Commentary on Proposition X.113

Back to Proposition X.113.

This proposition, which is companion to the preceding, gives us the equivalent of the rationalisation of the denominator of

$$\frac{c^2}{\sqrt A - \sqrt B} \quad\text{or}\quad \frac{c^2}{a \sim \sqrt B}.$$

Euclid (or the writer) proves that

$$\frac{\sigma^2}{\rho - \sqrt k \cdot \rho} = \lambda\rho + \lambda\sqrt k \cdot \rho, \qquad (k < 1)$$

and his method enables us to see that $\lambda = \sigma^2/(\rho^2 - k\rho^2)$.

Let

$$\frac{\sigma^2}{\rho - \sqrt k \cdot \rho} = u;$$

and it is proved that $u$ is the binomial straight line $(u - w) + w$, where $w$ is determined as shown below.

$$\left.\begin{aligned} u\, (\rho - \sqrt k \cdot \rho) &= \sigma^2 = \rho x\text{, say,} \\ \text{whence}\quad \rho : (\rho - \sqrt k \cdot \rho) &= u : x \end{aligned}\right\}\dots\dots\dots\dots\dots\dots(1),$$

so that $x < u$.

Let then

$$x = u - v.$$

Since $(u - v)\,\rho = \sigma^2$, a rational area,

$$(u - v) \text{ is rational and } \mathbin{\cap} \rho \dots\dots\dots\dots\dots\dots(2).$$

And [(1)]

$$\rho : (\rho - \sqrt k \cdot \rho) = u : (u - v),$$

so that, convertendo,

$$\rho : \sqrt k \cdot \rho = u : v.$$

Suppose that

$$u : v = w : (v - w),$$

so that [V. 19]

$$(u - w) : w = u : v = w : (v - w).$$

Thus, $w$ being a mean proportional between $(u - w)$, $(v - w)$,

$$(u - w)^2 : w^2 = (u - w) : (v - w).$$

But

$$\begin{aligned} (u - w)^2 : w^2 &= u^2 : v^2 \\ &= \rho^2 : k\rho^2 \dots\dots\dots\dots\dots\dots(3), \end{aligned}$$

so that $(u - w)^2 \mathbin{\cap} u^2.$

Therefore

$$\begin{aligned} (u - w) &\mathbin{\cap} (v - w) \\ &\mathbin{\cap} \{(u - w) - (v - w)\} \\ &\mathbin{\cap} (u - v). \end{aligned}$$

Therefore [(2)] $(u - w)$ is rational and $\mathbin{\cap} \rho \dots\dots\dots\dots\dots\dots(4).$

And, since

$$\rho : \sqrt k \cdot \rho = (u - w) : w,$$

$w$ is rational and $\mathbin{\cap} \sqrt k \cdot \rho \dots\dots\dots\dots\dots\dots(5).$

Hence [(4), (5)] $(u - w)$, $w$ are rational and $\mathbin{\sim}$, so that $(u - w) + w$ is a binomial straight line.

Now either

$$\begin{aligned} \text{(I)} &\quad \sqrt{\rho^2 - k\rho^2} \mathbin{\cap} \rho, \\ \text{or (II)} &\quad \sqrt{\rho^2 - k\rho^2} \mathbin{\cup} \rho. \end{aligned}$$

In case (I)

$$\sqrt{(u - w)^2 - w^2} \mathbin{\cap} (u - w),$$

and in case (II)

$$\sqrt{(u - w)^2 - w^2} \mathbin{\cup} (u - w). \qquad \text{[(3) and X. 14]}$$

And, if $\rho \mathbin{\cap} \sigma$, $(u - w) \mathbin{\cap} \sigma$; [(4)] if $\sqrt k \cdot \rho \mathbin{\cap} \sigma$, $w \mathbin{\cap} \sigma$; [5] while, if neither $\rho$ nor $\sqrt k \cdot \rho$ is $\mathbin{\cap} \sigma$, neither $(u - w)$ nor $w$ is $\mathbin{\cap} \sigma$.

Hence $(u - w) + w$ is a binomial straight line of the same order as the apotome $\rho - \sqrt k \cdot \rho$, its terms are proportional to those of the apotome [(3)], and commensurable with them respectively [(4), (5)].

To find $(u - w)$, $w$ algebraically we have

$$u = \frac{\sigma^2}{\rho - \sqrt{k} \cdot \rho},$$

$$\frac{u - w}{w} = \frac{\rho}{\sqrt{k} \cdot \rho}.$$

From the latter

$$\begin{aligned} w &= \frac{u \cdot \sqrt{k} \cdot \rho}{\rho + \sqrt{k} \cdot \rho} \\ &= \frac{\sigma^2 \cdot \sqrt{k} \cdot \rho}{\rho^2 - k\rho^2}. \end{aligned}$$

Thus

$$u - w = w \cdot \frac{1}{\sqrt{k}} = \frac{\sigma^2\rho}{\rho^2 - k\rho^2}.$$

Therefore

$$(u - w) + w = \sigma^2 \cdot \frac{\rho + \sqrt{k} \cdot \rho}{\rho^2 - k\rho^2}.$$