Commentary on Proposition X.17¶
Back to Proposition X.17.
This proposition gives the condition that the roots of the equation in $x$,
$$ax - x^2 = \beta \left(= \frac{b^2}{4}\text{, say}\right),$$
are commensurable with $a$, or that $x$ is expressible in terms of $a$ and integral numbers, i.e. is of the form $\frac{m}{n}a$. No better proof can be found for the fact that Euclid and the Greeks used their solutions of quadratic equations for numerical problems. On no other assumption could an elaborate discussion of the conditions of incommensurability of the roots with given lengths or with a given number of units of length be explained. In a purely geometrical solution the distinction between commensurable and incommensurable roots has no point, because each can equally easily be represented by straight lines. On the other hand, on the assumption that the numerical solution of quadratic equations was an important part of the system of the Greek geometers, the distinction between the cases where the roots are commensurable and incommensurable respectively with a given length or unit becomes of great importance. Since the Greeks had no means of expressing what we call an irrational number, the case of an equation with incommensurable roots could only be represented by them geometrically; and the geometrical representations had to serve instead of what we can express by formulae involving surds.
Euclid proves in this proposition and the next that, $x$ being determined from the equation
$$x(a - x) = \frac{b^2}{4} \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots (1),$$
$x$, $(a - x)$ are commensurable in length when $\sqrt{a^2 - b^2}$, $a$ are so, and incommensurable in length when $\sqrt{a^2 - b^2}$, $a$ are incommensurable; and conversely.
Observe the similarity of his proof to our algebraical method of solving the equation. $a$ being represented in the figure by BC, and $x$ by CD,
$$EF = ED = \frac{a}{2} - x$$
and
$$x(a - x) + \left(\frac{a}{2} - x\right)^2 = \frac{a^2}{4}, \qquad \text{by Eucl. II. 5.}$$
If we multiply throughout by 4,
$$4x(a - x) + 4\left(\frac{a}{2} - x\right)^2 = a^2,$$
whence, by (1),
$$b^2 + (a - 2x)^2 = a^2,$$
or
$$a^2 - b^2 = (a - 2x)^2,$$
and
$$\sqrt{a^2 - b^2} = a - 2x.$$
We have to prove in this proposition
that, if $x$, $(a - x)$ are commensurable in length, so are $a$, $\sqrt{a^2 - b^2}$,
that, if $a$, $\sqrt{a^2 - b^2}$ are commensurable in length, so are $x$, $(a - x)$.
To prove that $a$, $a - 2x$ are commensurable in length Euclid employs several successive steps, thus.
Since $(a - x) \mathrel{⌢} x,$ |
$a \mathrel{⌢} x.$ |
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But |
$x \mathrel{⌢} 2x.$ |
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Therefore |
$a \mathrel{⌢} 2x$ |
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$\mathrel{⌢} (a - 2x).$ |
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That is, |
$a \mathrel{⌢} \sqrt{a^2 - b^2}.$ |
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$a \mathrel{⌢} a - 2x,$ |
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whence |
$a \mathrel{⌢} 2x.$ |
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But |
$2x \mathrel{⌢} x;$ |
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therefore |
$a \mathrel{⌢} x,$ |
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and hence |
$(a - x) \mathrel{⌢} x.$ |
It is often more convenient to use the symmetrical form of equation in this and similar cases, viz.
$$\left.\begin{aligned} xy &= \frac{b^2}{4} \\ x + y &= a \end{aligned}\right\}$$
The result with this mode of expression is that
if $x \mathrel{⌢} y$, then $a \mathrel{⌢} \sqrt{a^2 - b^2}$; and
if $a \mathrel{⌢} \sqrt{a^2 - b^2}$, then $x \mathrel{⌢} y$. The truth of the proposition is even easier to see in this case, since $(x - y)^2 = (a^2 - b^2)$.