squares having to one another the ratio which (some)
ἀριθμόν, καὶ τὰς πλευρὰς ἕξει μήκει συμμέτρους. τὰ square number (has) to (some) square number will also
δὲ ἀπὸ τῶν μήκει ἀσυμμέτρων εὐθειῶν τετράγωνα πρὸς have sides (which are) commensurable in length. But
ἄλληλα λόγον οὐκ ἔχει, ὅνπερ τετράγωνος ἀριθμὸς πρὸς squares on straight-lines (which are) incommensurable
τετράγωνον ἀριθμόν· καὶ τὰ τετράγωνα τὰ πρὸς ἄλληλα in length do not have to one another the ratio which
λόγον μὴ ἔχοντα, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον (some) square number (has) to (some) square number.
ἀριθμόν, οὐδὲ τὰς πλευρὰς ἕξει μήκει συμμέτρους. And squares not having to one another the ratio which
(some) square number (has) to (some) square number
will not have sides (which are) commensurable in length
either.
Α Β A B
Γ ∆ C D
῎Εστωσαν γὰρ αἱ Α, Β μήκει σύμμετροι· λέγω, ὅτι τὸ For let A and B be (straight-lines which are) commen-
ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β τετράγωνον surable in length. I say that the square on A has to the
λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον square on B the ratio which (some) square number (has)
ἀριθμόν. to (some) square number.
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᾿Επεὶ γὰρ σύμμετρός ἐστιν ἡ Α τῇ Β μήκει, ἡ Α ἄρα πρὸς For since A is commensurable in length with B, A
τὴν Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ thus has to B the ratio which (some) number (has) to
Γ πρὸς τὸν Δ. ἐπεὶ οὖν ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ (some) number [Prop. 10.5]. Let it have (that) which
Γ πρὸς τὸν Δ, ἀλλὰ τοῦ μὲν τῆς Α πρὸς τὴν Β λόγου δι- C (has) to D. Therefore, since as A is to B, so C (is)
πλασίων ἐστὶν ὁ τοῦ ἀπὸ τῆς Α τετραγώνου πρὸς τὸ ἀπὸ τῆς to D. But the (ratio) of the square on A to the square
Β τετράγωνον· τὰ γὰρ ὅμοια σχήματα ἐν διπλασίονι λόγῳ on B is the square of the ratio of A to B. For similar
ἐστὶ τῶν ὁμολόγων πλευρῶν· τοῦ δὲ τοῦ Γ [ἀριθμοῦ] πρὸς figures are in the squared ratio of (their) corresponding
τὸν Δ [ἀριθμὸν] λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τοῦ Γ sides [Prop. 6.20 corr.]. And the (ratio) of the square
τετραγώνου πρὸς τὸν ἀπὸ τοῦ Δ τετράγωνον· δύο γὰρ τε- on C to the square on D is the square of the ratio of
τραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καί the [number] C to the [number] D. For there exits one
ὁ τετράγωνος πρὸς τὸν τετράγωνον [ἀριθμὸν] διπλασίονα number in mean proportion to two square numbers, and
λόγον ἔχει, ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν· ἔστιν ἄρα (one) square (number) has to the (other) square [num-
καὶ ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β ber] a squared ratio with respect to (that) the side (of the
τετράγωνον, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος [ἀριθμὸς] πρὸς former has) to the side (of the latter) [Prop. 8.11]. And,
τὸν ἀπὸ τοῦ Δ [ἀριθμοῦ] τετράγωνον [ἀριθμόν]. thus, as the square on A is to the square on B, so the
᾿Αλλὰ δὴ ἔστω ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ square [number] on the (number) C (is) to the square
ἀπὸ τῆς Β, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος πρὸς τὸν ἀπὸ [number] on the [number] D. †
τοῦ Δ [τετράγωνον]· λέγω, ὅτι σύμμετρός ἐστιν ἡ Α τῇ Β And so let the square on A be to the (square) on B as
μήκει. the square (number) on C (is) to the [square] (number)
᾿Επεὶ γάρ ἐστιν ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ on D. I say that A is commensurable in length with B.
ἀπὸ τῆς Β [τετράγωνον], οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος For since as the square on A is to the [square] on B, so
πρὸς τὸν ἀπὸ τοῦ Δ [τετράγωνον], ἀλλ᾿ ὁ μὲν τοῦ ἀπὸ τῆς the square (number) on C (is) to the [square] (number)
Α τετραγώνου πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] λόγος δι- on D. But, the ratio of the square on A to the (square)
πλασίων ἐστὶ τοῦ τῆς Α πρὸς τὴν Β λόγου, ὁ δὲ τοῦ ἀπὸ on B is the square of the (ratio) of A to B [Prop. 6.20
τοῦ Γ [ἀριθμοῦ] τετραγώνου [ἀριθμοῦ] πρὸς τὸν ἀπὸ τοῦ Δ corr.]. And the (ratio) of the square [number] on the
[ἀριθμοῦ] τετράγωνον [ἀριθμὸν] λόγος διπλασίων ἐστὶ τοῦ [number] C to the square [number] on the [number] D is
τοῦ Γ [ἀριθμοῦ] πρὸς τὸν Δ [ἀριθμὸν] λόγου, ἔστιν ἄρα the square of the ratio of the [number] C to the [number]
καὶ ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ Γ [ἀριθμὸς] πρὸς τὸν Δ D [Prop. 8.11]. Thus, as A is to B, so the [number] C
[ἀριθμόν]. ἡ Α ἄρα πρὸς τὴν Β λόγον ἔχει, ὃν ἀριθμὸς ὁ Γ also (is) to the [number] D. A, thus, has to B the ratio
πρὸς ἀριθμὸν τὸν Δ· σύμμετρος ἄρα ἐστὶν ἡ Α τῇ Β μήκει. which the number C has to the number D. Thus, A is
᾿Αλλὰ δὴ ἀσύμμετρος ἔστω ἡ Α τῇ Β μήκει· λέγω, ὅτι commensurable in length with B [Prop. 10.6]. ‡
τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] And so let A be incommensurable in length with B. I
λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον say that the square on A does not have to the [square] on
ἀριθμόν. B the ratio which (some) square number (has) to (some)
Εἰ γὰρ ἔχει τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ square number.
τῆς Β [τετράγωνον] λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς For if the square on A has to the [square] on B the ra-
τετράγωνον ἀριθμόν, σύμμετρος ἔσται ἡ Α τῇ Β. οὐκ ἔστι tio which (some) square number (has) to (some) square
δέ· οὐκ ἄρα τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς number then A will be commensurable (in length) with
Β [τετράγωνον] λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς B. But it is not. Thus, the square on A does not have
τετράγωνον ἀριθμόν. to the [square] on the B the ratio which (some) square
Πάλιν δὴ τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς number (has) to (some) square number.
Β [τετράγωνον] λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς So, again, let the square on A not have to the [square]
πρὸς τετράγωνον ἀριθμόν· λέγω, ὅτι ἀσύμμετρός ἐστιν ἡ Α on B the ratio which (some) square number (has) to
τῇ Β μήκει. (some) square number. I say that A is incommensurable
Εἰ γάρ ἐστι σύμμετρος ἡ Α τῇ Β, ἕξει τὸ ἀπὸ τῆς Α in length with B.
πρὸς τὸ ἀπὸ τῆς Β λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς For if A is commensurable (in length) with B then
τετράγωνον ἀριθμόν. οὐκ ἔχει δέ· οὐκ ἄρα σύμμετρός ἐστιν the (square) on A will have to the (square) on B the ra-
ἡ Α τῇ Β μήκει. tio which (some) square number (has) to (some) square
Τὰ ἄρα ἀπὸ τῶν μήκει συμμέτρων, καὶ τὰ ἑξῆς. number. But it does not have (such a ratio). Thus, A is
not commensurable in length with B.
Thus, (squares) on (straight-lines which are) com-
290
ST EW iþ. ìri sma ELEMENTS BOOK 10
.
Corollary
iþ (commensurable) in length.
And it will be clear, from (what) has been demon-
Καὶ φανερὸν ἐκ τῶν δεδειγμένων ἔσται, ὅτι αἱ μήκει mensurable in length, and so on . . . .
σύμμετροι πάντως καὶ δυνάμει, αἱ δὲ δυνάμει οὐ πάντως καὶ strated, that (straight-lines) commensurable in length
μήκει. (are) always also (commensurable) in square, but (straight-
lines commensurable) in square (are) not always also
2
2
2
† There is an unstated assumption here that if α : β :: γ : δ then α : β :: γ : δ .
2
2
2
2
2
‡ There is an unstated assumption here that if α : β :: γ : δ then α : β :: γ : δ.
Proposition 10
.
†
Τῇ προτεθείσῃ εὐθείᾳ προσευρεῖν δύο εὐθείας ἀσυμμέτ- To find two straight-lines incommensurable with a
ρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ δυνάμει. given straight-line, the one (incommensurable) in length
only, the other also (incommensurable) in square.
Α A
Ε E
∆ D
Β B
Γ C
῎Εστω ἡ προτεθεῖσα εὐθεῖα ἡ Α· δεῖ δὴ τῇ Α προσευρεῖν Let A be the given straight-line. So it is required to
δύο εὐθείας ἀσυμμέτρους, τὴν μὲν μήκει μόνον, τὴν δὲ καὶ find two straight-lines incommensurable with A, the one
δυνάμει. (incommensurable) in length only, the other also (incom-
᾿Εκκείσθωσαν γὰρ δύο αριθμοὶ οἱ Β, Γ πρὸς ἀλλήλους mensurable) in square.
λόγον μὴ ἔχοντες, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον For let two numbers, B and C, not having to one
ἀριθμόν, τουτέστι μὴ ὅμοιοι ἐπίπεδοι, καὶ γεγονέτω ὡς ὁ another the ratio which (some) square number (has) to
Β πρὸς τὸν Γ, οὕτως τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ (some) square number—that is to say, not (being) simi-
ἀπὸ τῆς Δ τετράγωνον· ἐμάθομεν γάρ· σύμμετρον ἄρα lar plane (numbers)—have been taken. And let it be con-
τὸ ἀπὸ τῆς Α τῷ ἀπὸ τῆς Δ. καὶ ἐπεὶ ὁ Β πρὸς τὸν Γ trived that as B (is) to C, so the square on A (is) to the
λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον square on D. For we learned (how to do this) [Prop. 10.6
ἀριθμόν, οὐδ᾿ ἄρα τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Δ λόγον corr.]. Thus, the (square) on A (is) commensurable with
ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· the (square) on D [Prop. 10.6]. And since B does not
ἀσύμμετρος ἄρα ἐστὶν ἡ Α τῇ Δ μήκει. εἰλήφθω τῶν Α, Δ have to C the ratio which (some) square number (has) to
μέση ἀνάλογον ἡ Ε· ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Δ, οὕτως (some) square number, the (square) on A thus does not
τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Ε. ἀσύμμετρος have to the (square) on D the ratio which (some) square
δέ ἐστιν ἡ Α τῇ Δ μήκει· ἀσύμμετρον ἄρα ἐστὶ καὶ τὸ ἀπὸ number (has) to (some) square number either. Thus, A
τῆς Α τετράγωνον τῷ ἀπὸ τῆς Ε τετραγώνῳ· ἀσύμμετρος is incommensurable in length with D [Prop. 10.9]. Let
ἄρα ἐστὶν ἡ Α τῇ Ε δυνάμει. the (straight-line) E (which is) in mean proportion to A
Τῂ ἄρα προτεθείσῃ εὐθείᾳ τῇ Α προσεύρηνται δύο and D have been taken [Prop. 6.13]. Thus, as A is to D,
εὐθεῖαι ἀσύμμετροι αἱ Δ, Ε, μήκει μὲν μόνον ἡ Δ, δυνάμει so the square on A (is) to the (square) on E [Def. 5.9].
δὲ καὶ μήκει δηλαδὴ ἡ Ε [ὅπερ ἔδει δεῖξαι]. And A is incommensurable in length with D. Thus, the
square on A is also incommensurble with the square on
E [Prop. 10.11]. Thus, A is incommensurable in square
with E.
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ST EW iþ. ELEMENTS BOOK 10
iaþ commensurable with the given straight-line A, have been
Thus, two straight-lines, D and E, (which are) in-
found, the one, D, (incommensurable) in length only, the
other, E, (incommensurable) in square, and, clearly, also
in length. [(Which is) the very thing it was required to
show.]
† This whole proposition is regarded by Heiberg as an interpolation into the original text.
Proposition 11
.
᾿Εὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ πρῶτον τῷ If four magnitudes are proportional, and the first is
δευτέρῳ σύμμετρον ᾖ, καὶ τὸ τρίτον τῷ τετάρτῳ σύμμετρον commensurable with the second, then the third will also
ἔσται· κἂν τὸ πρῶτον τῷ δευτέρῳ ἀσύμμετρον ᾖ, καὶ τὸ be commensurable with the fourth. And if the first is in-
τρίτον τῷ τετάρτῳ ἀσύμμετρον ἔσται. commensurable with the second, then the third will also
be incommensurable with the fourth.
Α Β A B
Γ ∆ C D
῎Εστωσαν τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, Let A, B, C, D be four proportional magnitudes,
ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, τὸ Α δὲ τῷ (such that) as A (is) to B, so C (is) to D. And let A
Β σύμμετρον ἔστω· λέγω, ὅτι καὶ τὸ Γ τῷ Δ σύμμετρον be commensurable with B. I say that C will also be com-
ἔσται. mensurable with D.
᾿Επεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Β, τὸ Α ἄρα πρὸς τὸ For since A is commensurable with B, A thus has to B
Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. καί ἐστιν ὡς τὸ Α the ratio which (some) number (has) to (some) number
πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ· καὶ τὸ Γ ἄρα πρὸς τὸ Δ [Prop. 10.5]. And as A is to B, so C (is) to D. Thus,
λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν· σύμμετρον ἄρα ἐστὶ C also has to D the ratio which (some) number (has)
τὸ Γ τῷ Δ. to (some) number. Thus, C is commensurable with D
᾿Αλλὰ δὴ τὸ Α τῷ Β ἀσύμμετρον ἔστω· λέγω, ὅτι καὶ τὸ [Prop. 10.6].
Γ τῷ Δ ἀσύμμετρον ἔσται. ἐπεὶ γὰρ ἀσύμμετρόν ἐστι τὸ Α And so let A be incommensurable with B. I say that
ibþ to (some) number either. Thus, C is incommensurable
τῷ Β, τὸ Α ἄρα πρὸς τὸ Β λόγον οὐκ ἔχει, ὃν ἀριθμὸς πρὸς C will also be incommensurable with D. For since A
ἀριθμόν. καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ is incommensurable with B, A thus does not have to B
Δ· οὐδὲ τὸ Γ ἄρα πρὸς τὸ Δ λόγον ἔχει, ὃν ἀριθμὸς πρὸς the ratio which (some) number (has) to (some) number
ἀριθμόν· ἀσύμμετρον ἄρα ἐστὶ τὸ Γ τῷ Δ. [Prop. 10.7]. And as A is to B, so C (is) to D. Thus, C
᾿Εὰν ἄρα τέσσαρα μεγέθη, καὶ τὰ ἑξῆς. does not have to D the ratio which (some) number (has)
with D [Prop. 10.8].
Thus, if four magnitudes, and so on . . . .
.
Proposition 12
Τὰ τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ (Magnitudes) commensurable with the same magni-
σύμμετρα. tude are also commensurable with one another.
῾Εκάτερον γὰρ τῶν Α, Β τῷ Γ ἔστω σύμμετρον. λέγω, For let A and B each be commensurable with C. I say
ὅτι καὶ τὸ Α τῷ Β ἐστι σύμμετρον. that A is also commensurable with B.
᾿Επεὶ γὰρ σύμμετρόν ἐστι τὸ Α τῷ Γ, τὸ Α ἄρα πρὸς For since A is commensurable with C, A thus has
τὸ Γ λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ Δ to C the ratio which (some) number (has) to (some)
πρὸς τὸν Ε. πάλιν, ἐπεὶ σύμμετρόν ἐστι τὸ Γ τῷ Β, τὸ Γ ἄρα number [Prop. 10.5]. Let it have (the ratio) which D
πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν (has) to E. Again, since C is commensurable with B,
ὁ Ζ πρὸς τὸν Η. καὶ λόγων δοθέντων ὁποσωνοῦν τοῦ τε, C thus has to B the ratio which (some) number (has)
ὃν ἔχει ὁ Δ πρὸς τὸν Ε, καὶ ὁ Ζ πρὸς τὸν Η εἰλήφθωσαν to (some) number [Prop. 10.5]. Let it have (the ratio)
ἀριθμοὶ ἑξῆς ἐν τοῖς δοθεῖσι λόγοις οἱ Θ, Κ, Λ· ὥστε εἶναι which F (has) to G. And for any multitude whatsoever
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ST EW iþ.
ELEMENTS BOOK 10
ὡς μὲν τὸν Δ πρὸς τὸν Ε, οὕτως τὸν Θ πρὸς τὸν Κ, ὡς δὲ of given ratios—(namely,) those which D has to E, and
τὸν Ζ πρὸς τὸν Η, οὕτως τὸν Κ πρὸς τὸν Λ. F to G—let the numbers H, K, L (which are) contin-
uously (proportional) in the(se) given ratios have been
taken [Prop. 8.4]. Hence, as D is to E, so H (is) to K,
and as F (is) to G, so K (is) to L.
Α Γ Β A C B
∆ Θ D H
Ε Κ E K
Ζ Λ F L
Η G
᾿Επεὶ οὖν ἐστιν ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Δ πρὸς Therefore, since as A is to C, so D (is) to E, but as
τὸν Ε, ἀλλ᾿ ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Θ πρὸς τὸν Κ, D (is) to E, so H (is) to K, thus also as A is to C, so H
ἔστιν ἄρα καὶ ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ. (is) to K [Prop. 5.11]. Again, since as C is to B, so F
πάλιν, ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Β, οὕτως ὁ Ζ πρὸς τὸν (is) to G, but as F (is) to G, [so] K (is) to L, thus also
igþ
Η, ἀλλ᾿ ὡς ὁ Ζ πρὸς τὸν Η, [οὕτως] ὁ Κ πρὸς τὸν Λ, καὶ as C (is) to B, so K (is) to L [Prop. 5.11]. And also as A
ὡς ἄρα τὸ Γ πρὸς τὸ Β, οὕτως ὁ Κ πρὸς τὸν Λ. ἔστι δὲ καὶ is to C, so H (is) to K. Thus, via equality, as A is to B,
ὡς τὸ Α πρὸς τὸ Γ, οὕτως ὁ Θ πρὸς τὸν Κ· δι᾿ ἴσου ἄρα so H (is) to L [Prop. 5.22]. Thus, A has to B the ratio
ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως ὁ Θ πρὸς τὸν Λ. τὸ Α ἄρα which the number H (has) to the number L. Thus, A is
πρὸς τὸ Β λόγον ἔχει, ὃν ἀριθμὸς ὁ Θ πρὸς ἀριθμὸν τὸν Λ· commensurable with B [Prop. 10.6].
σύμμετρον ἄρα ἐστὶ τὸ Α τῷ Β. Thus, (magnitudes) commensurable with the same
Τὰ ἄρα τῷ αὐτῷ μεγέθει σύμμετρα καὶ ἀλλήλοις ἐστὶ magnitude are also commensurable with one another.
σύμμετρα· ὅπερ ἔδει δεῖξαι. (Which is) the very thing it was required to show.
Proposition 13
.
᾿Εὰν ᾖ δύο μεγέθη σύμμετρα, τὸ δὲ ἕτερον αὐτῶν If two magnitudes are commensurable, and one of
μεγέθει τινὶ ἀσύμμετρον ᾖ, καὶ τὸ λοιπὸν τῷ αὐτῷ ἀσύμμετρ- them is incommensurable with some magnitude, then
ον ἔσται. the remaining (magnitude) will also be incommensurable
with it.
Α A
Γ C
Β B
῎Εστω δύο μεγέθη σύμμετρα τὰ Α, Β, τὸ δὲ ἕτερον Let A and B be two commensurable magnitudes, and
αὐτῶν τὸ Α ἄλλῳ τινὶ τῷ Γ ἀσύμμετρον ἔστω· λέγω, ὅτι let one of them, A, be incommensurable with some other
καὶ τὸ λοιπὸν τὸ Β τῷ Γ ἀσύμμετρόν ἐστιν. (magnitude), C. I say that the remaining (magnitude),
¨mma mensurable with C. Thus, (it is) incommensurable.
Εἰ γάρ ἐστι σύμμετρον τὸ Β τῷ Γ, ἀλλὰ καὶ τὸ Α τῷ B, is also incommensurable with C.
Β σύμμετρόν ἐστιν, καὶ τὸ Α ἄρα τῷ Γ σύμμετρόν ἐστιν. For if B is commensurable with C, but A is also com-
ἀλλὰ καὶ ἀσύμμετρον· ὅπερ ἀδύνατον. οὐκ ἄρα σύμμετρόν mensurable with B, A is thus also commensurable with
ἐστι τὸ Β τῷ Γ· ἀσύμμετρον ἄρα. C [Prop. 10.12]. But, (it is) also incommensurable (with
᾿Εὰν ἄρα ᾖ δύο μεγέθη σύμμετρα, καὶ τὰ ἑξῆς. C). The very thing (is) impossible. Thus, B is not com-
Thus, if two magnitudes are commensurable, and so
on . . . .
.
Lemma
Δύο δοθεισῶν εὐθειῶν ἀνίσων εὑρεῖν, τίνι μεῖζον For two given unequal straight-lines, to find by (the
δύναται ἡ μείζων τῆς ἐλάσσονος. square on) which (straight-line) the square on the greater
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ELEMENTS BOOK 10
(straight-line is) larger than (the square on) the lesser. †
Γ C
∆ D
Α Β A B
῎Εστωσαν αἱ δοθεῖσαι δύο ἄνισοι εὐθεῖαι αἱ ΑΒ, Γ, ὧν Let AB and C be the two given unequal straight-lines,
μείζων ἔστω ἡ ΑΒ· δεῖ δὴ εὑρεῖν, τίνι μεῖζον δύναται ἡ ΑΒ and let AB be the greater of them. So it is required to
τῆς Γ. find by (the square on) which (straight-line) the square
Γεγράφθω ἐπὶ τῆς ΑΒ ἡμικύκλιον τὸ ΑΔΒ, καὶ εἰς αὐτὸ on AB (is) greater than (the square on) C.
ἐνηρμόσθω τῇ Γ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΒ. φανερὸν Let the semi-circle ADB have been described on AB.
δή, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΑΔΒ γωνία, καὶ ὅτι ἡ ΑΒ τῆς And let AD, equal to C, have been inserted into it
ΑΔ, τουτέστι τῆς Γ, μεῖζον δύναται τῇ ΔΒ. [Prop. 4.1]. And let DB have been joined. So (it is) clear
῾Ομοίως δὲ καὶ δύο δοθεισῶν εὐθειῶν ἡ δυναμένη αὐτὰς that the angle ADB is a right-angle [Prop. 3.31], and
εὑρίσκεται οὕτως. that the square on AB (is) greater than (the square on)
῎Εστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ, καὶ δέον AD—that is to say, (the square on) C—by (the square
ἔστω εὑρεῖν τὴν δυναμένην αὐτάς. κείσθωσαν γάρ, ὥστε on) DB [Prop. 1.47].
ὀρθὴν γωνίαν περιέχειν τὴν ὑπὸ ΑΔ, ΔΒ, καὶ ἐπεζεύχθω And, similarly, the square-root of (the sum of the
ἡ ΑΒ· φανερὸν πάλιν, ὅτι ἡ τὰς ΑΔ, ΔΒ δυναμένη ἐστὶν ἡ squares on) two given straight-lines is also found likeso.
ΑΒ· ὅπερ ἔδει δεῖξαι. Let AD and DB be the two given straight-lines. And
let it be necessary to find the square-root of (the sum
of the squares on) them. For let them have been laid
idþ been joined. (It is) again clear that AB is the square-root
down such as to encompass a right-angle—(namely), that
(angle encompassed) by AD and DB. And let AB have
of (the sum of the squares on) AD and DB [Prop. 1.47].
(Which is) the very thing it was required to show.
† That is, if α and β are the lengths of two given straight-lines, with α being greater than β, to find a straight-line of length γ such that
2
2
2
2
2
2
α = β + γ . Similarly, we can also find γ such that γ = α + β .
Proposition 14
.
᾿Εὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, δύνηται δὲ ἡ If four straight-lines are proportional, and the square
πρώτη τῆς δευτέρας μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ on the first is greater than (the square on) the sec-
[μήκει], καὶ ἡ τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ ond by the (square) on (some straight-line) commen-
συμμέτρου ἑαυτῇ [μήκει]. καὶ ἐὰν ἡ πρώτη τῆς δευτέρας surable [in length] with the first, then the square on
μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ [μήκει], καὶ ἡ the third will also be greater than (the square on) the
τρίτη τῆς τετάρτης μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου fourth by the (square) on (some straight-line) commen-
ἑαυτῇ [μήκει]. surable [in length] with the third. And if the square on
῎Εστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, Δ, the first is greater than (the square on) the second by
ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν Δ, καὶ ἡ Α μὲν the (square) on (some straight-line) incommensurable
τῆς Β μεῖζον δυνάσθω τῷ ἀπὸ τῆς Ε, ἡ δὲ Γ τῆς Δ μεῖζον [in length] with the first, then the square on the third
δυνάσθω τῷ ἀπὸ τῆς Ζ· λέγω, ὅτι, εἴτε σύμμετρός ἐστιν will also be greater than (the square on) the fourth by
ἡ Α τῇ Ε, σύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός the (square) on (some straight-line) incommensurable
ἐστιν ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ὁ Γ τῇ Ζ. [in length] with the third.
Let A, B, C, D be four proportional straight-lines,
(such that) as A (is) to B, so C (is) to D. And let the
square on A be greater than (the square on) B by the
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ELEMENTS BOOK 10
(square) on E, and let the square on C be greater than
(the square on) D by the (square) on F. I say that A
is either commensurable (in length) with E, and C is
also commensurable with F, or A is incommensurable
(in length) with E, and C is also incommensurable with
F.
Α Β Ε Γ ∆ Ζ A B E C D F
᾿Επεὶ γάρ ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν For since as A is to B, so C (is) to D, thus as the
Δ, ἔστιν ἄρα καὶ ὡς τὸ ἀπὸ τῆς Α πρὸς τὸ ἀπὸ τῆς Β, οὕτως (square) on A is to the (square) on B, so the (square) on
τὸ ἀπὸ τῆς Γ πρὸς τὸ ἀπὸ τῆς Δ. ἀλλὰ τῷ μὲν ἀπὸ τῆς Α C (is) to the (square) on D [Prop. 6.22]. But the (sum
ἴσα ἐστὶ τὰ ἀπὸ τῶν Ε, Β, τῷ δὲ ἀπὸ τῆς Γ ἴσα ἐστὶ τὰ ἀπὸ of the squares) on E and B is equal to the (square) on
τῶν Δ, Ζ. ἔστιν ἄρα ὡς τὰ ἀπὸ τῶν Ε, Β πρὸς τὸ ἀπὸ τῆς A, and the (sum of the squares) on D and F is equal
Β, οὕτως τὰ ἀπὸ τῶν Δ, Ζ πρὸς τὸ ἀπὸ τῆς Δ· διελόντι ἄρα to the (square) on C. Thus, as the (sum of the squares)
ἐστὶν ὡς τὸ ἀπὸ τῆς Ε πρὸς τὸ ἀπὸ τῆς Β, οὕτως τὸ ἀπὸ on E and B is to the (square) on B, so the (sum of the
τῆς Ζ πρὸς τὸ ἀπὸ τῆς Δ· ἔστιν ἄρα καὶ ὡς ἡ Ε πρὸς τὴν squares) on D and F (is) to the (square) on D. Thus,
Β, οὕτως ἡ Ζ πρὸς τὴν Δ· ἀνάπαλιν ἄρα ἐστὶν ὡς ἡ Β πρὸς via separation, as the (square) on E is to the (square)
τὴν Ε, οὕτως ἡ Δ πρὸς τὴν Ζ. ἔστι δὲ καὶ ὡς ἡ Α πρὸς τὴν on B, so the (square) on F (is) to the (square) on D
Β, οὕτως ἡ Γ πρὸς τὴν Δ· δι᾿ ἴσου ἄρα ἐστὶν ὡς ἡ Α πρὸς [Prop. 5.17]. Thus, also, as E is to B, so F (is) to D
ieþ length) with E, and C is also commensurable with F, or
τὴν Ε, οὕτως ἡ Γ πρὸς τὴν Ζ. εἴτε οὖν σύμμετρός ἐστιν ἡ Α [Prop. 6.22]. Thus, inversely, as B is to E, so D (is)
τῇ Ε, συμμετρός ἐστι καὶ ἡ Γ τῇ Ζ, εἴτε ἀσύμμετρός ἐστιν to F [Prop. 5.7 corr.]. But, as A is to B, so C also (is)
ἡ Α τῇ Ε, ἀσύμμετρός ἐστι καὶ ἡ Γ τῇ Ζ. to D. Thus, via equality, as A is to E, so C (is) to F
᾿Εὰν ἄρα, καὶ τὰ ἑξῆς. [Prop. 5.22]. Therefore, A is either commensurable (in
A is incommensurable (in length) with E, and C is also
incommensurable with F [Prop. 10.11].
Thus, if, and so on . . . .
Proposition 15
.
᾿Εὰν δύο μεγέθη σύμμετρα συντεθῇ, καὶ τὸ ὅλον If two commensurable magnitudes are added together
ἑκατέρῳ αὐτῶν σύμμετρον ἔσται· κἂν τὸ ὅλον ἑνὶ αὐτῶν then the whole will also be commensurable with each of
σύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη σύμμετρα ἔσται. them. And if the whole is commensurable with one of
Συγκείσθω γὰρ δύο μεγέθη σύμμετρα τὰ ΑΒ, ΒΓ· λέγω, them then the original magnitudes will also be commen-
ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἐστι σύμμετρον. surable (with one another).
For let the two commensurable magnitudes AB and
BC be laid down together. I say that the whole AC is
also commensurable with each of AB and BC.
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Α Β Γ A ELEMENTS BOOK 10
C
B
∆ D
᾿Επεὶ γὰρ σύμμετρά ἐστι τὰ ΑΒ, ΒΓ, μετρήσει τι αὐτὰ For since AB and BC are commensurable, some mag-
μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, nitude will measure them. Let it (so) measure (them),
ΒΓ μετρεῖ, καὶ ὅλον τὸ ΑΓ μετρήσει. μετρεῖ δὲ καὶ τὰ ΑΒ, and let it be D. Therefore, since D measures (both) AB
ΒΓ. τὸ Δ ἄρα τὰ ΑΒ, ΒΓ, ΑΓ μετρεῖ· σύμμετρον ἄρα ἐστὶ and BC, it will also measure the whole AC. And it also
τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ. measures AB and BC. Thus, D measures AB, BC, and
᾿Αλλὰ δὴ τὸ ΑΓ ἔστω σύμμετρον τῷ ΑΒ· λέγω δή, ὅτι AC. Thus, AC is commensurable with each of AB and
καὶ τὰ ΑΒ, ΒΓ σύμμετρά ἐστιν. BC [Def. 10.1].
᾿Επεὶ γὰρ σύμμετρά ἐστι τὰ ΑΓ, ΑΒ, μετρήσει τι αὐτὰ And so let AC be commensurable with AB. I say that
μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΓΑ, AB and BC are also commensurable.
iþ And it also measures AB. Thus, D will measure (both)
ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει. μετρεῖ δὲ καὶ For since AC and AB are commensurable, some mag-
τὸ ΑΒ· τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρήσει· σύμμετρα ἄρα ἐστὶ nitude will measure them. Let it (so) measure (them),
τὰ ΑΒ, ΒΓ. and let it be D. Therefore, since D measures (both) CA
᾿Εὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς. and AB, it will thus also measure the remainder BC.
AB and BC. Thus, AB and BC are commensurable
[Def. 10.1].
Thus, if two magnitudes, and so on . . . .
.
Proposition 16
᾿Εὰν δύο μεγέθη ἀσύμμετρα συντεθῇ, καὶ τὸ ὅλον If two incommensurable magnitudes are added to-
ἑκατέρῳ αὐτῶν ἀσύμμετρον ἔσται· κἂν τὸ ὅλον ἑνὶ αὐτῶν gether then the whole will also be incommensurable with
ἀσύμμετρον ᾖ, καὶ τὰ ἐξ ἀρχῆς μεγέθη ἀσύμμετρα ἔσται. each of them. And if the whole is incommensurable with
one of them then the original magnitudes will also be in-
commensurable (with one another).
Α Β Γ A B C
∆ D
Συγκείσθω γὰρ δύο μεγέθη ἀσύμμετρα τὰ ΑΒ, ΒΓ· For let the two incommensurable magnitudes AB and
λέγω, ὅτι καὶ ὅλον τὸ ΑΓ ἑκατέρῳ τῶν ΑΒ, ΒΓ ἀσύμμετρόν BC be laid down together. I say that that the whole AC
ἐστιν. is also incommensurable with each of AB and BC.
Εἰ γὰρ μή ἐστιν ἀσύμμετρα τὰ ΓΑ, ΑΒ, μετρήσει τι For if CA and AB are not incommensurable then
[αὐτὰ] μέγεθος. μετρείτω, εἰ δυνατόν, καὶ ἔστω τὸ Δ. ἐπεὶ some magnitude will measure [them]. If possible, let it
οὖν τὸ Δ τὰ ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸ ΒΓ μετρήσει. (so) measure (them), and let it be D. Therefore, since
μετρεῖ δὲ καὶ τὸ ΑΒ· τὸ Δ ἄρα τὰ ΑΒ, ΒΓ μετρεῖ. σύμμετρα D measures (both) CA and AB, it will thus also mea-
ἄρα ἐστὶ τὰ ΑΒ, ΒΓ· ὑπέκειντο δὲ καὶ ἀσύμμετρα· ὅπερ sure the remainder BC. And it also measures AB. Thus,
ἐστὶν ἀδύνατον. οὐκ ἄρα τὰ ΓΑ, ΑΒ μετρήσει τι μέγεθος· D measures (both) AB and BC. Thus, AB and BC are
ἀσύμμετρα ἄρα ἐστὶ τὰ ΓΑ, ΑΒ. ὁμοίως δὴ δείξομεν, ὅτι καὶ commensurable [Def. 10.1]. But they were also assumed
τὰ ΑΓ, ΓΒ ἀσύμμετρά ἐστιν. τὸ ΑΓ ἄρα ἑκατέρῳ τῶν ΑΒ, (to be) incommensurable. The very thing is impossible.
ΒΓ ἀσύμμετρόν ἐστιν. Thus, some magnitude cannot measure (both) CA and
᾿Αλλὰ δὴ τὸ ΑΓ ἑνὶ τῶν ΑΒ, ΒΓ ἀσύμμετρον ἔστω. AB. Thus, CA and AB are incommensurable [Def. 10.1].
ἔστω δὴ πρότερον τῷ ΑΒ· λέγω, ὅτι καὶ τὰ ΑΒ, ΒΓ So, similarly, we can show that AC and CB are also
ἀσύμμετρά ἐστιν. εἰ γὰρ ἔσται σύμμετρα, μετρήσει τι αὐτὰ incommensurable. Thus, AC is incommensurable with
μέγεθος. μετρείτω, καὶ ἔστω τὸ Δ. ἐπεὶ οὖν τὸ Δ τὰ ΑΒ, each of AB and BC.
ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸ ΑΓ μετρήσει. μετρεῖ δὲ καὶ τὸ And so let AC be incommensurable with one of AB
ΑΒ· τὸ Δ ἄρα τὰ ΓΑ, ΑΒ μετρεῖ. σύμμετρα ἄρα ἐστὶ τὰ and BC. So let it, first of all, be incommensurable with
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ELEMENTS BOOK 10
ΓΑ, ΑΒ· ὑπέκειτο δὲ καὶ ἀσύμμετρα· ὅπερ ἐστὶν ἀδύνατον. AB. I say that AB and BC are also incommensurable.
οὐκ ἄρα τὰ ΑΒ, ΒΓ μετρήσει τι μέγεθος· ἀσύμμετρα ἄρα For if they are commensurable then some magnitude will
ἐστὶ τὰ ΑΒ, ΒΓ. measure them. Let it (so) measure (them), and let it be
᾿Εὰν ἄρα δύο μεγέθη, καὶ τὰ ἑξῆς. D. Therefore, since D measures (both) AB and BC, it
¨mma CA and AB are commensurable [Def. 10.1]. But they
will thus also measure the whole AC. And it also mea-
sures AB. Thus, D measures (both) CA and AB. Thus,
were also assumed (to be) incommensurable. The very
thing is impossible. Thus, some magnitude cannot mea-
sure (both) AB and BC. Thus, AB and BC are incom-
mensurable [Def. 10.1].
Thus, if two. . . magnitudes, and so on . . . .
.
Lemma
†
᾿Εὰν παρά τινα εὐθεῖαν παραβληθῇ παραλληλόγραμμον If a parallelogram, falling short by a square figure, is
ἐλλεῖπον εἴδει τετραγώνῳ, τὸ παραβληθὲν ἴσον ἐστὶ τῷ ὑπὸ applied to some straight-line then the applied (parallelo-
τῶν ἐκ τῆς παραβολής γενομένων τμημάτων τῆς εὐθείας. gram) is equal (in area) to the (rectangle contained) by
the pieces of the straight-line created via the application
(of the parallelogram).
∆ D
Α Γ Β A C B
Παρὰ γὰρ εὐθεῖαν τὴν ΑΒ παραβεβλήσθω παραλ- For let the parallelogram AD, falling short by the
ληλόγραμμον τὸ ΑΔ ἐλλεῖπον εἴδει τετραγώνῳ τῷ ΔΒ· square figure DB, have been applied to the straight-line
λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΔ τῷ ὑπὸ τῶν ΑΓ, ΓΒ. AB. I say that AD is equal to the (rectangle contained)
izþ CB. Thus, if . . . to some straight-line, and so on . . . .
Καί ἐστιν αὐτόθεν φανερόν· ἐπεὶ γὰρ τετράγωνόν ἐστι by AC and CB.
τὸ ΔΒ, ἴση ἐστὶν ἡ ΔΓ τῇ ΓΒ, καί ἐστι τὸ ΑΔ τὸ ὑπὸ τῶν And it is immediately obvious. For since DB is a
ΑΓ, ΓΔ, τουτέστι τὸ ὑπὸ τῶν ΑΓ, ΓΒ. square, DC is equal to CB. And AD is the (rectangle
᾿Εὰν ἄρα παρά τινα εὐθεῖαν, καὶ τὰ ἑξῆς. contained) by AC and CD—that is to say, by AC and
† Note that this lemma only applies to rectangular parallelograms.
.
Proposition 17
†
᾿Εὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετράτῳ μέρει If there are two unequal straight-lines, and a (rect-
τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ angle) equal to the fourth part of the (square) on the
ἐλλεῖπον εἴδει τετραγώνῳ καὶ εἰς σύμμετρα αὐτὴν διαιρῇ lesser, falling short by a square figure, is applied to the
μήκει, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ greater, and divides it into (parts which are) commen-
συμμέτου ἑαυτῇ [μήκει]. καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος surable in length, then the square on the greater will be
μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ [μήκει], τῷ δὲ larger than (the square on) the lesser by the (square)
τετράρτῳ τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα on (some straight-line) commensurable [in length] with
παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν the greater. And if the square on the greater is larger
διαιρεῖ μήκει. than (the square on) the lesser by the (square) on
῎Εστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ (some straight-line) commensurable [in length] with the
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ΒΓ, τῷ δὲ τετράρτῳ μέρει τοῦ ἀπὸ ἐλάσσονος τῆς Α, greater, and a (rectangle) equal to the fourth (part) of the
τουτέστι τῷ ἀπὸ τῆς ἡμισείας τῆς Α, ἴσον παρὰ τὴν ΒΓ (square) on the lesser, falling short by a square figure, is
παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ applied to the greater, then it divides it into (parts which
τῶν ΒΔ, ΔΓ, σύμμετρος δὲ ἔστω ἡ ΒΔ τῇ ΔΓ μήκει· λέγω, are) commensurable in length.
ὃτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. Let A and BC be two unequal straight-lines, of which
(let) BC (be) the greater. And let a (rectangle) equal to
the fourth part of the (square) on the lesser, A—that is,
(equal) to the (square) on half of A—falling short by a
square figure, have been applied to BC. And let it be
the (rectangle contained) by BD and DC [see previous
lemma]. And let BD be commensurable in length with
DC. I say that that the square on BC is greater than
the (square on) A by (the square on some straight-line)
commensurable (in length) with (BC).
Α A
Β Ζ Ε ∆ Γ B F E D C
Τετμήσθω γὰρ ἡ ΒΓ δίχα κατὰ τὸ Ε σημεῖον, καὶ κείσθω For let BC have been cut in half at the point E [Prop.
τῇ ΔΕ ἴση ἡ ΕΖ. λοιπὴ ἄρα ἡ ΔΓ ἴση ἐστὶ τῇ ΒΖ. καὶ ἐπεὶ 1.10]. And let EF be made equal to DE [Prop. 1.3].
εὐθεῖα ἡ ΒΓ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Ε, εἰς δὲ ἄνισα Thus, the remainder DC is equal to BF. And since the
κατὰ τὸ Δ, τὸ ἄρα ὑπὸ ΒΔ, ΔΓ περειχόμενον ὀρθογώνιον straight-line BC has been cut into equal (pieces) at E,
μετὰ τοῦ ἀπὸ τῆς ΕΔ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς and into unequal (pieces) at D, the rectangle contained
ΕΓ τετραγώνῳ· καὶ τὰ τετραπλάσια· τὸ ἄρα τετράκις ὑπὸ by BD and DC, plus the square on ED, is thus equal to
τῶν ΒΔ, ΔΓ μετὰ τοῦ τετραπλασίου τοῦ ἀπὸ τῆς ΔΕ ἴσον the square on EC [Prop. 2.5]. (The same) also (for) the
ἐστὶ τῷ τετράκις ἀπὸ τῆς ΕΓ τετραγώνῳ. ἀλλὰ τῷ μὲν quadruples. Thus, four times the (rectangle contained)
τετραπλασίῳ τοῦ ὑπὸ τῶν ΒΔ, ΔΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς by BD and DC, plus the quadruple of the (square) on
Α τετράγωνον, τῷ δὲ τετραπλασίῳ τοῦ ἀπὸ τῆς ΔΕ ἴσον DE, is equal to four times the square on EC. But, the
ἐστὶ τὸ ἀπὸ τῆς ΔΖ τετράγωνον· διπλασίων γάρ ἐστιν ἡ ΔΖ square on A is equal to the quadruple of the (rectangle
τῆς ΔΕ. τῷ δὲ τετραπλασίῳ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τὸ contained) by BD and DC, and the square on DF is
ἀπὸ τῆς ΒΓ τετράγωνον· διπλασίων γάρ ἐστι πάλιν ἡ ΒΓ equal to the quadruple of the (square) on DE. For DF
τῆς ΓΕ. τὰ ἄρα ἀπὸ τῶν Α, ΔΖ τετράγωνα ἴσα ἐστὶ τῷ ἀπὸ is double DE. And the square on BC is equal to the
τῆς ΒΓ τετράγωνῳ· ὥστε τὸ ἀπὸ τῆς ΒΓ τοῦ ἀπὸ τῆς Α quadruple of the (square) on EC. For, again, BC is dou-
μεῖζόν ἐστι τῷ ἀπὸ τῆς ΔΖ· ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται ble CE. Thus, the (sum of the) squares on A and DF is
τῇ ΔΖ. δεικτέον, ὅτι καὶ σύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ. equal to the square on BC. Hence, the (square) on BC
ἐπεὶ γὰρ σύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, σύμμετρος is greater than the (square) on A by the (square) on DF.
ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει. ἀλλὰ ἡ ΓΔ ταῖς ΓΔ, ΒΖ Thus, BC is greater in square than A by DF. It must
ἐστι σύμμετρος μήκει· ἴση γάρ ἐστιν ἡ ΓΔ τῇ ΒΖ. καὶ ἡ ΒΓ also be shown that BC is commensurable (in length)
ἄρα σύμμετρός ἐστι ταῖς ΒΖ, ΓΔ μήκει· ὥστε καὶ λοιπῇ τῇ with DF. For since BD is commensurable in length
ΖΔ σύμμετρός ἐστιν ἡ ΒΓ μήκει· ἡ ΒΓ ἄρα τῆς Α μεῖζον with DC, BC is thus also commensurable in length with
δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ. CD [Prop. 10.15]. But, CD is commensurable in length
᾿Αλλὰ δὴ ἡ ΒΓ τῆς Α μεῖζον δυνάσθω τῷ ἀπὸ συμμέτρου with CD plus BF. For CD is equal to BF [Prop. 10.6].
ἑαυτῇ, τῷ δὲ τετράτρῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ Thus, BC is also commensurable in length with BF plus
παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ CD [Prop. 10.12]. Hence, BC is also commensurable
τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι σύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ in length with the remainder FD [Prop. 10.15]. Thus,
μήκει. the square on BC is greater than (the square on) A by
Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, the (square) on (some straight-line) commensurable (in
ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. δύναται δὲ ἡ length) with (BC).
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ΒΓ τῆς Α μεῖζον τῷ ἀπὸ συμμέτρου ἑαυτῇ. σύμμετρος ἄρα And so let the square on BC be greater than the
ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει· ὥστε καὶ λοιπῇ συναμφοτέρῳ τῇ (square on) A by the (square) on (some straight-line)
ΒΖ, ΔΓ σύμμετρός ἐστιν ἡ ΒΓ μήκει. ἀλλὰ συναμφότερος commensurable (in length) with (BC). And let a (rect-
ἡ ΒΖ, ΔΓ σύμμετρός ἐστι τῇ ΔΓ [μήκει]. ὥστε καὶ ἡ ΒΓ angle) equal to the fourth (part) of the (square) on A,
τῇ ΓΔ σύμμετρός ἐστι μήκει· καὶ διελόντι ἄρα ἡ ΒΔ τῇ ΔΓ falling short by a square figure, have been applied to BC.
ἐστι σύμμετρος μήκει. And let it be the (rectangle contained) by BD and DC. It
᾿Εὰν ἄρα ὦσι δύο εὐθεῖαι ἄνισοι, καὶ τὰ ἑξῆς. must be shown that BD is commensurable in length with
DC.
For, similarly, by the same construction, we can show
that the square on BC is greater than the (square on) A
by the (square) on FD. And the square on BC is greater
than the (square on) A by the (square) on (some straight-
line) commensurable (in length) with (BC). Thus, BC
is commensurable in length with FD. Hence, BC is
also commensurable in length with the remaining sum
of BF and DC [Prop. 10.15]. But, the sum of BF and
DC is commensurable [in length] with DC [Prop. 10.6].
ihþ surable in length with DC [Prop. 10.15].
Hence, BC is also commensurable in length with CD
[Prop. 10.12]. Thus, via separation, BD is also commen-
Thus, if there are two unequal straight-lines, and so
on . . . .
2
2
2
† This proposition states that if α x − x = β /4 (where α = BC, x = DC, and β = A) then α and p α − β are commensurable when α − x
2
are x are commensurable, and vice versa.
Proposition 18
.
†
᾿Εὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει If there are two unequal straight-lines, and a (rect-
τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ angle) equal to the fourth part of the (square) on the
ἐλλεῖπον εἴδει τετραγώνῳ, καὶ εἰς ἀσυμμετρα αὐτὴν διαιρῇ lesser, falling short by a square figure, is applied to the
[μήκει], ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ greater, and divides it into (parts which are) incom-
ἀσυμμέτρου ἑαυτῇ. καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος μεῖζον mensurable [in length], then the square on the greater
δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετράρτῳ τοῦ ἀπὸ will be larger than the (square on the) lesser by the
τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον (square) on (some straight-line) incommensurable (in
εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διαιρεῖ [μήκει]. length) with the greater. And if the square on the
῎Εστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ ΒΓ, greater is larger than the (square on the) lesser by the
τῷ δὲ τετάρτῳ [μέρει] τοῦ ἀπὸ τῆς ἐλάσσονος τῆς Α ἴσον (square) on (some straight-line) incommensurable (in
παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ length) with the greater, and a (rectangle) equal to the
ἔστω τὸ ὑπὸ τῶν ΒΔΓ, ἀσύμμετρος δὲ ἔστω ἡ ΒΔ τῇ fourth (part) of the (square) on the lesser, falling short by
ΔΓ μήκει· λέγω, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ a square figure, is applied to the greater, then it divides it
ἀσυμμέτρου ἑαυτῇ. into (parts which are) incommensurable [in length].
Let A and BC be two unequal straight-lines, of which
(let) BC (be) the greater. And let a (rectangle) equal to
the fourth [part] of the (square) on the lesser, A, falling
short by a square figure, have been applied to BC. And
let it be the (rectangle contained) by BDC. And let BD
be incommensurable in length with DC. I say that that
the square on BC is greater than the (square on) A by
the (square) on (some straight-line) incommensurable
(in length) with (BC).
299
ST EW iþ.
Α A ELEMENTS BOOK 10
Β Ζ Ε ∆ Γ B F E D C
Τῶν γὰρ αὐτῶν κατασκευασθέντων τῷ πρότερον ὁμοίως For, similarly, by the same construction as before, we
δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. can show that the square on BC is greater than the
δεικτέον [οὖν], ὅτι ἀσύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ μήκει. (square on) A by the (square) on FD. [Therefore] it
ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, ἀσύμμετρος must be shown that BC is incommensurable in length
ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει. ἀλλὰ ἡ ΔΓ σύμμετρός ἐστι with DF. For since BD is incommensurable in length
συναμφοτέραις ταῖς ΒΖ, ΔΓ· καὶ ἡ ΒΓ ἄρα ἀσύμμετρός with DC, BC is thus also incommensurable in length
ἐστι συναμφοτέραις ταῖς ΒΖ, ΔΓ. ὥστε καὶ λοιπῇ τῇ ΖΔ with CD [Prop. 10.16]. But, DC is commensurable (in
ἀσύμμετρός ἔστιν ἡ ΒΓ μήκει. καὶ ἡ ΒΓ τῆς Α μεῖζον length) with the sum of BF and DC [Prop. 10.6]. And,
δύναται τῷ ἀπὸ τῆς ΖΔ· ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῷ thus, BC is incommensurable (in length) with the sum of
ἀπὸ ἀσυμμέτρου ἑαυτῇ. BF and DC [Prop. 10.13]. Hence, BC is also incommen-
Δυνάσθω δὴ πάλιν ἡ ΒΓ τῆς Α μεῖζον τῷ ἀπὸ ἀσυμμέτρ- surable in length with the remainder FD [Prop. 10.16].
ου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ And the square on BC is greater than the (square on)
παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ A by the (square) on FD. Thus, the square on BC is
τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ greater than the (square on) A by the (square) on (some
μήκει. straight-line) incommensurable (in length) with (BC).
Τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, So, again, let the square on BC be greater than the
ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. ἀλλὰ (square on) A by the (square) on (some straight-line) in-
ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ. commensurable (in length) with (BC). And let a (rect-
ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει· ὥστε καὶ λοιπῇ angle) equal to the fourth [part] of the (square) on A,
συναμφοτέρῳ τῇ ΒΖ, ΔΓ ἀσύμμετρός ἐστιν ἡ ΒΓ. ἀλλὰ συ- falling short by a square figure, have been applied to BC.
ναμφότερος ἡ ΒΖ, ΔΓ τῇ ΔΓ σύμμετρός ἐστι μήκει· καὶ ἡ And let it be the (rectangle contained) by BD and DC.
ΒΓ ἄρα τῇ ΔΓ ἀσύμμετρός ἐστι μήκει· ὥστε καὶ διελόντι ἡ It must be shown that BD is incommensurable in length
ΒΔ τῇ ΔΓ ἀσύμμετρός ἐστι μήκει. with DC.
᾿Εὰν ἄρα ὦσι δύο εὐθεῖαι, καὶ τὰ ἑξῆς. For, similarly, by the same construction, we can show
that the square on BC is greater than the (square) on
A by the (square) on FD. But, the square on BC is
greater than the (square) on A by the (square) on (some
straight-line) incommensurable (in length) with (BC).
Thus, BC is incommensurable in length with FD. Hence,
BC is also incommensurable (in length) with the re-
maining sum of BF and DC [Prop. 10.16]. But, the
sum of BF and DC is commensurable in length with
ijþ tion, BD is also incommensurable in length with DC
DC [Prop. 10.6]. Thus, BC is also incommensurable
in length with DC [Prop. 10.13]. Hence, via separa-
[Prop. 10.16].
Thus, if there are two . . . straight-lines, and so on . . . .
2
† This proposition states that if α x − x = β /4 (where α = BC, x = DC, and β = A) then α and p α − β are incommensurable when
2
2
2
α − x are x are incommensurable, and vice versa.
.
Proposition 19
Τὸ ὑπὸ ῥητῶν μήκει συμμέτρων εὐθειῶν περιεχόμενον The rectangle contained by rational straight-lines
ὀρθογώνιον ῥητόν ἐστιν. (which are) commensurable in length is rational.
῾Υπὸ γὰρ ῥητῶν μήκει συμμέτρων εὐθειῶν τῶν ΑΒ, ΒΓ For let the rectangle AC have been enclosed by the
300
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