Commentary on Proposition X.59

Back to Proposition X.59.

[Folio 0135 | p. 131]

Euclid here finds the square root of the expression [cf. X. 53]

$$\rho(\sqrt{k}\cdot\rho + \sqrt{\lambda}\cdot\rho).$$

As usual, we solve the equations

$$\left.\begin{aligned} u+v &= \sqrt{k}\cdot\rho \\ uv &= \frac{1}{4}\lambda\rho^2 \end{aligned}\right\}\quad\quad\quad\quad\dots\dots(1);$$

then, $u$, $v$ being found, we put

$$\left.\begin{aligned} x^2 &= \rho u \\ y^2 &= \rho v \end{aligned}\right\}\quad\quad\quad\quad\dots\dots(2),$$

and $(x+y)$ is the square root required.

Euclid proves that $(x+y)$ is the side of (the sum of) two medial areas, as follows.

As in the last two propositions, $x$, $y$ are proved to be incommensurable in square.

Now $\sqrt{k}\cdot\rho$, $\rho$ are commensurable in square only; therefore $\rho(u+v)$, or $(x^2+y^2)$, is a medial area $\dots\dots(3).$

Next, $xy = \frac{1}{2}\sqrt{\lambda}\cdot\rho^2$, which is again a medial area $\dots\dots(4).$

Lastly,

$$\sqrt{k}\cdot\rho \smile \frac{1}{2}\sqrt{\lambda}\cdot\rho,$$

$$\sqrt{k}\cdot\rho^2 \smile \frac{1}{2}\sqrt{\lambda}\cdot\rho^2;$$

that is,

$$(x^2+y^2) \smile xy \dots\dots(5).$$

Hence [(3), (4), (5)] $(x+y)$ is the side of the sum of two medial areas.

Solving the equations algebraically, we have

$$\begin{aligned} u &= \frac{\rho}{2}(\sqrt{k} + \sqrt{k-\lambda}), \\ v &= \frac{\rho}{2}(\sqrt{k} - \sqrt{k-\lambda}), \end{aligned}$$

and

$$x+y = \rho\sqrt{\frac{1}{2}(\sqrt{k}+\sqrt{k-\lambda})} + \rho\sqrt{\frac{1}{2}(\sqrt{k}-\sqrt{k-\lambda})}.$$

The corresponding square root found in X. 96 is $x-y$, where $x$, $y$ are the same as here.

The two square roots are (cf. note on X. 53) the roots of the biquadratic equation

$$x^4 - 2\sqrt{k}\cdot\rho^2 x^2 + (k-\lambda)\rho^4 = 0.$$

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[LEMMA.

If a straight line be cut into unequal parts, the squares on the unequal parts are greater than twice the rectangle contained by the unequal parts.

Let AB be a straight line, and let it be cut into unequal parts at C, and let AC be the greater ; I say that the squares on AC, CB are greater than twice the rectangle AC, CB.

[Folio 0136 | p. 132]

For let AB be bisected at D.

Since then a straight line has been cut into equal parts at D, and into unequal parts at C,

therefore the rectangle AC, CB together with the square on CD is equal to the square on AD, [II. 5] so that the rectangle AC, CB is less than the square on AD ; therefore twice the rectangle AC, CB is less than double of the square on AD.

But the squares on AC, CB are double of the squares on AD, DC ; [II. 9]

therefore the squares on AC, CB are greater than twice the rectangle AC, CB. Q. E. D.]

We have already remarked (note on X. 44) that the Lemma here proving that

$$x^2 + y^2 > 2xy$$

can hardly be genuine, since the result is used in X. 44.