But AF is commensurable with KM; therefore AB is also commensurable with KL.

And, as AB is to KL, so is the rectangle CD, AB to rectangle CD, KL;

[v

therefore the rectangle CD, AB is also commensurable v the rectangle CD, KL.

[x

But the rectangle CD, KL is equal to the square on therefore the rectangle CD, AB is commensurable with square on H.

But the square on G is equal to the rectangle CD, Al therefore the square on G is commensurable with the squ on H.

But the square on H is rational;

therefore the square on G is also rational;

therefore G is rational.

And it is the "side" of the rectangle CD, AB.
Therefore etc.

PORISM. And it is made manifest to us by this also it is possible for a rational area to be contained by irrati straight lines.

and

This theorem is equivalent to the proof of the fact that √(JA – JB) (X JA + λ √B) = √λ (A – B),

√(a ~ √B) (λa + λ √√ B) = √√λ (a2 ~ B).

Q. E. D.

The result of the theorem X. 112 is used for the purpose thus.
We have to prove that

is rational.

√(p −√k. p) (^p + λ √k . p)

By x. 112 we have, if σ is a rational straight line,

therefore

√(p −√k. p) (λp + λk. p) is rational.

PROPOSITION 115.

From a medial straight line there arise irrational straight es infinite in number, and none of them is the same as any the preceding.

that which is contained by an irrational and a rational

aight line is irrational.

[deduction from x. 20]

And it is not the same with any of the preceding;

the square on none of the preceding, if applied to a rational raight line produces as breadth a medial straight line.

Again, let the square on D be equal to the rectangle B, C; erefore the square on D is irrational. [deduction from x. 20]

Therefore D is irrational;

[x. Def. 4]

d it is not the same with any of the preceding, for the uare on none of the preceding, if applied to a rational raight line, produces C as breadth.

Similarly, if this arrangement proceeds ad infinitum, it manifest that from the medial straight line there arise ational straight lines infinite in number, and none is the me with any of the preceding.

Q. E. D.

x. 115]

PROPOSITIONS 114, 115

255

Heiberg is clearly right in holding that this proposition, at all events, is alien to the general scope of Book x, and is therefore probably an interpolation, made however before Theon's time. It is of the same character as a scholium at the end of the Book, which is (along with the interpolated proposition proving, in two ways, the incommensurability of the diagonal of a square with its side) relegated by August as well as Heiberg to an Appendix.

The proposition amounts to this.

The straight line p being medial, if o be a rational straight line, √#po is a new irrational straight line. So is the mean proportional between this and another rational straight line σ', and so on indefinitely.

ANCIENT EXTENSIONS OF THE THEORY OF BOOK X.

From the hints given by the author of the commentary found in Arabic by Woepcke (cf. pp. 3-4 above) it would seem probable that Apollonius' extensions of the theory of irrationals took two directions: (1) generalising the medial straight line of Euclid, and (2) forming compound irrationals by the addition and subtraction of more than two terms of the sort composing the binomials, apotomes, etc. The commentator writes (Woepcke's article, pp. 694 sqq.):

"It is also necessary that we should know that, not only when we join together two straight lines rational and commensurable in square do we obtain the binomial straight line, but three or four lines produce in an analogous manner the same thing. In the first case, we obtain the trinomial straight line, since the whole line is irrational; and in the second case we obtain the quadrinomial, and so on ad infinitum. The proof of the (irrationality of the) line composed of three lines rational and commensurable in square is exactly the same as the proof relating to the combination of two lines.

"But we must start afresh and remark that not only can we take one sole medial line between two lines commensurable in square, but we can take three or four of them and so on ad infinitum, since we can take, between any two given straight lines, as many lines as we wish in continued proportion.

"Likewise, in the lines formed by addition not only can we construct the binomial straight line, but we can also construct the trinomial, as well as the first and second trimedial; and, further, the line composed of three straight lines incommensurable in square and such that the one of them gives with each of the two others a sum of squares (which is) rational, while the rectangle contained by the two lines is medial, so that there results a major (irrational) composed of three lines.

And, in an analogous manner, we obtain the straight line which is the 'side' of a rational plus a medial area, composed of three straight lines, and, likewise, that which is the 'side' of (the sum of) two medials."

The generalisation of the medial is apparently after the following manner. Let x, y be two straight lines rational and commensurable in square only and suppose that m means are interposed, so that

ich is the generalised medial.

We now pass to the trinomial etc., with the commentator's further remarks >ut them.

The trinomial.

"Suppose three rational straight lines commensurable in are only. The line composed of two of these lines, that is, the binomial aight line, is irrational, and, in consequence, the area contained by this line d the remaining line is irrational, and, likewise, the double of the area ntained by these two lines will be irrational. Thus the square on the ole line composed of three lines is irrational and consequently the line is ational, and it is called a trinomial straight line."

It is easy to see that this "proof" is not conclusive as stated. Nor does oepcke seem to show how the proposition can be proved on Euclidean ies. But I think it would be somewhat as follows.

Suppose x, y, z to be rational and.

Then x2, y2, z2 are rational, and 2yz, 2zx, 2xy are all medial.

First, (2yz+22x + 2xy) cannot be rational.

For suppose this sum equal to a rational area, say σ2.

Since

2yz + 2zx + 2xy = σ2,

2%x + 2xy = σ2 – 2yz,

the sum of two medial areas incommensurable with one another is equal to e difference between a rational area and a medial area.

But the "side" of the sum of the two medial areas must [x. 72] be one of vo irrationals with a positive sign; and the "side" of the difference between a tional area and a medial area must [x. 108] be one of two irrationals with a egative sign.

And the first "side" cannot be the same as the second [x. 111 and exlanation following].

o that (x + y + z)2, and therefore also (x + y + z), is irrational.

The commentator goes on:

"And, if we have four lines commensurable in square, as we have said, the rocedure will be exactly the same; and we shall treat the succeeding lines in analogous manner."

Without speculating further as to how the extension was made to the adrinomial etc., we may suppose with Woepcke that Apollonius probabl vestigated the multinomial

p+ √k.p+ √λ. p + √μ. p + ...

ANCIENT EXTENSIONS

(2) The first trimedial straight line.

257

The commentator here says: "Suppose we have three medial lines commensurable in square [only], one of which contains with each of the two others a rational rectangle; then the straight line composed of the two lines is irrational and is called the first bimedial; the remaining line is medial, and the area contained by these two lines is irrational. Consequently the square on the whole line is irrational."

To begin with, the conditions here given are incompatible. If x, y, z be medial straight lines such that xy, xz are both rational,

y: z = xy: xz = m : n,

and y, z are commensurable in length and not in square only.

Hence it seems that we must, with Woepcke, understand "three medial straight lines such that one is commensurable with each of the other two in square only and makes with it a rational rectangle."

If x, y, z be the three medial straight lines,

so that (x2 + y2 + z2) is medial.

(x2 + y2 + 22) ~ x2,

Also we have 2xy, 2xz both rational and 2yz medial.

Now (x2 + y2 + 22) + 2yz + 2xy + 2xz cannot be rational, for, if it were, the sum of two medial areas, (x2 + y2+z2), 2y%, would be rational: which is impossible. [Cf. x. 72.]

Hence

(x+y+2) is irrational.

(3) The second trimedial straight line.

Suppose x, y, z to be medial straight lines commensurable in square only and containing with each other medial rectangles.

Then

Also

(x2 + y2 + z2) ~ x2, and is medial.

2yz, 2zx, 2xy are all medial areas.

To prove the irrationality in this case I presume that the method would be like that of x. 38 about the second bimedial.

Suppose σ to be a rational straight line and let

u, v, w are commensurable in square only.

Also, since

t is incommensurable with w.

H. E. III.

x: Z = Wu,