EUCLID S ELEMENTS IN GREEK VOLUME II BOOKS 5 9

EUCLID S ELEMENTS IN GREEK VOLUME II BOOKS 5 9 The Greek text of J.L. Heiberg (1884) from Euclidis Elementa, edidit et Latine interpretatus est I.L. Heiberg, Uol. II, Libros V-IX continens, Lipsiae, in aedibus B.G. Teubneri, 1884 with an accompanying English translation by Richard Fitzpatrick

For Faith

Preface Euclid s Elements is by far the most famous mathematical work of classical antiquity, and also has the distinction of being the world s oldest continuously used mathematical textbook. Little is known about the author, beyond the fact that he lived in Alexandria around 300 BCE. The main subject of this work is Geometry, which was something of an obsession for the Ancient Greeks. Most of the theorems appearing in Euclid s Elements were not discovered by Euclid himself, but were the work of earlier Greek mathematicians such as Pythagoras (and his school), Hippocrates of Chios, Theaetetus, and Eudoxus of Cnidos. However, Euclid is generally credited with arranging these theorems in a logical manner, so as to demonstrate (admittedly, not always with the rigour demanded by modern mathematics) that they necessarily follow from five simple axioms. Euclid is also credited with devising a number of particularly ingenious proofs of previously discovered theorems: e.g., Theorem 48 in Book 1. It is natural that anyone with a knowledge of Ancient Greek, combined with a general interest in Mathematics, would wish to read Euclid s Elements in its original form. It is therefore extremely surprizing that, whilst translations of this work into modern languages are easily available, the Greek text has been completely unobtainable (as a book) for many years. This purpose of this publication is to make the definitive Greek text of Euclid s Elements i.e., that edited by J.L. Heiberg (1883-1888) again available to the general public in book form. The Greek text is accompanied by my own English translation. The aim of my translation is to be as literal as possible, whilst still (approximately) remaining within the bounds of idiomatic English. Text within square parenthesis (in both Greek and English) indicates material identified by Heiberg as being later interpolations to the original text (some particularly obvious or unhelpful interpolations are omitted altogether). Text within round parenthesis (in English) indicates material which is implied, but not actually present, in the Greek text. My thanks goes to Mariusz Wodzicki for advice regarding the typesetting of this work. Richard Fitzpatrick; Austin, Texas; September, 2005. References Euclidus Opera Ominia, J.L. Heiberg & H. Menge (editors), Teubner (1883-1916). Euclid in Greek, Book 1, T.L. Heath (translator), Cambridge (1920). Euclid s Elements, T.L. Heath (translator), Dover (1956). History of Greek Mathematics, T.L. Heath, Dover (1981).

ΣΤΟΙΧΕΙΩΝ ε

ELEMENTS BOOK 5 Proportion 1 1 The theory of proportion set out in this book is generally attributed to Eudoxus of Cnidus. The novel feature of this theory is its ability to deal with irrational magnitudes, which had hitherto been a major stumbling block for Greek mathematicians. Throughout the footnotes in this book, α, β, γ, etc., denote general (possibly irrational) magnitudes, whereas m, n, l, etc., denote positive integers.

ΣΤΟΙΧΕΙΩΝ ε Οροι α Μέρος στ µέγεθος µεγέθους τ λασσον το µείζονος, ταν καταµετρ τ µε ζον. β Πολλαπλάσιον δ τ µε ζον το λάττονος, ταν καταµετρ ται π το λάττονος. γ Λόγος στ δύο µεγεθ ν µογεν ν κατ πηλικότητά ποια σχέσις. δ Λόγον χειν πρ ς λληλα µεγέθη λέγεται, δύναται πολλαπλασιαζόµενα λλήλων περέχειν. ε Εν τ α τ λόγ µεγέθη λέγεται ε ναι πρ τον πρ ς δεύτερον κα τρίτον πρ ς τέταρτον, ταν τ το πρώτου καί τρίτου σάκις πολλαπλάσια τ ν το δευτέρου κα τετάρτου σάκις πολλαπλασίων καθ ποιονο ν πολλαπλασιασµ ν κάτερον κατέρου µα περέχ µα σα µα λλείπ ληφθέντα κατάλληλα. Τ δ τ ν α τ ν χοντα λόγον µεγέθη νάλογον καλείσθω. ζ Οταν δ τ ν σάκις πολλαπλασίων τ µ ν το πρώτου πολλαπλάσιον περέχ το το δευτέρου πολλαπλασίου, τ δ το τρίτου πολλαπλάσιον µ περέχ το το τετάρτου πολλαπλασίου, τότε τ πρ τον πρ ς τ δεύτερον µείζονα λόγον χειν λέγεται, περ τ τρίτον πρ ς τ τέταρτον. η Αναλογία δ ν τρισ ν ροις λαχίστη στίν. θ Οταν δ τρία µεγέθη νάλογον, τ πρ τον πρ ς τ τρίτον διπλασίονα λόγον χειν λέγεται περ πρ ς τ δεύτερον. ι Οταν δ τέσσαρα µεγέθη νάλογον, τ πρ τον πρ ς τ τέταρτον τριπλασίονα λόγον χειν λέγεται περ πρ ς τ δεύτερον, κα ε ξ ς µοίως, ς ν ναλογία πάρχ. 6

ELEMENTS BOOK 5 Definitions 1 A magnitude is a part of a(nother) magnitude, the lesser of the greater, when it measures the greater. 2 2 And the greater (magnitude is) a multiple of the lesser when it is measured by the lesser. 3 A ratio is a certain type of condition with respect to size of two magnitudes of the same kind. 3 4 (Those) magnitudes are said to have a ratio with respect to one another which, being multiplied, are capable of exceeding one another. 4 5 Magnitudes are said to be in the same ratio, the first to the second, and the third to the fourth, when equal multiples of the first and the third either both exceed, are both equal to, or are both less than, equal multiples of the second and the fourth, respectively, being taken in corresponding order, according to any kind of multiplication whatever. 5 6 And let magnitudes having the same ratio be called proportional. 6 7 And when for equal multiples (as in Def. 5), the multiple of the first (magnitude) exceeds the multiple of the second, and the multiple of the third (magnitude) does not exceed the multiple of the fourth, then the first (magnitude) is said to have a greater ratio to the second than the third (magnitude has) to the fourth. 8 And a proportion in three terms is the smallest (possible). 7 9 And when three magnitudes are proportional, the first is said to have a squared 8 ratio to the third with respect to the second. 9 10 And when four magnitudes are (continuously) proportional, the first is said to have a cubed 10 ratio to the fourth with respect to the second. 11 And so on, similarly, in successive order, whatever the (continuous) proportion might be. 2 In other words, α is said to be a part of β if β = mα. 3 In modern notation, the ratio of two magnitudes, α and β, is denoted α : β. 4 In other words, α has a ratio with respect to β if mα > β and nβ > α, for some m and n. 5 In other words, α : β :: γ : δ if and only if mα > nβ whenever mγ > nδ, and mα = nβ whenever mγ = nδ, and mα < nβ whenever mγ < nδ, for all m and n. This definition is the kernel of Eudoxus theory of proportion, and is valid even if α, β, etc., are irrational. 6 Thus if α and β have the same ratio as γ and δ then they are proportional. In modern notation, α : β :: γ : δ. 7 In modern notation, a proportion in three terms α, β, and γ is written: α : β :: β : γ. 8 Literally, double. 9 In other words, if α : β :: β : γ then α : γ :: α 2 : β 2. 10 Literally, triple. 11 In other words, if α : β :: β : γ :: γ : δ then α : δ :: α 3 : β 3. 7

ΣΤΟΙΧΕΙΩΝ ε ιβ Οµόλογα µεγέθη λέγεται τ µ ν γούµενα το ς γουµένοις τ δ πόµενα το ς ποµένοις. ιγ Εναλλ ξ λόγος στ λ ψις το γουµένου πρ ς τ γούµενον κα το ποµένου πρ ς τ πόµενον. ιδ Ανάπαλιν λόγος στ λ ψις το ποµένου ς γουµένου πρ ς τ γούµενον ς πόµενον. ιε Σύνθεσις λόγου στ λ ψις το γουµένου µετ το ποµένου ς ν ς πρ ς α τ τ πόµενον. ι ιαίρεσις λόγου στ λ ψις τ ς περοχ ς, περέχει τ γούµενον το ποµένου, πρ ς α τ τ πόµενον. ιζ Αναστροφ λόγου στ λ ψις το γουµένου πρ ς τ ν περοχήν, περέχει τ γούµενον το ποµένου. ιη ι σου λόγος στ πλειόνων ντων µεγεθ ν κα λλων α το ς σων τ πλ θος σύνδυο λαµβανοµένων κα ν τ α τ λόγ, ταν ς ν το ς πρώτοις µεγέθεσι τ πρ τον πρ ς τ σχατον, ο τως ν το ς δευτέροις µεγέθεσι τ πρ τον πρ ς τ σχατον λλως Λ ψις τ ν κρων καθ πεξαίρεσιν τ ν µέσων. ιθ Τεταραγµένη δ ναλογία στίν, ταν τρι ν ντων µεγεθ ν κα λλων α το ς σων τ πλ θος γίνηται ς µ ν ν το ς πρώτοις µεγέθεσιν γούµενον πρ ς πόµενον, ο τως ν το ς δευτέροις µεγέθεσιν γούµενον πρ ς πόµενον, ς δ ν το ς πρώτοις µεγέθεσιν πόµενον πρ ς λλο τι, ο τως ν το ς δευτέροις λλο τι πρ ς γούµενον. 8

ELEMENTS BOOK 5 12 These magnitudes are said to be corresponding (magnitudes): the leading to the leading (of two ratios), and the following to the following. 13 An alternate ratio is a taking of the (ratio of the) leading (magnitude) to the leading (of two equal ratios), and (setting it equal to) the (ratio of the) following (magnitude) to the following. 12 14 An inverse ratio is a taking of the (ratio of the) following (magnitude) as the leading and the leading (magnitude) as the following. 13 15 A composition of a ratio is a taking of the (ratio of the) leading plus the following (magnitudes), as one, to the same following (magnitude). 14 16 A separation of a ratio is a taking of the (ratio of the) excess by which the leading (magnitude) exceeds the following to the same following (magnitude). 15 17 A conversion of a ratio is a taking of the (ratio of the) leading (magnitude) to the excess by which the leading (magnitude) exceeds the following. 16 18 There being several magnitudes, and other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, a ratio via equality (or ex aequali) occurs when as the first is to the last in the first (set of) magnitudes, so the first (is) to the last in the second (set of) magnitudes. Or alternately, (it is) a taking of the (ratio of the) outer (magnitudes) by the removal of the inner (magnitudes). 17 19 There being three magnitudes, and other (magnitudes) of equal number to them, a perturbed proportion occurs when as the leading is to the following in the first (set of) magnitudes, so the leading (is) to the following in the second (set of) magnitudes, and as the following (is) to some other (i.e., the remaining magnitude) in the first (set of) magnitudes, so some other (is) to the leading in the second (set of) magnitudes. 18 12 In other words, if α : β :: γ : δ then the alternate ratio corresponds to α : γ :: β : δ. 13 In other words, if α : β then the inverse ratio corresponds to β : α. 14 In other words, if α : β then the composed ratio corresponds to α + β : β. 15 In other words, if α : β then the separated ratio corresponds to α β : β. 16 In other words, if α : β then the converted ratio corresponds to α : α β. 17 In other words, if α,β,γ are the first set of magnitudes, and δ,ɛ,ζ the second set, and α : β : γ :: δ : ɛ : ζ, then the ratio via equality (or ex aequali) corresponds to α : γ :: δ : ζ. 18 In other words, if α,β,γ are the first set of magnitudes, and δ,ɛ,ζ the second set, and α : β :: δ : ɛ and β : γ :: ζ : δ, then the proportion is said to be perturbed. 9

ΣΤΟΙΧΕΙΩΝ ε α Ε Α Η Β Γ Θ Ζ Ε ν ποσαο ν µεγέθη ποσωνο ν µεγεθ ν σων τ πλ θος καστον κάστου σάκις πολλαπλάσιον, σαπλάσιόν στιν ν τ ν µεγεθ ν νός, τοσαυταπλάσια σται κα τ πάντα τ ν πάντων. Εστω ποσαο ν µεγέθη τ ΑΒ, Γ ποσωνο ν µεγεθ ν τ ν Ε, Ζ σων τ πλ θος καστον κάστου σάκις πολλαπλάσιον λέγω, τι σαπλάσιόν στι τ ΑΒ το Ε, τοσαυταπλάσια σται κα τ ΑΒ, Γ τ ν Ε, Ζ. Επε γ ρ σάκις στ πολλαπλάσιον τ ΑΒ το Ε κα τ Γ το Ζ, σα ρα στ ν ν τ ΑΒ µεγέθη σα τ Ε, τοσα τα κα ν τ Γ σα τ Ζ. δι ρήσθω τ µ ν ΑΒ ε ς τ τ Ε µεγέθη σα τ ΑΗ, ΗΒ, τ δ Γ ε ς τ τ Ζ σα τ ΓΘ, Θ σται δ σον τ πλ θος τ ν ΑΗ, ΗΒ τ πλήθει τ ν ΓΘ, Θ. κα πε σον στ τ µ ν ΑΗ τ Ε, τ δ ΓΘ τ Ζ, σον ρα τ ΑΗ τ Ε, κα τ ΑΗ, ΓΘ το ς Ε, Ζ. δι τ α τ δ σον στ τ ΗΒ τ Ε, κα τ ΗΒ, Θ το ς Ε, Ζ σα ρα στ ν ν τ ΑΒ σα τ Ε, τοσα τα κα ν το ς ΑΒ, Γ σα το ς Ε, Ζ σαπλάσιον ρα στ τ ΑΒ το Ε, τοσαυταπλάσια σται κα τ ΑΒ, Γ τ ν Ε, Ζ. Ε ν ρα ποσαο ν µεγέθη ποσωνο ν µεγεθ ν σων τ πλ θος καστον κάστου σάκις πολλαπλάσιον, σαπλάσιόν στιν ν τ ν µεγεθ ν νός, τοσαυταπλάσια σται κα τ πάντα τ ν πάντων περ δει δε ξαι. 10

ELEMENTS BOOK 5 E Proposition 1 19 A G B C H D F If there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second). Let there be any number of magnitudes whatsoever, AB, CD, (which are) equal multiples, respectively, of some (other) magnitudes, E, F, of equal number (to them). I say that as many times as AB is (divisible) by E, so many times will AB, CD also be (divisible) by E, F. For since AB, CD are equal multiples of E, F, thus as many magnitudes as (there) are in AB equal to E, so many (are there) also in CD equal to F. Let AB have been divided into magnitudes AG, GB, equal to E, and CD into (magnitudes) CH, HD, equal to F. So, the number of (divisions) AG, GB will be equal to the number of (divisions) CH, HD. And since AG is equal to E, and CH to F, AG (is) thus equal to E, and AG, CH to E, F. So, for the same (reasons), GB is equal to E, and GB, HD to E, F. Thus, as many (magnitudes) as (there) are in AB equal to E, so many (are there) also in AB, CD equal to E, F. Thus, as many times as AB is (divisible) by E, so many times will AB, CD also be (divisible) by E, F. Thus, if there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second). (Which is) the very thing it was required to show. 19 In modern notation, this proposition reads mα + mβ + = m(α + β + ). 11

Γ Ζ ΣΤΟΙΧΕΙΩΝ ε β Α Β Η Ε Θ Ε ν πρ τον δευτέρου σάκις πολλαπλάσιον κα τρίτον τετάρτου, δ κα πέµπτον δευτέρου σάκις πολλαπλάσιον κα κτον τετάρτου, κα συντεθ ν πρ τον κα πέµπτον δευτέρου σάκις σται πολλαπλάσιον κα τρίτον κα κτον τετάρτου. Πρ τον γ ρ τ ΑΒ δευτέρου το Γ σάκις στω πολλαπλάσιον κα τρίτον τ Ε τετάρτου το Ζ, στω δ κα πέµπτον τ ΒΗ δευτέρου το Γ σάκις πολλαπλάσιον κα κτον τ ΕΘ τετάρτου το Ζ λέγω, τι κα συντεθ ν πρ τον κα πέµπτον τ ΑΗ δευτέρου το Γ σάκις σται πολλαπλάσιον κα τρίτον κα κτον τ Θ τετάρτου το Ζ. Επε γ ρ σάκις στ πολλαπλάσιον τ ΑΒ το Γ κα τ Ε το Ζ, σα ρα στ ν ν τ ΑΒ σα τ Γ, τοσα τα κα ν τ Ε σα τ Ζ. δι τ α τ δ κα σα στ ν ν τ ΒΗ σα τ Γ, τοσα τα κα ν τ ΕΘ σα τ Ζ σα ρα στ ν ν λ τ ΑΗ σα τ Γ, τοσα τα κα ν λ τ Θ σα τ Ζ σαπλάσιον ρα στ τ ΑΗ το Γ, τοσαυταπλάσιον σται κα τ Θ το Ζ. κα συντεθ ν ρα πρ τον κα πέµπτον τ ΑΗ δευτέρου το Γ σάκις σται πολλαπλάσιον κα τρίτον κα κτον τ Θ τετάρτου το Ζ. Ε ν ρα πρ τον δευτέρου σάκις πολλαπλάσιον κα τρίτον τετάρτου, δ κα πέµπτον δευτέρου σάκις πολλαπλάσιον κα κτον τετάρτου, κα συντεθ ν πρ τον κα πέµπτον δευτέρου σάκις σται πολλαπλάσιον κα τρίτον κα κτον τετάρτου περ δει δε ξαι. 12

ELEMENTS BOOK 5 Proposition 2 20 A B G C F D E H If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magnitude) and the fifth, being added together, and the third and the sixth, (being added together), will also be equal multiples of the second (magnitude) and the fourth (respectively). For let a first (magnitude) AB and a third DE be equal multiples of a second C and a fourth F (respectively). And let a fifth (magnitude) BG and a sixth EH also be (other) equal multiples of the second C and the fourth F (respectively). I say that the first (magnitude) and the fifth, being added together, (to give) AG, and the third (magnitude) and the sixth, (being added together, to give) DH, will also be equal multiples of the second (magnitude) C and the fourth F (respectively). For since AB and DE are equal multiples of C and F (respectively), thus as many (magnitudes) as (there) are in AB equal to C, so many (are there) also in DE equal to F. And so, for the same (reasons), as many (magnitudes) as (there) are in BG equal to C, so many (are there) also in EH equal to F. Thus, as many (magnitudes) as (there) are in the whole of AG equal to C, so many (are there) also in the whole of DH equal to F. Thus, as many times as AG is (divisible) by C, so many times will DH also be divisible by F. Thus, the first (magnitude) and the fifth, being added together, (to give) AG, and the third (magnitude) and the sixth, (being added together, to give) DH, will also be equal multiples of the second (magnitude) C and the fourth F (respectively). Thus, if a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and a fifth (magnitude) and a sixth (are) also equal multiples of the second and fourth (respectively), then the first (magnitude) and the fifth, being added together, and the third and sixth, (being added together), will also be equal multiples of the second (magnitude) and the fourth (respectively). (Which is) the very thing it was required to show. 20 In modern notation, this propostion reads mα + nα = (m + n)α. 13

ΣΤΟΙΧΕΙΩΝ ε γ Α Β Ε Κ Ζ Γ Η Λ Θ Ε ν πρ τον δευτέρου σάκις πολλαπλάσιον κα τρίτον τετάρτου, ληφθ δ σάκις πολλαπλάσια το τε πρώτου κα τρίτου, κα δι σου τ ν ληφθέντων κάτερον κατέρου σάκις σται πολλαπλάσιον τ µ ν το δευτέρου τ δ το τετάρτου. Πρ τον γ ρ τ Α δευτέρου το Β σάκις στω πολλαπλάσιον κα τρίτον τ Γ τετάρτου το, κα ε λήφθω τ ν Α, Γ σάκις πολλαπλάσια τ ΕΖ, ΗΘ λέγω, τι σάκις στ πολλαπλάσιον τ ΕΖ το Β κα τ ΗΘ το. Επε γ ρ σάκις στ πολλαπλάσιον τ ΕΖ το Α κα τ ΗΘ το Γ, σα ρα στ ν ν τ ΕΖ σα τ Α, τοσα τα κα ν τ ΗΘ σα τ Γ. δι ρήσθω τ µ ν ΕΖ ε ς τ τ Α µεγέθη σα τ ΕΚ, ΚΖ, τ δ ΗΘ ε ς τ τ Γ σα τ ΗΛ, ΛΘ σται δ σον τ πλ θος τ ν ΕΚ, ΚΖ τ πλήθει τ ν ΗΛ, ΛΘ. κα πε σάκις στ πολλαπλάσιον τ Α το Β κα τ Γ το, σον δ τ µ ν ΕΚ τ Α, τ δ ΗΛ τ Γ, σάκις ρα στ πολλαπλάσιον τ ΕΚ το Β κα τ ΗΛ το. δι τ α τ δ σάκις στ πολλαπλάσιον τ ΚΖ το Β κα τ ΛΘ τ. πε ο ν πρ τον τ ΕΚ δευτέρου το Β σάκις στ πολλαπλάσιον κα τρίτον τ ΗΛ τετάρτου το, στι δ κα πέµπτον τ ΚΖ δευτέρου το Β σάκις πολλαπλάσιον κα κτον τ ΛΘ τετάρτου το, κα συντεθ ν ρα πρ τον κα πέµπτον τ ΕΖ δευτέρου το Β σάκις στ πολλαπλάσιον κα τρίτον κα κτον τ ΗΘ τετάρτου το. Ε ν ρα πρ τον δευτέρου σάκις πολλαπλάσιον κα τρίτον τετάρτου, ληφθ δ το πρώτου κα τρίτου σάκις πολλαπλάσια, κα δι σου τ ν ληφθέντων κάτερον κατέρου σάκις σται πολλαπλάσιον τ µ ν το δευτέρου τ δ το τετάρτου περ δει δε ξαι. 14

ELEMENTS BOOK 5 Proposition 3 21 A B E K F C D G L H If a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and equal multiples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal multiples of the second (magnitude) and the fourth, respectively. For let a first (magnitude) A and a third C be equal multiples of a second B and a fourth D (respectively), and let the equal multiples EF and GH have been taken of A and C (respectively). I say that EF and GH are equal multiples of B and D (respectively). For since EF and GH are equal multiples of A and C (respectively), thus as many (magnitudes) as (there) are in EF equal to A, so many (are there) also in GH equal to C. Let EF have been divided into magnitudes EK, KF equal to A, and GH into (magnitudes) GL, LH equal to C. So, the number of (magnitudes) EK, KF will be equal to the number of (magnitudes) GL, LH. And since A and C are equal multiples of B and D (respectively), and EK (is) equal to A, and GL to C, EK and GL are thus equal multiples of B and D (respectively). So, for the same (reasons), KF and LH are equal multiples of B and D (respectively). Therefore, since the first (magnitude) EK and the third GL are equal multiples of the second B and the fourth D (respectively), and the fifth (magnitude) KF and the sixth LH are also equal multiples of the second B and the fourth D (respectively), then the first (magnitude) and fifth, being added together, (to give) EF, and the third (magnitude) and sixth, (being added together, to give) GH, are thus also equal multiples of the second (magnitude) B and the fourth D (respectively) [Prop. 5.2]. Thus, if a first (magnitude) and a third are equal multiples of a second and a fourth (respectively), and equal multiples are taken of the first and the third, then, via equality, the (magnitudes) taken will also be equal multiples of the second (magnitude) and the fourth, respectively. (Which is) the very thing it was required to show. 21 In modern notation, this proposition reads m(nα) = (mn)α. 15

ΣΤΟΙΧΕΙΩΝ ε δ Α Β Ε Η Κ Μ Γ Ζ Θ Λ Ν Ε ν πρ τον πρ ς δεύτερον τ ν α τ ν χ λόγον κα τρίτον πρ ς τέταρτον, κα τ σάκις πολλαπλάσια το τε πρώτου κα τρίτου πρ ς τ σάκις πολλαπλάσια το δευτέρου κα τετάρτου καθ ποιονο ν πολλαπλασιασµ ν τ ν α τ ν ξει λόγον ληφθέντα κατάλληλα. Πρ τον γ ρ τ Α πρ ς δεύτερον τ Β τ ν α τ ν χέτω λόγον κα τρίτον τ Γ πρ ς τέταρτον τ, κα ε λήφθω τ ν µ ν Α, Γ σάκις πολλαπλάσια τ Ε, Ζ, τ ν δ Β, λλα, τυχεν, σάκις πολλαπλάσια τ Η, Θ λέγω, τι στ ν ς τ Ε πρ ς τ Η, ο τως τ Ζ πρ ς τ Θ. Ε λήφθω γ ρ τ ν µ ν Ε, Ζ σάκις πολλαπλάσια τ Κ, Λ, τ ν δ Η, Θ λλα, τυχεν, σάκις πολλαπλάσια τ Μ, Ν. [Κα ] πε σάκις στ πολλαπλάσιον τ µ ν Ε το Α, τ δ Ζ το Γ, κα ε ληπται τ ν Ε, Ζ σάκις πολλαπλάσια τ Κ, Λ, σάκις ρα στ πολλαπλάσιον τ Κ το Α κα τ Λ το Γ. δι τ α τ δ σάκις στ πολλαπλάσιον τ Μ το Β κα τ Ν το. κα πεί στιν ς τ Α πρ ς τ Β, ο τως τ Γ πρ ς τ, κα ε ληπται τ ν µ ν Α, Γ σάκις πολλαπλάσια τ Κ, Λ, τ ν δ Β, λλα, τυχεν, σάκις πολλαπλάσια τ Μ, Ν, ε ρα περέχει τ Κ το Μ, περέχει κα τ Λ το Ν, κα ε σον, σον, κα ε λαττον, λαττον. καί στι τ µ ν Κ, Λ τ ν Ε, Ζ σάκις πολλαπλάσια, τ δ Μ, Ν τ ν Η, Θ λλα, τυχεν, σάκις πολλαπλάσια στιν ρα ς τ Ε πρ ς τ Η, ο τως τ Ζ πρ ς τ Θ. Ε ν ρα πρ τον πρ ς δεύτερον τ ν α τ ν χ λόγον κα τρίτον πρ ς τέταρτον, κα τ σάκις πολλαπλάσια το τε πρώτου κα τρίτου πρ ς τ σάκις πολλαπλάσια το δευτέρου κα τετάρτου τ ν α τ ν ξει λόγον καθ ποιονο ν πολλαπλασιασµ ν ληφθέντα κατάλληλα περ δει δε ξαι. 16

ELEMENTS BOOK 5 Proposition 4 22 A B E G K M C D F H L N If a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever. For let a first (magnitude) A have the same ratio to a second B that a third C (has) to a fourth D. And let equal multiples E and F have been taken of A and C (respectively), and other random equal multiples G and H of B and D (respectively). I say that as E (is) to G, so F (is) to H. For let equal multiples K and L have been taken of E and F (respectively), and other random equal multiples M and N of G and H (respectively). [And] since E and F are equal multiples of A and C (respectively), and the equal multiples K and L have been taken of E and F (respectively), K and L are thus equal multiples of A and C (respectively) [Prop. 5.3]. So, for the same (reasons), M and N are equal multiples of B and D (respectively). And since as A is to B, so C (is) to D, and the equal multiples K and L have been taken of A and C (respectively), and the other random equal multiples M and N of B and D (respectively), then if K exceeds M then L also exceeds N, and if (K is) equal (to M then L is also) equal (to N), and if (K is) less (than M then L is also) less (than N) [Def. 5.5]. And K and L are equal multiples of E and F (respectively), and M and N other random equal multiples of G and H (respectively). Thus, as E (is) to G, so F (is) to H [Def. 5.5]. Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever. (Which is) the very thing it was required to show. 22 In modern notation, this proposition reads that if α : β :: γ : δ then mα : nβ :: mγ : nδ, for all m and n. 17

ΣΤΟΙΧΕΙΩΝ ε Α ε Ε Β Η Γ Ζ Ε ν µέγεθος µεγέθους σάκις πολλαπλάσιον, περ φαιρεθ ν φαιρεθέντος, κα τ λοιπ ν το λοιπο σάκις σται πολλαπλάσιον, σαπλάσιόν στι τ λον το λου. Μέγεθος γ ρ τ ΑΒ µεγέθους το Γ σάκις στω πολλαπλάσιον, περ φαιρεθ ν τ ΑΕ φαιρεθέντος το ΓΖ λέγω, τι κα λοιπ ν τ ΕΒ λοιπο το Ζ σάκις σται πολλαπλάσιον, σαπλάσιόν στιν λον τ ΑΒ λου το Γ. Οσαπλάσιον γάρ στι τ ΑΕ το ΓΖ, τοσαυταπλάσιον γεγονέτω κα τ ΕΒ το ΓΗ. Κα πε σάκις στ πολλαπλάσιον τ ΑΕ το ΓΖ κα τ ΕΒ το ΗΓ, σάκις ρα στ πολλαπλάσιον τ ΑΕ το ΓΖ κα τ ΑΒ το ΗΖ. κε ται δ σάκις πολλαπλάσιον τ ΑΕ το ΓΖ κα τ ΑΒ το Γ. σάκις ρα στ πολλαπλάσιον τ ΑΒ κατέρου τ ν ΗΖ, Γ σον ρα τ ΗΖ τ Γ. κοιν ν φ ρήσθω τ ΓΖ λοιπ ν ρα τ ΗΓ λοιπ τ Ζ σον στίν. κα πε σάκις στ πολλαπλάσιον τ ΑΕ το ΓΖ κα τ ΕΒ το ΗΓ, σον δ τ ΗΓ τ Ζ, σάκις ρα στ πολλαπλάσιον τ ΑΕ το ΓΖ κα τ ΕΒ το Ζ. σάκις δ πόκειται πολλαπλάσιον τ ΑΕ το ΓΖ κα τ ΑΒ το Γ σάκις ρα στ πολλαπλάσιον τ ΕΒ το Ζ κα τ ΑΒ το Γ. κα λοιπ ν ρα τ ΕΒ λοιπο το Ζ σάκις σται πολλαπλάσιον, σαπλάσιόν στιν λον τ ΑΒ λου το Γ. Ε ν ρα µέγεθος µεγέθους σάκις πολλαπλάσιον, περ φαιρεθ ν φαιρεθέντος, κα τ λοιπ ν το λοιπο σάκις σται πολλαπλάσιον, σαπλάσιόν στι κα τ λον το λου περ δει δε ξαι. 18

ELEMENTS BOOK 5 A Proposition 5 23 E B G C F D If a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively). For let the magnitude AB be the same multiple of the magnitude CD that the (part) taken away AE (is) of the (part) taken away CF (respectively). I say that the remainder EB will also be the same multiple of the remainder FD as that which the whole AB (is) of the whole CD (respectively). For as many times as AE is (divisible) by CF, so many times let EB also have been made (divisible) by CG. And since AE and EB are equal multiples of CF and GC (respectively), AE and AB are thus equal multiples of CF and GF (respectively) [Prop. 5.1]. And AE and AB are assumed (to be) equal multiples of CF and CD (respectively). Thus, AB is an equal multiple of each of GF and CD. Thus, GF (is) equal to CD. Let CF have been subtracted from both. Thus, the remainder GC is equal to the remainder FD. And since AE and EB are equal multiples of CF and GC (respectively), and GC (is) equal to DF, AE and EB are thus equal multiples of CF and FD (respectively). And AE and AB are assumed (to be) equal multiples of CF and CD (respectively). Thus, EB and AB are equal multiples of F D and CD (respectively). Thus, the remainder EB will also be the same multiple of the remainder FD as that which the whole AB (is) of the whole CD (respectively). Thus, if a magnitude is the same multiple of a magnitude that a (part) taken away (is) of a (part) taken away (respectively) then the remainder will also be the same multiple of the remainder as that which the whole (is) of the whole (respectively). (Which is) the very thing it was required to show. 23 In modern notation, this proposition reads mα mβ = m(α β). 19

ΣΤΟΙΧΕΙΩΝ ε Α Η Β Ε Κ Γ Θ Ζ Ε ν δύο µεγέθη δύο µεγεθ ν σάκις πολλαπλάσια, κα φαιρεθέντα τιν τ ν α τ ν σάκις πολλαπλάσια, κα τ λοιπ το ς α το ς τοι σα στ ν σάκις α τ ν πολλαπλάσια. ύο γ ρ µεγέθη τ ΑΒ, Γ δύο µεγεθ ν τ ν Ε, Ζ σάκις στω πολλαπλάσια, κα φαιρεθέντα τ ΑΗ, ΓΘ τ ν α τ ν τ ν Ε, Ζ σάκις στω πολλαπλάσια λέγω, τι κα λοιπ τ ΗΒ, Θ το ς Ε, Ζ τοι σα στ ν σάκις α τ ν πολλαπλάσια. Εστω γ ρ πρότερον τ ΗΒ τ Ε σον λέγω, τι κα τ Θ τ Ζ σον στίν. Κείσθω γ ρ τ Ζ σον τ ΓΚ. πε σάκις στ πολλαπλάσιον τ ΑΗ το Ε κα τ ΓΘ το Ζ, σον δ τ µ ν ΗΒ τ Ε, τ δ ΚΓ τ Ζ, σάκις ρα στ πολλαπλάσιον τ ΑΒ το Ε κα τ ΚΘ το Ζ. σάκις δ πόκειται πολλαπλάσιον τ ΑΒ το Ε κα τ Γ το Ζ σάκις ρα στ πολλαπλάσιον τ ΚΘ το Ζ κα τ Γ το Ζ. πε ο ν κάτερον τ ν ΚΘ, Γ το Ζ σάκις στ πολλαπλάσιον, σον ρα στ τ ΚΘ τ Γ. κοιν ν φ ρήσθω τ ΓΘ λοιπ ν ρα τ ΚΓ λοιπ τ Θ σον στίν. λλ τ Ζ τ ΚΓ στιν σον κα τ Θ ρα τ Ζ σον στίν. στε ε τ ΗΒ τ Ε σον στίν, κα τ Θ σον σται τ Ζ. Οµοίως δ δείξοµεν, τι, κ ν πολλαπλάσιον τ ΗΒ το Ε, τοσαυταπλάσιον σται κα τ Θ το Ζ. Ε ν ρα δύο µεγέθη δύο µεγεθ ν σάκις πολλαπλάσια, κα φαιρεθέντα τιν τ ν α τ ν σάκις πολλαπλάσια, κα τ λοιπ το ς α το ς τοι σα στ ν σάκις α τ ν πολλαπλάσια περ δει δε ξαι. 20

ELEMENTS BOOK 5 Proposition 6 24 A G B E K C H D F If two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples of them (respectively). For let two magnitudes AB and CD be equal multiples of two magnitudes E and F (respectively). And let the (parts) taken away (from the former) AG and CH be equal multiples of E and F (respectively). I say that the remainders GB and HD are also either equal to E and F (respectively), or (are) equal multiples of them. For, let GB be, first of all, equal to E. I say that HD is also equal to F. For let CK be made equal to F. Since AG and CH are equal multiples of E and F (respectively), and GB (is) equal to E, and KC to F, AB and KH are thus equal multiples of E and F (respectively) [Prop. 5.2]. And AB and CD are assumed (to be) equal multiples of E and F (respectively). Thus, KH and CD are equal multiples of F and F (respectively). Therefore, KH and CD are each equal multiples of F. Thus, KH is equal to CD. Let CH have be taken away from both. Thus, the remainder KC is equal to the remainder HD. But, F is equal to KC. Thus, HD is also equal to F. Hence, if GB is equal to E then HD will also be equal to F. So, similarly, we can show that even if GB is a multiple of E then HD will be the same multiple of F. Thus, if two magnitudes are equal multiples of two (other) magnitudes, and some (parts) taken away (from the former magnitudes) are equal multiples of the latter (magnitudes, respectively), then the remainders are also either equal to the latter (magnitudes), or (are) equal multiples of them (respectively). (Which is) the very thing it was required to show. 24 In modern notation, this proposition reads mα nα = (m n)α. 21

ΣΤΟΙΧΕΙΩΝ ε ζ Α Β Γ Ε Ζ Τ σα πρ ς τ α τ τ ν α τ ν χει λόγον κα τ α τ πρ ς τ σα. Εστω σα µεγέθη τ Α, Β, λλο δέ τι, τυχεν, µέγεθος τ Γ λέγω, τι κάτερον τ ν Α, Β πρ ς τ Γ τ ν α τ ν χει λόγον, κα τ Γ πρ ς κάτερον τ ν Α, Β. Ε λήφθω γ ρ τ ν µ ν Α, Β σάκις πολλαπλάσια τ, Ε, το δ Γ λλο, τυχεν, πολλαπλάσιον τ Ζ. Επε ο ν σάκις στ πολλαπλάσιον τ το Α κα τ Ε το Β, σον δ τ Α τ Β, σον ρα κα τ τ Ε. λλο δέ, τυχεν, τ Ζ. Ε ρα περέχει τ το Ζ, περέχει κα τ Ε το Ζ, κα ε σον, σον, κα ε λαττον, λαττον. καί στι τ µ ν, Ε τ ν Α, Β σάκις πολλαπλάσια, τ δ Ζ το Γ λλο, τυχεν, πολλαπλάσιον στιν ρα ς τ Α πρ ς τ Γ, ο τως τ Β πρ ς τ Γ. Λέγω [δή], τι κα τ Γ πρ ς κάτερον τ ν Α, Β τ ν α τ ν χει λόγον. Τ ν γ ρ α τ ν κατασκευασθέντων µοίως δείξοµεν, τι σον στ τ τ Ε λλο δέ τι τ Ζ ε ρα περέχει τ Ζ το, περέχει κα το Ε, κα ε σον, σον, κα ε λαττον, λαττον. καί στι τ µ ν Ζ το Γ πολλαπλάσιον, τ δ, Ε τ ν Α, Β λλα, τυχεν, σάκις πολλαπλάσια στιν ρα ς τ Γ πρ ς τ Α, ο τως τ Γ πρ ς τ Β. Τ σα ρα πρ ς τ α τ τ ν α τ ν χει λόγον κα τ α τ πρ ς τ σα. Πόρισµα Εκ δ τούτου φανερόν, τι ν µεγέθη τιν νάλογον, κα νάπαλιν νάλογον σται. περ δει δε ξαι. 22

ELEMENTS BOOK 5 Proposition 7 A B C D E F Equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ratio) to the equal (magnitudes). Let A and B be equal magnitudes, and C some other random magnitude. I say that A and B each have the same ratio to C, and (that) C (has the same ratio) to each of A and B. For let the equal multiples D and E have been taken of A and B (respectively), and the other random multiple F of C. Therefore, since D and E are equal multiples of A and B (respectively), and A (is) equal to B, D (is) thus also equal to E. And F (is) different, at random. Thus, if D exceeds F then E also exceeds F, and if (D is) equal (to F then E is also) equal (to F), and if (D is) less (than F then E is also) less (than F). And D and E are equal multiples of A and B (respectively), and F another random multiple of C. Thus, as A (is) to C, so B (is) to C [Def. 5.5]. [So] I say that C 25 also has the same ratio to each of A and B. For, similarly, we can show, by the same construction, that D is equal to E. And F (has) some other (value). Thus, if F exceeds D then it also exceeds E, and if (F is) equal (to D then it is also) equal (to E), and if (F is) less (than D then it is also) less (than E). And F is a multiple of C, and D and E other random equal multiples of A and B. Thus, as C (is) to A, so C (is) to B [Def. 5.5]. Thus, equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ratio) to the equal (magnitudes). Corollary 26 So (it is) clear from this that if some magnitudes are proportional then they will also be proportional inversely. (Which is) the very thing it was required to show. 25 The Greek text has E, which is obviously a mistake. 26 In modern notation, this corollary reads that if α : β :: γ : δ then β : α :: δ : γ. 23

ΣΤΟΙΧΕΙΩΝ ε η Γ Α Ε Β Α Ε Β Ζ Η Θ Ζ Η Θ Γ Κ Λ Μ Ν Κ Λ Μ Ν Τ ν νίσων µεγεθ ν τ µε ζον πρ ς τ α τ µείζονα λόγον χει περ τ λαττον. κα τ α τ πρ ς τ λαττον µείζονα λόγον χει περ πρ ς τ µε ζον. Εστω νισα µεγέθη τ ΑΒ, Γ, κα στω µε ζον τ ΑΒ, λλο δέ, τυχεν, τ λέγω, τι τ ΑΒ πρ ς τ µείζονα λόγον χει περ τ Γ πρ ς τ, κα τ πρ ς τ Γ µείζονα λόγον χει περ πρ ς τ ΑΒ. Επε γ ρ µε ζόν στι τ ΑΒ το Γ, κείσθω τ Γ σον τ ΒΕ τ δ λασσον τ ν ΑΕ, ΕΒ πολλαπλασιαζόµενον σται ποτ το µε ζον. στω πρότερον τ ΑΕ λαττον το ΕΒ, κα πεπολλαπλασιάσθω τ ΑΕ, κα στω α το πολλαπλάσιον τ ΖΗ µε ζον ν το, κα σαπλάσιόν στι τ ΖΗ το ΑΕ, τοσαυταπλάσιον γεγονέτω κα τ µ ν ΗΘ το ΕΒ τ δ Κ το Γ κα ε λήφθω το διπλάσιον µ ν τ Λ, τριπλάσιον δ τ Μ, κα ξ ς ν πλε ον, ως ν τ λαµβανόµενον πολλαπλάσιον µ ν γένηται το, πρώτως δ µε ζον το Κ. ε λήφθω, κα στω τ Ν τετραπλάσιον µ ν το, πρώτως δ µε ζον το Κ. Επε ο ν τ Κ το Ν πρώτως στ ν λαττον, τ Κ ρα το Μ ο κ στιν λαττον. κα πε σάκις στ πολλαπλάσιον τ ΖΗ το ΑΕ κα τ ΗΘ το ΕΒ, σάκις ρα στ πολλαπλάσιον τ ΖΗ το ΑΕ κα τ ΖΘ το ΑΒ. σάκις δέ στι πολλαπλάσιον τ ΖΗ το ΑΕ κα τ Κ το Γ σάκις ρα στ πολλαπλάσιον τ ΖΘ το ΑΒ κα τ Κ το Γ. τ ΖΘ, Κ ρα τ ν ΑΒ, Γ σάκις στ πολλαπλάσια. πάλιν, πε σάκις στ πολλαπλάσιον τ ΗΘ το ΕΒ κα τ Κ το Γ, σον δ τ ΕΒ τ Γ, σον ρα κα τ ΗΘ τ Κ. τ δ Κ το Μ ο κ στιν λαττον ο δ ρα τ ΗΘ το Μ λαττόν στιν. µε ζον δ τ ΖΗ το λον ρα τ ΖΘ συναµφοτέρων τ ν, Μ µε ζόν στιν. λλ συναµφότερα τ, Μ τ Ν στιν σα, πειδήπερ τ Μ το τριπλάσιόν στιν, συναµφότερα δ τ Μ, το στι τετραπλάσια, στι δ κα τ Ν το τετραπλάσιον συναµφότερα ρα τ Μ, τ Ν σα στίν. λλ τ ΖΘ τ ν Μ, µε ζόν στιν τ ΖΘ ρα το Ν περέχει τ δ Κ το Ν ο χ περέχει. καί στι τ µ νζθ,κ τ ν ΑΒ, Γ σάκις πολλα- 24

ELEMENTS BOOK 5 Proposition 8 A E B A E B C F G H C F G H K K D D L L M M N N For unequal magnitudes, the greater (magnitude) has a greater ratio than the lesser to the same (magnitude). And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater. Let AB and C be unequal magnitudes, and let AB be the greater (of the two), and D another random magnitude. I say that AB has a greater ratio to D than C (has) to D, and (that) D has a greater ratio to C than (it has) to AB. For since AB is greater than C, let BE be made equal to C. So, the lesser of AE and EB, being multiplied, will sometimes be greater than D [Def. 5.4]. First of all, let AE be less than EB, and let AE have been multiplied, and let FG be a multiple of it which (is) greater than D. And as many times as FG is (divisible) by AE, so many times let GH also have become (divisible) by EB, and K by C. And let the double multiple L of D have been taken, and the triple multiple M, and several more, (each increasing) in order by one, until the (multiple) taken becomes the first multiple of D (which is) greater than K. Let it have been taken, and let it also be the quadruple multiple N of D the first (multiple) greater than K. Therefore, since K is less than N first, K is thus not less than M. And since FG and GH are equal multiples of AE and EB (respectively), FG and FH are thus equal multiples of AE and AB (respectively) [Prop. 5.1]. And FG and K are equal multiples of AE and C (respectively). Thus, FH and K are equal multiples of AB and C (respectively). Thus, FH, K are equal multiples of AB, C. Again, since GH and K are equal multiples of EB and C, and EB (is) equal to C, GH (is) thus also equal to K. And K is not less than M. Thus, GH not less than M either. And FG (is) greater than D. Thus, the whole of FH is greater than D and M (added) together. But, D and M (added) together is equal to N, inasmuch as M is three times D, and M and D (added) together is four times D, and N is also four times D.Thus, M and D (added) together is equal to 25

ΣΤΟΙΧΕΙΩΝ ε η πλάσια, τ δ Ν το λλο, τυχεν, πολλαπλάσιον τ ΑΒ ρα πρ ς τ µείζονα λόγον χει περ τ Γ πρ ς τ. Λέγω δή, τι κα τ πρ ς τ Γ µείζονα λόγον χει περ τ πρ ς τ ΑΒ. Τ ν γ ρ α τ ν κατασκευασθέντων µοίως δείξοµεν, τι τ µ ν Ν το Κ περέχει, τ δ Ν το ΖΘ ο χ περέχει. καί στι τ µ ν Ν το πολλαπλάσιον, τ δ ΖΘ, Κ τ ν ΑΒ, Γ λλα, τυχεν, σάκις πολλαπλάσια τ ρα πρ ς τ Γ µείζονα λόγον χει περ τ πρ ς τ ΑΒ. Αλλ δ τ ΑΕ το ΕΒ µε ζον στω. τ δ λαττον τ ΕΒ πολλαπλασιαζόµενον σται ποτ το µε ζον. πεπολλαπλασιάσθω, κα στω τ ΗΘ πολλαπλάσιον µ ν το ΕΒ, µε ζον δ το κα σαπλασιόν στι τ ΗΘ το ΕΒ, τοσαυταπλάσιον γεγονέτω κα τ µ ν ΖΗ το ΑΕ, τ δ Κ το Γ. µοίως δ δείξοµεν, τι τ ΖΘ, Κ τ ν ΑΒ, Γ σάκις στ πολλαπλάσια κα ε λήφθω µοίως τ Ν πολλαπλάσιον µ ν το, πρώτως δ µε ζον το ΖΗ στε πάλιν τ ΖΗ το Μ ο κ στιν λασσον. µε ζον δ τ ΗΘ το λον ρα τ ΖΘ τ ν, Μ, τουτέστι το Ν, περέχει. τ δ Κ το Ν ο χ περέχει, πειδήπερ κα τ ΖΗ µε ζον ν το ΗΘ, τουτέστι το Κ, το Ν ο χ περέχει. κα σαύτως κατακολουθο ντες το ς πάνω περαίνοµεν τ ν πόδειξιν. Τ ν ρα νίσων µεγεθ ν τ µε ζον πρ ς τ α τ µείζονα λόγον χει περ τ λαττον κα τ α τ πρ ς τ λαττον µείζονα λόγον χει περ πρ ς τ µε ζον περ δει δε ξαι. 26

ELEMENTS BOOK 5 Proposition 8 N. But, FH is greater than M and D. Thus, FH exceeds N. And K does not exceed N. And FH, K are equal multiples of AB, C, and N another random multiple of D. Thus, AB has a greater ratio to D than C (has) to D [Def. 5.7]. So, I say that D also has a greater ratio to C than D (has) to AB. For, similarly, by the same construction, we can show that N exceeds K, and N does not exceed FH. And N is a multiple of D, and FH and K other random equal multiples of AB and C (respectively). Thus, D has a greater ratio to C than D (has) to AB [Def. 5.5]. And so let AE be greater than EB. So, the lesser EB, being multiplied, will sometimes be greater than D. Let it have been multiplied, and let GH be a multiple of EB (which is) greater than D. And as many times as GH is (divisible) by EB, so many times let FG also have become (divisible) by AE, and K by C. So, similarly (to the above), we can show that FH and K are equal multiples of AB and C (respectively). And, similarly (to the above), let the multiple N of D, (which is) the first (multiple) greater than FG, have been taken. So, FG is again not less than M. And GH (is) greater than D. Thus, the whole of FH exceeds D and M, that is to say N. And K does not exceed N, inasmuch as FG, which (is) greater than GH that is to say, K also does not exceed N. And, following the above (arguments), we (can) complete the proof in the same manner. Thus, for unequal magnitudes, the greater (magnitude) has a greater ratio than the lesser to the same (magnitude). And the latter (magnitude) has a greater ratio to the lesser (magnitude) than to the greater. (Which is) the very thing it was required to show. 27

ΣΤΟΙΧΕΙΩΝ ε Α Γ θ Β Τ πρ ς τ α τ τ ν α τ ν χοντα λ γον σα λλήλοις στίν κα πρ ς τ α τ τ ν α τ ν χει λόγον, κε να σα στίν. Εχέτω γ ρ κάτερον τ ν Α, Β πρ ς τ Γ τ ν α τ ν λόγον λέγω, τι σον στ τ Α τ Β. Ε γ ρ µή, ο κ ν κάτερον τ ν Α, Β πρ ς τ Γ τ ν α τ ν ε χε λόγον χει δέ σον ρα στ τ Α τ Β. Εχέτω δ πάλιν τ Γ πρ ς κάτερον τ ν Α, Β τ ν α τ ν λόγον λέγω, τι σον στ τ Α τ Β. Ε γ ρ µή, ο κ ν τ Γ πρ ς κάτερον τ ν Α, Β τ ν α τ ν ε χε λόγον χει δέ σον ρα στ τ Α τ Β. Τ ρα πρ ς τ α τ τ ν α τ ν χοντα λόγον σα λλήλοις στίν κα πρ ς τ α τ τ ν α τ ν χει λόγον, κε να σα στίν περ δει δε ξαι. 28

A C ELEMENTS BOOK 5 Proposition 9 (Magnitudes) having the same ratio to the same (magnitude) are equal to one another. And those (magnitudes) to which the same (magnitude) has the same ratio are equal. For let A and B each have the same ratio to C. I say that A is equal to B. For if not, A and B would not each have the same ratio to C [Prop. 5.8]. But they do. Thus, A is equal to B. So, again, let C have the same ratio to each of A and B. I say that A is equal to B. For if not, C would not have the same ratio to each of A and B [Prop. 5.8]. But it does. Thus, A is equal to B. Thus, (magnitudes) having the same ratio to the same (magnitude) are equal to one another. And those (magnitudes) to which the same (magnitude) has the same ratio are equal. (Which is) the very thing it was required to show. B 29

ΣΤΟΙΧΕΙΩΝ ε ι Α Γ Β Τ ν πρ ς τ α τ λόγον χόντων τ µείζονα λόγον χον κε νο µε ζόν στιν πρ ς δ τ α τ µείζονα λόγον χει, κε νο λαττόν στιν. Εχέτω γ ρ τ Α πρ ς τ Γ µείζονα λόγον περ τ Β πρ ς τ Γ λέγω, τι µε ζόν στι τ Α το Β. Ε γ ρ µή, τοι σον στ τ Α τ Β λασσον. σον µ ν ο ν ο κ στ τ Α τ Β κάτερον γ ρ ν τ ν Α, Β πρ ς τ Γ τ ν α τ ν ε χε λόγον. ο κ χει δέ ο κ ρα σον στ τ Α τ Β. ο δ µ ν λασσόν στι τ Α το Β τ Α γ ρ ν πρ ς τ Γ λάσσονα λόγον ε χεν περ τ Β πρ ς τ Γ. ο κ χει δέ ο κ ρα λασσόν στι τ Α το Β. δείχθη δ ο δ σον µε ζον ρα στ τ Α το Β. Εχέτω δ πάλιν τ Γ πρ ς τ Β µείζονα λόγον περ τ Γ πρ ς τ Α λέγω, τι λασσόν στι τ Β το Α. Ε γ ρ µή, τοι σον στ ν µε ζον. σον µ ν ο ν ο κ στι τ Β τ Α τ Γ γ ρ ν πρ ς κάτερον τ ν Α, Β τ ν α τ ν ε χε λόγον. ο κ χει δέ ο κ ρα σον στ τ Α τ Β. ο δ µ ν µε ζόν στι τ Β το Α τ Γ γ ρ ν πρ ς τ Β λάσσονα λόγον ε χεν περ πρ ς τ Α. ο κ χει δέ ο κ ρα µε ζον στι τ Β το Α. δείχθη δέ, τι ο δ σον λαττον ρα στ τ Β το Α. Τ ν ρα πρ ς τ α τ λόγον χόντων τ µείζονα λόγον χον µε ζόν στιν κα πρ ς τ α τ µείζονα λόγον χει, κε νο λαττόν στιν περ δει δε ξαι. 30

A ELEMENTS BOOK 5 C Proposition 10 For (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater. And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser. For let A have a greater ratio to C than B (has) to C. I say that A is greater than B. For if not, A is surely either equal to or less than B. In fact, A is not equal to B. For (then) A and B would each have the same ratio to C [Prop. 5.7]. But they do not. Thus, A is not equal to B. Neither, indeed, is A less than B. For (then) A would have a lesser ratio to C than B (has) to C [Prop. 5.8]. But it does not. Thus, A is not less than B. And it was shown not (to be) equal either. Thus, A is greater than B. So, again, let C have a greater ratio to B than C (has) to A. I say that B is less than A. For if not, (it is) surely either equal or greater. In fact, B is not equal to A. For (then) C would have the same ratio to each of A and B [Prop. 5.7]. But it does not. Thus, A is not equal to B. Neither, indeed, is B greater than A. For (then) C would have a lesser ratio to B than (it has) to A [Prop. 5.8]. But it does not. Thus, B is not greater than A. And it was shown that (it is) not equal (to A) either. Thus, B is less than A. Thus, for (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater. And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser. (Which is) the very thing it was required to show. B 31

ΣΤΟΙΧΕΙΩΝ ε ια Α Β Η Λ Γ Θ Μ Ε Ζ Κ Ν Ο τ α τ λόγ ο α το κα λλήλοις ε σ ν ο α τοί. Εστωσαν γ ρ ς µ ν τ Α πρ ς τ Β, ο τως τ Γ πρ ς τ, ς δ τ Γ πρ ς τ, ο τως τ Ε πρ ς τ Ζ λέγω, τι στ ν ς τ Α πρ ς τ Β, ο τως τ Ε πρ ς τ Ζ. Ε λήφθω γ ρ τ ν Α, Γ, Ε σάκις πολλαπλάσια τ Η, Θ, Κ, τ ν δ Β,, Ζ λλα, τυχεν, σάκις πολλαπλάσια τ Λ, Μ, Ν. Κα πεί στιν ς τ Α πρ ς τ Β, ο τως τ Γ πρ ς τ, κα ε ληπται τ ν µ ν Α, Γ σάκις πολλαπλάσια τ Η, Θ, τ ν δ Β, λλα, τυχεν, σάκις πολλαπλάσια τ Λ, Μ, ε ρα περέχει τ Η το Λ, περέχει κα τ Θ το Μ, κα ε σον στίν, σον, κα ε λλείπει, λλείπει. πάλιν, πεί στιν ς τ Γ πρ ς τ, ο τως τ Ε πρ ς τ Ζ, κα ε ληπται τ ν Γ, Ε σάκις πολλαπλάσια τ Θ, Κ, τ ν δ, Ζ λλα, τυχεν, σάκις πολλαπλάσια τ Μ, Ν, ε ρα περέχει τ Θ το Μ, περέχει κα τ Κ το Ν, κα ε σον, σον, κα ε λλατον, λαττον. λλ ε περε χε τ Θ το Μ, περε χε κα τ Η το Λ, κα ε σον, σον, κα ε λαττον, λαττον στε κα ε περέχει τ Η το Λ, περέχει κα τ Κ το Ν, κα ε σον, σον, κα ε λαττον, λαττον. καί στι τ µ ν Η, Κ τ ν Α, Ε σάκις πολλαπλάσια, τ δ Λ, Ν τ ν Β, Ζ λλα, τυχεν, σάκις πολλαπλάσια στιν ρα ς τ Α πρ ς τ Β, ο τως τ Ε πρ ς τ Ζ. Ο ρα τ α τ λόγ ο α το κα λλήλοις ε σ ν ο α τοί περ δει δε ξαι. 32

ELEMENTS BOOK 5 Proposition 11 27 A B G L C D H M E F K N (Ratios which are) the same with the same ratio are also the same with one another. For let it be that as A (is) to B, so C (is) to D, and as C (is) to D, so E (is) to F. I say that as A is to B, so E (is) to F. For let the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively). And since as A is to B, so C (is) to D, and the equal multiples G and H have been taken of A and C (respectively), and the other random equal multiples L and M of B and D (respectively), thus if G exceeds L then H also exceeds M, and if (G is) equal (to L then H is also) equal (to M), and if (G is) less (than L then H is also) less (than M) [Def. 5.5]. Again, since as C is to D, so E (is) to F, and the equal multiples H and K have been taken of C and E (respectively), and the other random equal multiples M and N of D and F (respectively), thus if H exceeds M then K also exceeds N, and if (H is) equal (to M then K is also) equal (to N), and if (H is) less (than M then K is also) less (than N) [Def. 5.5]. But if H was exceeding M then G was also exceeding L, and if (H was) equal (to M then G was also) equal (to L), and if (H was) less (than M then G was also) less (than L). And, hence, if G exceeds L then K also exceeds N, and if (G is) equal (to L then K is also) equal (to N), and if (G is) less (than L then K is also) less (than N). And G and K are equal multiples of A and E (respectively), and L and N other random equal multiples of B and F (respectively). Thus, as A is to B, so E (is) to F [Def. 5.5]. Thus, (ratios which are) the same with the same ratio are also the same with one another. (Which is) the very thing it was required to show. 27 In modern notation, this proposition reads that if α : β :: γ : δ and γ : δ :: ɛ : ζ then α : β :: ɛ : ζ. 33

ΣΤΟΙΧΕΙΩΝ ε ιβ Α Γ Ε Β Ζ Η Θ Κ Λ Μ Ν Ε ν ποσαο ν µεγέθη νάλογον, σται ς ν τ ν γουµένων πρ ς ν τ ν ποµένων, ο τως παντα τ γούµενα πρ ς παντα τ πόµενα. Εστωσαν ποσαο ν µεγέθη νάλογον τ Α, Β, Γ,, Ε, Ζ, ς τ Α πρ ς τ Β, ο τως τ Γ πρ ς τ, κα τ Ε πρ ς το Ζ λέγω, τι στ ν ς τ Α πρ ς τ Β, ο τως τ Α, Γ, Ε πρ ς τ Β,, Ζ. Ε λήφθω γ ρ τ ν µ ν Α, Γ, Ε σάκις πολλαπλάσια τ Η, Θ, Κ, τ ν δ Β,, Ζ λλα, τυχεν, σάκις πολλαπλάσια τ Λ, Μ, Ν. Κα πεί στιν ς τ Α πρ ς τ Β, ο τως τ Γ πρ ς τ, κα τ Ε πρ ς τ Ζ, κα ε ληπται τ ν µ ν Α, Γ, Ε σάκις πολλαπλάσια τ Η, Θ, Κ τ ν δ Β,, Ζ λλα, τυχεν, σάκις πολλαπλάσια τ Λ, Μ, Ν, ε ρα περέχει τ Η το Λ, περέχει κα τ Θ το Μ, κα τ Κ το Ν, κα ε σον, σον, κα ε λαττον, λαττον. στε κα ε περέχει τ Η το Λ, περέχει κα τ Η, Θ, Κ τ ν Λ, Μ, Ν, κα ε σον, σα, κα ε λαττον, λαττονα. καί στι τ µ ν Η κα τ Η, Θ, Κ το Α κα τ ν Α, Γ, Ε σάκις πολλαπλάσια, πειδήπερ ν ποσαο ν µεγέθη ποσωνο ν µεγεθ ν σων τ πλ θος καστον κάστου σάκις πολλαπλάσιον, σαπλάσιόν στιν ν τ ν µεγεθ ν νός, τοσαυταπλάσια σται κα τ πάντα τ ν πάντων. δι τ α τ δ κα τ Λ κα τ Λ, Μ, Ν το Β κα τ ν Β,, Ζ σάκις στ πολλαπλάσια στιν ρα ς τ Α πρ ς τ Β, ο τως τ Α, Γ, Ε πρ ς τ Β,, Ζ. Ε ν ρα ποσαο ν µεγέθη νάλογον, σται ς ν τ ν γουµένων πρ ς ν τ ν ποµένων, ο τως παντα τ γούµενα πρ ς παντα τ πόµενα περ δει δε ξαι. 34

ELEMENTS BOOK 5 Proposition 12 28 A C E B D F G H K If there are any number of magnitudes whatsoever (which are) proportional then as one of the leading (magnitudes is) to one of the following, so will all of the leading (magnitudes) be to all of the following. Let there be any number of magnitudes whatsoever, A, B, C, D, E, F, (which are) proportional, (so that) as A (is) to B, so C (is) to D, and E to F. I say that as A is to B, so A, C, E (are) to B, D, F. For let the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively). And since as A is to B, so C (is) to D, and E to F, and the equal multiples G, H, K have been taken of A, C, E (respectively), and the other random equal multiples L, M, N of B, D, F (respectively), thus if G exceeds L then H also exceeds M, and K (exceeds) N, and if (G is) equal (to L then H is also) equal (to M, and K to N), and if (G is) less (than L then H is also) less (than M, and K than N) [Def. 5.5]. And, hence, if G exceeds L then G, H, K also exceed L, M, N, and if (G is) equal (to L then G, H, K are also) equal (to L, M, N) and if (G is) less (than L then G, H, K are also) less (than L, M, N). And G and G, H, K are equal multiples of A and A, C, E (respectively), inasmuch as if there are any number of magnitudes whatsoever (which are) equal multiples, respectively, of some (other) magnitudes, of equal number (to them), then as many times as one of the (first) magnitudes is (divisible) by one (of the second), so many times will all (of the first magnitudes) also (be divisible) by all (of the second) [Prop. 5.1]. So, for the same (reasons), L and L, M, N are also equal multiples of B and B, D, F (respectively). Thus, as A is to B, so A, C, E (are) to B, D, F (respectively). Thus, if there are any number of magnitudes whatsoever (which are) proportional then as one of the leading (magnitudes is) to one of the following, so will all of the leading (magnitudes) be to all of the following. (Which is) the very thing it was required to show. 28 In modern notation, this proposition reads that if α : α :: β : β :: γ : γ etc. then α : α :: (α + β + γ + ) : (α + β + γ + ). 35 L M N