translated by Henry Mendell (Cal. State U., L.A.)

Appendix Prop. 27: Let it be proposed by us to show that in the case of square figures the diameter is incommensurable in length with the side. (diagram 1) Let there be a square, ABGD, and its diameter, AG. I say that GA is incommensurable in length with AB.
For if it is possible, let it be commensurable. I say that it will follow that the same arithmos is even and odd. And so it is obvious that the square from AG is double the square from AB [I 47]. And since GA is commensurable with AB, therefore, GA has a ratio to AB that an arithmos has to an arithmos [X 6]. (diagram 2) Let it have what EZ has to H, and let EZ, H be smallest of those having the same ratio as them. Therefore, EZ is not a unit.1 For if EZ will be a unit, it has a ratio to H that AG has to AB, and AG is larger than AB. Therefore, EZ is also larger than arithmos H [V 14], which is absurd. Therefore, EZ is not a unit. Therefore, it is an arithmos. And since it is: as GA to AB, so EZ to H, therefore it is: as the square from GA to the square from AB so the square from EZ to the square from H [VI 20 cor.; VIII 11]. But the square from GA is double that from AB. Therefore, the square from EZ is also double that from H. Therefore, the square from EZ is even. Thus too EZ itself is even. For if it were odd, then the square from it would be odd, since, in fact, if odd numbers are composed however-many-times, but the plethos of them is odd, the whole is odd [IX 23]. Therefore, EZ is even. (diagram 3 = general diagram) Let it be bisected at Q. And since EZ, H are smallest of those having the same ratio as them, they are prime relative to one another [VII 21]. And EZ is even. Therefore, H is odd. For if it were even, two would measure EZ, H (for every even arithmos has a half part VII def. 6), although they are prime relative to one another, which is impossible. Therefore, H is not even. Therefore, it is odd. And since EZ is double EQ, therefore, the square from EZ is four-times that from EQ [VIII 11]. But, the square from EZ is double that from H. Therefore, the square from H is double that from EQ. Therefore, the square from H is even. Therefore, H is for the mentioned reasons. But it is also odd, which is impossible. Therefore, GA is not commensurable in length with AB, just what it was required to show.

1. Euclid (or the author) needs to show that EZ cannot be a unit, because the unit is neither odd nor even, as only arithmoi (i.e., pluralities of units) are and the unit is not an arithmos.

Analysis:
Note that T(x) is the square on a line x.
T(AG) = 2 T(AB)
GA and AB are commensurate ⇒ there are arithmoi n, m: GA : AB = n : m.
Let GA : AB = EZ : H (arithmoi in smallest ratio and hence relatively prime)
Possibilities:
1. EZ is a unit
2. EZ odd
3. EZ is even

1. EZ is a unit ⇒ EZ : H = AG : AB & AG > AB ⇒ EZ > H, so that a unit is larger than a unit or number.
2. EZ is odd and AG : AB = EZ : H and T(AG) = 2 T(AB) ⇒ T(AG) = T(AB) = EZ2 : H2 ⇒ EZ2 = 2 H2 ⇒ EZ2 is even ⇒ EZ is even (since EZ is odd ⇒ EZ2 = EZ + ... + EZ odd times = an odd arithmos).
3. EZ is even so that there is a half of it, EQ, and EZ2 = 2 H2 ⇒ 2 EQ2 = H2 ⇒ H2 is even ⇒ H is even (see the argument against 2) or a unit. It cannot be even since, again, 2 would measure EZ and H, and (Euclid seems not to consider this) it cannot be a unit since H is larger than EQ (see the argument against 1).

Differently

One must also show differently that the diameter of a square is incommensurable with the side. (diagram 1) Let there be instead of the diameter A, and instead of the side B. I say that A is incommensurable in length with B.
For if it is possible, let it be commensurable; (diagram 1 = general diagram) and let it come-about again as A to B so arithmos EZ to H [X 6], and let EZ, H be smallest of those having the same ratio as them [VII 33]. Therefore, EZ, H are prime relative to one another [VII 21]. I say first that H is not a unit. For if it is possible, let it be a unit. And since it is: as A to B so EZ to H, therefore also as the square from A to that from B so the square from EZ to that from H [VI 20 cor.; VIII 11]. But the square from A is double that from B [I 47]. Therefore, the square from EZ is also double that from H. And H is a unit. Therefore, the square from EZ is two, which is impossible. Therefore H is not a unit. Therefore, it is an arithmos. And since it is: as the square from A to that from B so the square from EZ to that from H, and inversely as the square from B to that from A so the square from H to that from EZ [V 7], but the square from B measures that from A, therefore also the square from H measures that from EZ. Thus the side itself, H, also measures EZ. But H also measures itself. Therefore, H measures EZ, H, although they are prime relative to one another, which is impossible. Therefore, A is not commensurable in length with B. Therefore, it is incommensurable, just what it was required to show.

Analysis:
Note that T(x) is the square on a line x.
and that Meas(x,y) is: x measures y
, where A is the side and B the diagonal.
A and B are commensurate ⇒ there are arithmoi n, m: A : B = n : m.
Let A : B = EZ : H (arithmoi in smallest ratio and hence relatively prime)
T(B) = 2 T(A) ⇒ EZ2 = 2 H2
Possibilities: H is unit or H is an arithmos:
H is a unit ⇒ 2 H2 = 2 12 = EZ2 = 2
and H2 = 2 ⇒ 1 < H < 2, which is impossible.

H is an arithmos and 2 H2 = EZ2 ⇒ Meas(H2,EZ2) ⇒ Meas(H,EZ) and Meas(H,H) ⇒ H and EZ are not relatively prime, which is impossible
Lemma: Meas(m2,n2) ⇒ Meas(m,,n)
Meas(m2,n2) ⇒ there is a p, s.t. n2 = p m2 ⇒ p is a square arithmos (otherwise, p m2 won't be a square arithmos and n2 won't be a square arithmos, either) ⇒ there is an arithmos k, where k2 = p ⇒ n2 = k2 m2 = (k m)2 ⇒ n = k m ⇒ Meas(m,n).