Euclid, Elements X 9 (with scholia)©
translated by Henry Mendell (Cal. State U., L.A.)

Go to prop. 8

Prop. 9. Squares from straight-lines commensurable in length have a ratio to one another that a square arithmos has to a square arithmos; and squares having to one another a ratio that a square arithmos has to a square arithmos also have their sides commensurable in length. But squares from straight-lines incommensurable in length do not have a ratio to one another which a square arithmos has to a square arithmos; and squares not having a ratio to one another that a square arithmos has to a square arithmos will not have their sides commensurable in length either.

Corollary: Possible combinations of commensurable in length, incommensurable in length, commensurable in power, incommensurable in power, etc.

Alternative proof:

Scholia

The lemma associated with this theorem is placed with Theorem 10 (generally regarded as a later addition), for which it is useful. (diagram 1) For let A, B be commensurable in length. I say that the square from A has a ratio to the square from B that a square arithmos has to a square arithmos. For since A is commensurable with B in length, therefore, A has a ratio to B that an arithmos has to an arithmos. (diagram 2) Let it have what G has to D. And so, since it is: as A to B, so G to D, (diagram 3) but the ratio of the square from A to the square from B is duplicate the ratio of A to B. For similar figures are in duplicate ratio of their corresponding sides [VI 20 cor.]. (anachronistic diagram 4) But the ratio the square from G to the square from D is duplicate the ratio of arithmos G to arithmos D. For there is one mean in-ratio arithmos of two square arithmoi, and the square arithmos to the square arithmos has ratio duplicate of the side to the side. Therefore, it is: as the square from A to the square from B, so the square arithmos from G to the square arithmos from arithmos D.

(anachronistic diagram 4) But, in fact, let it be as the square from A to that from B so the square arithmos from G to the square from D. I say that A is commensurable with B in length. For since it is: as the square from A to the square from B, so the square arithmos from G to the square from D, but the ratio of the square from A to the square from B is duplicate that from A to that from B, while the ratio of the square from arithmos G to the square arithmos from arithmos D is duplicate the ratio of arithmos G to arithmos D, therefore it is also: as A to B, so arithmos G to arithmos D. Therefore, A has a ratio to B which an arithmos, G, arithmos to an arithmos, D. Therefore, A is commensurable with B in length.

(diagram 1) But, let A, in fact, be incommensurable with B in length. I say that the square from A to the square from B does not have a ratio that a square arithmos has to a square arithmos. (diagram 4) For if the square from A has a ratio to the square from B that a square arithmos has to a square arithmos, A will be commensurable with B. But it is not. Therefore, the square from A does not have a ratio to the square from B that a square arithmos has to a square arithmos.

(diagram 4) Again, in fact, let the square from A not have a ratio to the square from B that a square arithmos has to a square arithmos. I say that A is incommensurable with B in length. For if A is commensurable with B, that from A will have a ratio to that from B that a square arithmos has to a square arithmos. But it does not have it. Therefore, A is not commensurable with B in length.

Therefore, those from commensurables in length, etc

Provided (Corollary): And it is obvious from what have been shown that lines commensurable in length altogether are also commensurable in power, while those commensurable in power are not altogether also commensurable in length, if squares from straight-lines commensurable in length have a ratio that a square arithmos has to a square arithmos, but things having a ratio that an arithmos has to an arithmos are commensurable. Thus, straight-lines commensurable in length are not merely commensurable in length, but also in power.

Again, since it was shown that as many squares as have a ratio to one another that a square arithmos has to a square arithmos, are commensurable in length and commensurable in power by their squares having a ratio which an arithmos has to an arithmos, therefore, as many squares as do not have a ratio that a square arithmos has to a square arithmos, but simply that an arithmos has to an arithmos, their squares will themselves be commensurable in power, but no longer also in length. Thus magnitudes commensurable in length are altogether also commensurable in power, while those commensurable in power are not altogether also commensurable in length, unless they should also have a ratio that a square arithmos has to a square arithmos.

I say, in fact, that lines incommensurable in length are also not altogether also incommensurable in power, since, in fact, lines commensurable in power are able not to have a ratio that a square arithmos has has to a square arithmos, and due to this while being commensurable in power they are incommensurable in length. Thus, lines incommensurable in length are not altogether also incommensurable in power, but while being incommensurable in length, they can be both incommensurable and commensurable in power.

But lines incommensurable in power are also incommensurable in length. For if they are commensurable in length, they will also be commensurable in power. But that were also supposed as incommensuravble, which is absurd. Therefore, lines incommensurable in power are altogether also incommensurable in length.

Differently the 9th
(general diagram) (diagram 1) For since A is commensurable with B, it has a ratio to it that an arithmos that an arithmos has to an arithmos. (diagram 2) Let it have what G has to D. (diagram 3) And let G, by multiplying itself, make E, let G, by multiplying D, make Z, and D, by multiplying itself, make H. And so, since G by multiplying itself, made E, while D by multiplying itself, made Z, therefore, it is: as G to D, that is, as A to B, so E to Z. (genreral diagram = diagram 4) But as A to B, so the square from A to that enclosed by A, B. (anachronistic diagram 5) Therefore, it is: as the square from A to that enclosed by A, B, so E to Z. Again, since D, by multiplying itself, made H, while G, by multiplying D, made Z, therefore, it is: as G to D, that is, as A to B, so Z to H. But as A to B, so that enclosed by A, B to that from B. Therefore, it is: as that enclosed by A, B to that from B, so Z to H. But as that from A to that enclosed by A, B, so was E to Z. Therefore, ex aequali, as that from A to that from B, so E to H. But each of E, H is a square. For E is from G, while H that is from D. Therefore, that from A has a ratio to that from B what a square arithmos has to a square arithmos, just what it was required to show.

(diagram 6) But, in fact, let the square from A have to the square from B a ratio that a square arithmos, E, has to a square arithmos H. I say that A is commensurable with B. (diagram 7) For let there be a side of E, namely G, and of H, namely D, (diagram 8) and let G, by multiplying D, make Z. Therefore E, Z, H are successively in-ratio in the ratio of G to D [VIII 11].  (general diagram = diagram 8 or anachronistic diagram 5) And since that enclosed by A, B is a mean in-ratio of the squares from A, B, but Z is a mean in-ratio of E, H, therefore, it is: as the square from A to that enclosed by A, B, so E to Z. But as that enclosed by A, B to that from B, so Z to H, but as that from A to that enclosed by A, B, so A to B. Therefore, A, B are commensurable. For they have a ratio that an arithmos, E, has to an arithmos, Z, that is what G has to D. For as G to D, E to Z. For G by multiplying itself made E and by multiplying D made Z. Therefore, it is: as G to D, E to Z.

Analysis:
.1 Let A : B = G : D, where G, D are arithmoi
2. Let G * G = E
3. Let D * G = Z
4. Let D * D = H
5. Hence, A : B = G : D = E : Z = Z : H (line 1, x : y = x2 : xy = xy : y2)
6. and T(A) : O(A, B) = A : B = E : Z (lines 1, 5, similar reason)
7. O(A, B) : T(B) = A : B = Z : H (5, similar reason)
8. Ex aequali, T(A) : T(B) = E : H = E : H (lines 6, 7)

1. Let T(A) : T(B) = E : H, with E, H square arithmoi
2. Let G * G = E
3. Let D * D = H
4. Let D * G = Z
5. Hence, E : Z = Z : H = G : D ((lines 2, 3, 4 and x : y = x2 : xy = xy : y2)
6. Hence, T(A) : O(A,B) = O(A,B) : T(B) = A : B ((line 1, x : y = x2 : xy = xy : y)
7. Hence, T(A) : O(A,B) = E : Z (see below)
8. O(A,B) : T(B) = Z : H (see below)
9. Hence, A : B = E : Z = G : D (6, 7, 5)

The justification for 7, 8 will be the same. But the reasoning is obscure, to say the least. Cf. the discussion in Heath, on VI 22 and Data 24: If three straight-lines are in-ratio, but the first to the third has a given ratio, it to the second also will have a given ratio. So, if A : C is given, it follows that there is only one ratio, A : B = B : C. We also know the same for arithmoi (VIII 11), if p2, r2 are given, there is one mean, p*r. What we don't know is: if A : C = p2 : r2, whether p2 : r*p, that is, p : r = A : B. For this one needs a sort of proof that relies on the definition of same ratio, such as Heath provides, iwith a proviso that one can always construct the fourth proportional to a line.

Scholia (numbering from Heiberg, Euclidis opera, vol. 5)

61. He shows here that commensurables in length are also commensurable in power.

62. This theorem is a Theaetetan discovery, and Plato mentions it in the Theaetetus, but there it is displayed more partially, while here universally. For there he says that the squares measured by square arithmoi also have their sides commensurable, but this proposition is partial. For it does not encompass all commensurable areas whose sides are also commensurable. For among square commensurable areas the sides of 18 and of 8, unless they are also found according to the measure of arithmoi, are at all events otherwise commensurable [note: at 147D3-148AB2, Theaetetus is only concerned with the square feet of the square and whether the side is commensurable with the side of a unit squares]. Nevertheless, the areas have not been measured by square arithmoi, although it is possible for them to be measured. And so, he reasonably did not defined it here in this manner, but says they have a ratio that a square arithmos has to a square arithmos, but the mention of square arithmos did not here come about vacuously. For if he had merely defined it as 'what an arithmos has to an arithmos', the definition1 would have over-extended. For it would have required that squares having a duplicate ratio to each other have their sides commensurable. But they do not. For the side of the larger square is also a diagonal of the smaller parallogram.2 If then through saying 'that an arithmos has to an arithmos' the term would have over-extended by encompassing even those having non-commensurable sides, while through saying 'measured by square arithmoi' it would have been deficient in not containing those that have commensurable sides as they were were not measured by square arithmoi, but with the arithmoi having a ratio that a square has to a square arithmos, 'that a square has to a square' is reasonably added. For it encompasses all areas, those that, although they are not measured by squares, at all events being commensurable also have their sides commensurable. Anyhow, with 18 and 8 being commensurable, due to the fact that they are inscribed up also from commensurable sides, you will find the sides commensurable because they have a ratio that a square arithmos has to a square arithmos. For as 8 to 4 so 18 to 9. But taking the sides of 9 and of 4 divide equal-times the sides of the displayed squares and have commensurability. For as the squares to the squares, so the sides to the sides.

1. The word horos means boundary, whence it means 'definition'. It can mean rule, but the language here is of a definition of 'commensurable'. It is possible that the scholiast is thinking of demonstrated definitions in Aristotle's sense of revealing the essence.

2. Mss: καὶ γὰρ τῆς τοῦ παράλλης διαγώνιός ἐστιν is difficult to make sense of. I propose reading: καὶ γὰρ τοῦ ἥττονος παραλληλογράμμου διαγώνιός ἐστιν. This is still awkward, yet possible on the assumption that the original uncial manuscript used extensive abbreviation, which was elsewhere clear and from which a misreadind could occur, e.g., Τ˅ ΗτΤˆˊ ΠΑΡΑΛΛΗ. Cf. Acerbi, Il Silenzio delle Sirene (Roma, 2010), 340, for the issue, and the Bobbio math. frag. (Ambros. L 99) in Heiberg, Math. Graeci Minores, 90-91, and Wattenbach, Scriptorae graecae specimina (1883), Table 8, for the abbreviations.

63. Squares from straight-lines commensurable in length have to one another a ratio that a square arithmos has to a square arithmos. The mention of square arithmos did not here come about vacuously. For if he had merely said 'that an arithmos has to an arithmos, the definition would have over-extended. For squares having duplicate ratio would have had their sides commensurable. But they do not, as it holds in the case of the diameter and the side.

64. One must know that squares from straight-lines commensurable in length have a ratio that a square arithmos has to a square arithmos, nevertheless it also converts, so that if the squares have a ratio that a square arithmos has to a square arithmos, the straight-lines empowering the squares are also commensurable in length. For 18 to 8 has a square-causing ratio double-sesquiquartan, which is square 9 to square 4.3 And yet the side of 8 is not commensurable in length with the side of 18. Buτ the side of one is 2;49,42 while that of 18 is 4;14,33.4 65. E.g., on a diagram let there be commensurable straight-lines having 6 and 4 hand-spans. And the squares from them, 36, 16, have a ratio to one another that a square arithmos, 9, has to 4. For arithmos 9 has a ratio to 4 double-sesquiquartan, just as 36 to 16. (In Vat. gr. 190 (P) 139r, the diagram, on which this diagram is based, is the only one near scholion 65, but it would appear to go with scholion 64)

66. A square arithmos to another square arithmos is said to have a ratio when their sides by being multiplied with respect to one another make another arithmos that's mean in-ratio. e.g., the square-making sides of 16 and 36 are 4 and 6, which, by being multiplied with respect to one another come to be24, a mean in-ratio of 16 and 36. For 36 to 24 has a sesquialter (half-again) ratio and 24 to 15 has a sesquialter (half-again) ratio. And so, the sides had a sesquialter ratio to another, but 36 and 24 and 15 have a ratio 2-times sesquialter.

67. Let A be four-foot, B six-foot, and the squares from them sixteen-foot and 36 feet. And so, it is clear that the four-foot is commensurable with the six-foot in length. But also it is not unclear that 36 to 16 has a ratio that square 9 has to square 4. For the ratios, both these and those, are double-sesquiquartan.3

68. One must further understand: therefore, that from A to that from B also has a ratio duplicate the ratio that G has to D. But ratios that are the same as the same ratio are also the same as one another. (v 11)

69. Due to the provided (corollary) of 20 of book 6 that from A to that from B also has a ratio duplicate of the ratio that G has to D.

70. (on: but the ratio of the square from A to the square from B is duplicate the ratio of A to B) Or rather of the duplicate ratio that 16 has to 8, that of the square from A to the square from B. is duplicate. For 256 to 64 is quadruplicate and has the ratio that A has to B or 16 to 8 twice. For twice thw duplicate is quadruplicate. Thus, the ratio of the square from A to the square from B is duplicate or twice twice the ratio that A has to B, 16 to 8, just what he clarified when he said, "For similar figures are in duplicate ratio of their corresponding sides," just what was shown in the 11th theorem of the 6th book.

71. (on: but the ratio of the square from A to the square from B is duplicate the ratio of A to B) Therefore, I say, the ratio from G to the square from A is also duplicate the ratio of A to B. For quals have the same ratio to the same. (v 7) For the ratios are both the same and equal.

72. What's being said can also be of this sort: lines commensurable in power, if they have a ratio that a square has to a square arithmos will also be commensurable in length, but if they don't have it, they will be commensurable in power but not in length.

73. For example, 5 and 7, while being commensurable, are also in power. For 25 and 49 are not measured by a common measure.

74. For example 12 and 16 are commensurable in length but also in power. For the squares from them 144 and 256 are measured by the same area 4.

75. For example, 5 and 15 are commensurable in pare. For the squares from them, 25 and 225 are measured by the same area. But 5 and 15 are incommensurable in length. For they do not have a ratio that a square arithmos has to a square arithmos. For 15 is triple 5, and this is not found a square arithmos to a square arithmos having the same ratio. For example, 16 and 25 are arithmoi that have a ratio that arithmos 9 has to arithmos 4, the double-sesquiquartan.3

76. For example, 25 and 225 are arthmoi that do not have a ratio that a square arithmos has to a square arithmos, but simply, what an arithmos has to an arithmos. And so, they are commensurable in power but no longer also in length. For their sides, 5 and 15 do not have a ratio that a square arithmos has to a square arithmos. (It is difficult to know where the confusion is in some of these examples, except that they get repeated.)

77. Differently. For example 30 and 60. For 60 to 30 does not have a ratio that a square arithmos has to a square arithmos, but simply that an arithmos has to an arithmos. And so, they are commensurable in power, but no longer also in length. For their sides, 5 and 15, do not have a ratio that a square arithmos has to a square arithmos.

78. The clause, "unless they should also have a ratio that a square arithmos has to a square arithmos," was not said the sides, but about the squares. For there's no necessity that sides commensurable in length have a ratio that a square arithmos has to a square, but merely what an arithmos has to an arithmos. But it is one thing "what an arithmos has to an arithmos" and another "what a square arithmos has to a square arithmos." For things having a ratio that a square arithmos has to a square arithmos necessarily also have what an arithmos has to an arithmos, but there's no necessity that things having a ratio that an arithmos has to an arithmos also have a ratio that a square arithmos has to a square arithmos. For arithmos extends beyond square arithmos. Thus. if squares of certain straight-lines have a ratio that a square arithmos has to a square, those straight-lines will necessarily be commensurable in length; however, there is no necessity that those lines also have a ratio that a square arithmos has to a square, but it is possible both that they have and that they not have it.

79. For example, 5 and 7 being incommensurable in length are also in power. For the squares from them, 25 and 49 are not measured by a common area. [strangely in error. It is correct that 25 and 49 are relatively prime, so that they are not measured by a common arithmos (with the unit not being an arithmos), but these are areas, not arithmoi.]

80. One must know that whenever the sides of squares have a ratio to one another that a square arithmos has to a square arithmos, that is the double in length, then the square to the square is also quadruplicate, as in the case of 4 and 16 and 9 and 36. For 2 is a side of 4, but 4 of 16, and 3 of 9, 6 of 36. And so such sides are in duplicate ratio, that is in a ratio of a square arithmos to a square arithmos, and for this reason the square areass that come-about from them are observed in quadruplicate ratio according to the axiom (i.e., principle) that states that the doubles in length are quadruple in power. But if the side has a certain ratio to the side, half-again as it were, or third-again, or any other of the epimorics (i.e., superparticulars, n+1 : n, cf. Nicomachus, Arith. i 19) or of the epimerics (i.e., superpartients, n+m : n for 1 < m < n, cf. Nicomachus, Arith. i 20), some squares that come-about from them have a ratio to one another that a square arithmos has to a square arithmos, yet not the quadrupicate, as in the case of 9 and 4, whose sides have a ratio that an arithmos has to an arithmos, without being a square to a square. For two and three, which are sides of 4 and 9 have half-again ratio. Wherefore 9 cannot also be quadruplicate of 4, as 16 is of 4 and 36 of 9.

Scholia on the alternative proof.

84. Is someone should say, from where is it clear that as A to B so that from A to that by A, B, we will speak in this way: let there be positioned straight-lines AB so that they are in a straight-line, and there be AB, BG, and let there be described up from AB a square, AD, and let parallelogram AZ be filled in. And since BZ is that by AB, BG (for BD is equal to AB), and BD is the common height of AD, BZ, therefore it is: as AB to BG so that from AB to that by AB, BG, as he will also show through the lemma in 21. (The diagram for scholion 84 below is based on Vind. gr. 031 131v) <

85. For through this theorem it is shown that if there are two straight-lines, it is: as one of these to the remaining, so the square from that to the rectangle by this and the remaining line. For both are parallelograms and equiangular, and the ratio of the sides, being composed, remains the same as the initial ratio due to the fact that both in the case of the square the same side is taken twice and in the case of the rectangle the same line is taken one, e. let there be two lines, A of 4 cubits and B of 2 cubits. The square from A, being equiangular with the parallelgoram by A, B, has a ratio to that composed from the sides. But that composed from the ratios of 4 to 4 and 4 to 2 is the initial ratio of 4 to 2.

The 23 of the 6th book is that the equiangular parallelograms have a ratio that is composed from the sides.

86 For since it is: as that from A to that by A, B, so A to B, but as A to B, so that by A, B to that from B, therefore as that from A to that by A, B, so too that by A, B to that from B. For ratios that are the same as the same ratio are also the same as one another. (v 11)

3. The epimoric or superparticular or n-again is 1 + 1/n = (n+1)/n or rather n+1 : n, so that the double-epirmoric is 2 + 1/n = (2*n + 1)/n or rather 2*n + 1 : n. In general, the m-tuple-epimoric is m*n + 1 : n, or more generally for p ≥ 1, m*(p*n) + p : p*n, whence 9 : 4 and 36 : 16 will be double-sesquiquartan (double fourth-again). Cf. Nicomachus, Arithmetica i 22.
4. It is easy to verify that these two numbers are sexigessimal in format. The square of 2;49,42 is 7;59,58,5,24, with the square of 2;49,42 being 8;0,3,44,49 (or less close to 8) while the square of 4;14,33 is 17;59,55,42,9, with that of 4;14,34 being 18;0,4,11,16 (or less close to 18). However, even taking these approximations, 4;14,33 / 2;49,42 = 1;30 = 3/2, as is √18 / √8. So the scholion is in error.