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 $m$ means are interposed, so that
$$x : x_1 = x_1 : x_2 = x_2 : x_3 = \dots = x_{m-1} : x_m = x_m : y.$$
We easily derive herefrom
$$\frac{x}{x_r} = \left(\frac{x}{x_1}\right)^r,$$
$$\frac{x}{y} = \left(\frac{x}{x_1}\right)^{m+1},$$
and hence
$$x_1^r = x_r \cdot x^{r-1},$$
$$x_1^{m+1} = y \cdot x^m,$$
so that
$$(x_r \cdot x^{r-1})^{m+1} = (y \cdot x^m)^r,$$
and therefore
$$x_r^{m+1} = x^{m-r+1} \cdot y^r,$$
or
$$x_r = (x^{m-r+1} \cdot y^r)^{\frac{1}{m+1}},$$
which is the generalised medial.
We now pass to the trinomial etc., with the commentator’s further remarks about them.
The trinomial. “Suppose three rational straight lines commensurable in square only. The line composed of two of these lines, that is, the binomial straight line, is irrational, and, in consequence, the area contained by this line and the remaining line is irrational, and, likewise, the double of the area contained by these two lines will be irrational. Thus the square on the whole line composed of three lines is irrational and consequently the line is irrational, and it is called a trinomial straight line.”
It is easy to see that this “proof” is not conclusive as stated. Nor does Woepcke seem to show how the proposition can be proved on Euclidean lines. But I think it would be somewhat as follows.
Suppose $x$, $y$, $z$ to be rational and $\sim$. Then $x^2$, $y^2$, $z^2$ are rational, and $2yz$, $2zx$, $2xy$ are all medial. First, $(2yz + 2zx + 2xy)$ cannot be rational. For suppose this sum equal to a rational area, say $\sigma^2$. Since
$$2yz + 2zx + 2xy = \sigma^2,$$
$$2zx + 2xy = \sigma^2 - 2yz,$$
or the sum of two medial areas incommensurable with one another is equal to the 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 two irrationals with a positive sign; and the “side” of the difference between a rational area and a medial area must [X. 108] be one of two irrationals with a negative sign. And the first “side” cannot be the same as the second [X. 111 and explanation following]. Therefore
$$2zx + 2xy \ne \sigma^2 - 2yz,$$
and
$$2yz + 2zx + 2xy \text{ is consequently \emph{irrational}.}$$
Therefore
$$(x^2 + y^2 + z^2) \mathbin{\cup} (2yz + 2zx + 2xy),$$
whence
$$(x + y + z)^2 \mathbin{\cup} (x^2 + y^2 + z^2),$$
so 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 procedure will be exactly the same; and we shall treat the succeeding lines in an analogous manner.”
Without speculating further as to how the extension was made to the quadrinomial etc., we may suppose with Woepcke that Apollonius probably investigated the multinomial
$$\rho + \sqrt{k}\cdot\rho + \sqrt{\lambda}\cdot\rho + \sqrt{\mu}\cdot\rho + \dots$$
The first trimedial straight line.
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,
$$(x^2 + y^2 + z^2) \mathbin{\cap} x^2,$$
so that $(x^2 + y^2 + z^2)$ is medial. Also we have $2xy$, $2xz$ both rational and $2yz$ medial. Now $(x^2 + y^2 + z^2) + 2yz + 2xy + 2xz$ cannot be rational, for, if it were, the sum of two medial areas, $(x^2 + y^2 + z^2)$, $2yz$, would be rational : which is impossible. [Cf. X. 72] Hence
$$(x + y + z) \text{ is irrational.}$$
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
$$(x^2 + y^2 + z^2) \mathbin{\cap} x^2\text{, and is medial.}$$
Also
$$2yz\text{, } 2zx\text{, } 2xy \text{ are all medial areas.}$$
To prove the irrationality in this case I presume that the metnod would be like that of X. 38 about the second bimedial. Suppose $\sigma$ to be a rational straight line and let
$$\left. \begin{aligned} (x^2 + y^2 + z^2) &= \sigma t \\ 2yz &= \sigma u \\ 2zx &= \sigma v \\ 2xy &= \sigma w \end{aligned} \right\}$$
Here, since, e.g.,
$$xz : xy = v : w,$$
or
$$z : y = v : w,$$
and similarly
$$x : z = w : u,$$
$u$, $v$, $w$ are commensurable in square only.
Also, since
$$\begin{aligned} (x^2 + y^2 + z^2) &\mathbin{\cap} x^2 \\ &\mathbin{\cup} xy, \end{aligned}$$
$t$ is incommensurable with $w$.
Similarly $t$ is incommensurable with $u$, $v$. But $t$, $u$, $v$, $w$ are all rational and $\sim \sigma$. Therefore $(t + u + v + w)$ is a quadrinomial and therefore irrational. Therefore $\sigma(t + u + v + w)$, or $(x + y + z)^2$, is irrational, whence $(x + y + z)$ is irrational.
The major made up of three straight lines.
The commentator describes this as “the line composed of three straight lines incommensurable in square and such that one of them gives with each of the other two a sum of squares (which is) rational, while the rectangle contained by the two lines is medial.” If $x$, $y$, $z$ are the three straight lines, this would indicate
$$\begin{aligned} (x^2 + y^2) &\text{ rational,} \\ (x^2 + z^2) &\text{ rational,} \\ 2yz &\text{ medial.} \end{aligned}$$
Woepcke points out (pp. 696—8, note) the difficulties connected with this supposition or the supposition of
$$\begin{aligned} (x^2 + y^2) &\text{ rational,} \\ (x^2 + z^2) &\text{ rational,} \\ 2xy \text{ (or } 2xz) &\text{ medial,} \end{aligned}$$
and concludes that what is meant is the supposition
$$\left. \begin{aligned} (x^2 + y^2) &\text{ rational} \\ xy &\text{ medial} \\ xz &\text{ medial} \end{aligned} \right\}$$
(though the text is against this). The assumption of $(x^2 + y^2)$ and $(x^2 + z^2)$ being concurrently rational is certainly further removed from Euclid, for X. 33 only enables us to find one pair of lines having the property, as $x$, $y$. But we will not pursue these speculations further.
As regards further irrationals formed by subtraction the commentator writes as follows. “Again, it is not necessary that, in the irrational straight lines formed by means of subtraction, we should confine ourselves to making one subtraction only, so as to obtain the apotome, or the first apotome of the medial, or the second apotome of the medial, or the minor, or the straight line which produces with a rational area a medial whole, or that which produces with a medial area a medial whole; but we shall be able here to make two or three or four subtractions.
“When we do that, we show in manner analogous to the foregoing that the lines which remain are irrational and that each of them is one of the lines formed by subtraction. That is to say that, if from a rational line we cut off another rational line commensurable with the whole line in square, we obtain, for remainder, an apotome; and, if we subtract from this line (which is) cut off and rational—that which Euclid calls the annex ($\pi\rho\omicron\sigma\alpha\rho\mu\acute{o}\zeta\omicron\upsilon\sigma\alpha$)— another rational line which is commensurable with it in square, we obtain, as the remainder, an apotome; likewise, if we cut off from the rational line cut off from this line (i.e. the annex of the apotome last arrived at) another line which is commensurable with it in square, the remainder is an apotome. The same thing occurs in the subtraction of the other lines.”
As Woepcke remarks, the idea is the formation of the successive apotomes $\sqrt{a} - \sqrt{b}$, $\sqrt{b} - \sqrt{c}$, $\sqrt{c} - \sqrt{d}$, etc. We should naturally have expected to see the writer form and discuss the following expressions
$$(\sqrt{a} - \sqrt{b}) - \sqrt{c},$$
$$\{(\sqrt{a} - \sqrt{b}) - \sqrt{c}\} - \sqrt{d}\text{, etc.}$$