Commentary on Proposition X.114

Back to Proposition X.114.

[Folio 0257 | p. 253]

This theorem is equivalent to the proof of the fact that

$$\sqrt{(\sqrt{A} - \sqrt{B})(\lambda\sqrt{A} + \lambda\sqrt{B})} = \sqrt{\lambda(A - B)},$$

and

$$\sqrt{(a \sim \sqrt{B})(\lambda a + \lambda\sqrt{B})} = \sqrt{\lambda(a^2 \sim B)}.$$

The result of the theorem X. 112 is used for the purpose thus.

We have to prove that

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

is rational.

By X. 112 we have, if $\sigma$ is a rational straight line,

$$\frac{\sigma^2}{\lambda\rho + \lambda\sqrt{k}\cdot\rho} = \lambda'\rho - \lambda'\sqrt{k}\cdot\rho \dots\dots\dots\dots\dots\dots(1).$$

[Folio 0258 | p. 254]

Now $\rho : \lambda'\rho = \sqrt{k}\cdot\rho : \lambda'\sqrt{k}\cdot\rho = (\rho - \sqrt{k}\cdot\rho) : (\lambda'\rho - \lambda'\sqrt{k}\cdot\rho)$, so that

$$(\rho - \sqrt{k}\cdot\rho) \mathbin{\cap} (\lambda'\rho - \lambda'\sqrt{k}\cdot\rho).$$

Multiplying each by $(\lambda\rho + \lambda\sqrt{k}\cdot\rho)$, we have

$$(\rho - \sqrt{k}\cdot\rho)(\lambda\rho + \lambda\sqrt{k}\cdot\rho) \mathbin{\cap} (\lambda\rho + \lambda\sqrt{k}\cdot\rho)(\lambda'\rho - \lambda'\sqrt{k}\cdot\rho)$$

$$\mathbin{\cap} \sigma^2\text{, by (1).}$$

That is, $(\rho - \sqrt{k}\cdot\rho)(\lambda\rho + \lambda\sqrt{k}\cdot\rho)$ is a rational area, and therefore $\sqrt{(\rho - \sqrt{k}\cdot\rho)(\lambda\rho + \lambda\sqrt{k}\cdot\rho)}$ is rational.