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(vi. 1] [x. 11]

And, as BD is to BC, so is DA to AC.
Therefore DA is commensurable with AC.

But DA is rational ; therefore AC is also rational.

Therefore etc.

[x. Def. 4)

There is a difficulty in the text of the enunciation of this proposition. The Greek runs το υπό ρητων μήκει συμμέτρων κατά τινα των προειρημένων τρόπων ευθειών περιεχόμενον ορθογώνιον ρητόν έστιν, where the rectangle is said to be contained by “rational straight lines commensurable in length in any of the aforesaid ways.” Now straight lines can only be commensurable in length in one way, the degrees of commensurability being commensurability in length and commensurability in square only. But a straight line may be rational in two ways in relation to a given rational straight line, since it may be either commensurable in length, or commensurable in square only, with the latter. Hence Billingsley takes κατά τινα των προειρημένων τρόπων with ρητων, translating “straight lines commensurable in length and rational in any of the aforesaid ways,” and this agrees with the expression in the next proposition “ a straight line once more rational in any of the aforesaid ways”; but the order of words in the Greek seems to be fatal to this way of translating the passage.

The best solution of the difficulty seems to be to reject the words “in any of the aforesaid ways" altogether. They have reference to the Lemma which immediately precedes and which is itself open to the gravest suspicion. It is very prolix, and cannot be called necessary; it appears moreover in connexion with an addition clearly spurious and therefore relegated by Heiberg to the Appendix. The addition does not even pretend to be "Euclid's, for it begins with the words “for he calls rational straight lines those....” Hence we should no doubt relegate the Lemma itself to the Appendix. August does so and leaves out the suspected words in the enunciation, as I have done.

Exactly the same arguments apply to the Lemma added (without the heading “ Lemma ") to x. 23 and the same words “in any of the aforesaid ways ” used with “redial straight lines commensurable in length” in the enunciation of x. 24. The said Lemma must stand or fall with that now in question, since it refers to it in terms: “And in the same way as was explained in the case of rationals...."

Hence I have bracketed the Lemma added to x. 23 and left out the objectionable words in the enunciation of x. 24. If p be one of the given rational straight lines (rational of course in the

p sense of x. Def. 3), the other can be denoted by kp, where k is, as usual, of the form min (where m, n are integers). Thus the rectangle is kp”, which is obviously rational since it is commensurable with p. (x. Def

. 4.] A rational rectangle may have any of the forms ab, ka, kA or A, where a, b are commensurable with the unit of length, and A with the unit of area.

Since Euclid is not able to use kp as a symbol for a straight line commensurable in length with p, he has to put his proof in a form corresponding to

: kp = p : kp, whence, p, kp being commensurable, p, kp' are so also.

[x. 1] PROPOSITION 20.

D

А

If a rational area be applied to a rational straight line, it produces as breadth a straight line rational and commensurable in length with the straight line to which it is applied.

For let the rational area AC be applied to AB, a straight line once more rational in any of the aforesaid ways, producing BC as breadth ; I say that BC is rational and commensurable in length with BA. For on AB let the square AD be described ; therefore AD is rational.

[x. Def. 4] But AC is also rational ; therefore DA is commensurable with AC. And, as DA is to AC, so is DB to BC.

[vi. 1] Therefore DB is also commensurable with BC;

BC:

(x. 11] and DB is equal to BA; therefore AB is also commensurable with BC.

But AB is rational ; therefore BC is also rational and commensurable in length with AB.

Therefore etc.

с

The converse of the last. If p is a rational straight line, any rational area is of the form kp. If this be “applied” to p, the breadth is kp commensurable in length with

р

and therefore rational. We should reach the same result if we applied the area to another rational straight line o. The breadth is then kp2 kp?

ko or k'o, say. o?

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m =

n

PROPOSITION 21.

The rectangle contained by rational straight lines commensurable in square only is irrational, and the side of the square equal to it is irrational. Let the latter be called mediai.

For let the rectangle AC be contained by the rational straight lines AB, BC commensurable in square only ;

H. E. III.

4

B

A

с

I say that AC is irrational, and the side of the square equal to it is irrational ; and let the latter be called medial.

D For on AB let the square AD be described ; therefore AD is rational.

[x. Def. 4] And, since AB is incommensurable in length with BC, for by hypothesis they are commensurable in square only, while AB is equal to BD, therefore DB is also incommensurable in length with BC. And, as DB is to BC, so is AD to AC;

[vi. 1] therefore DA is incommensurable with AC.

[x. 11) But DA is rational ; therefore AC is irrational, so that the side of the square equal to AC is also irrational.

[x. Def. 4) And let the latter be called medial.

Q. E. D. A medial straight line, now defined for the first time, is so called because it is a mean proportional between two rational straight lines commensurable in square only. Such straight lines can be denoted by p, pk. A medial straight line is therefore of the form Up Ik or kip. Euclid's proof that this is irrational is equivalent to the following. Take p, pVk commensurable in square only, so that they are incommensurable in length. Now

P:PVk=po: pak, whence [x. ] p Jk is incommensurable with på and therefore irrational [x. Def. 4], so that IpIk is also irrational (ibid.).

A medial straight line may evidently take either of the forms va JB or VAB, where of course B is not of the form kA.

,

F

E

G

LEMMA. If there be two straight lines, then, as the first is to the second, so is the square on the first to the rectangle contained by the two straight lines.

Let FE, EG be two straight lines.

I
say

that, as FE is to EG, so is the square on FE to the rectangle FE, EG.

D

For on FE let the square DF be described, and let GD be completed.

Since then, as FE is to EG, so is FD to DG, (vi. 1] and FD is the square on FE, and DG the rectangle DE, EG, that is, the rectangle FE, EG, therefore, as FE is to EG, so is the square on FE to the rectangle FE, EG.

Similarly also, as the rectangle GE, EF is to the square on EF, that is, as GD is to FD, so is GE to EF.

Q. E. D. If a, b be two straight lines,

a: b = a: ab.

PROPOSITION 22.

B

The square on a medial straight line, if applied to a rational straight line, produces as breadth a straight line rational and incommensurable in length with that to which it is applied.

Let A be medial and CB rational, and let a rectangular area BD equal to the square on A be applied to BC, producing CD as breadth; I say

that CD is rational and incommensurable in length with CB.

For, since A is medial, the square on it is equal to a rectangular area contained by rational straight lines commensurable in square only.

[x. 21]
Let the square on it be equal to GF.

But the square on it is also equal to BD; therefore BD is equal to GF.

But it is also equiangular with it; and in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional ; (vi. 14] therefore, proportionally, as BC is to EG, so is EF to CD.

Therefore also, as the square on BC is to the square on EG, so is the square on EF to the square on CD. [vi. 22]

E

F

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But the square on CB is commensurable with the square on EG, for each of these straight lines is rational ; therefore the square on EF is also commensurable with the square on CD.

[x. 11] But the square on EF is rational ; therefore the square on CD is also rational; [x. Def. 4) therefore CD is rational.

And, since EF is incommensurable in length with EG, for they are commensurable in square only, and, as EF is to EG, so is the square on EF to the rectangle FE, EG,

[Lemma] therefore the square.on EF is incommensurable with the rectangle FE, EG.

[x. 11] But the square on CD is commensurable with the square on EF, for the straight lines are rational in square; and the rectangle DC, CB is commensurable with the rectangle FE, EG, for they are equal to the square on A ; therefore the square on CD is also incommensurable with the rectangle DC, CB.

(x. 13] But, as the square on CD is to the rectangle DC, CB, so is DC to CB;

[Lemma] therefore DC is incommensurable in length with CB. (x. 11]

Therefore CD is rational and incommensurable in length with CB.

Q. E. D.

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m

O, n

Our algebraical notation makes the result of this proposition almost selfevident. We have seen that the square of a medial straight line is of the form Jk.p? If we “apply" this area to another rational straight line o, the

sk. p2 breadth is

sk.p2 This is equal to

where m, n are integers. The latter

o?
straight line, which we may express, if we please, in the form k'.o, is clearly
commensurable with o in square only, and therefore rational but incom-
mensurable in length with o.

Euclid's proof, necessarily longer, is in two parts.
Suppose that the rectangle Jk.p=0.x.
Then (1)
o: p= Jk.p:x,

[vi. 147
whence
o?:p*= kp: x".

[Vi. 22] But o’ pʻ, and therefore kp~ x”.

[x. 11]

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