Commentary on Proposition X.41

Back to Proposition X.41.

In this case $x$, $y$ are of the form

$$\frac{\rho\lambda^{\frac{1}{4}}}{\sqrt{2}}\sqrt{1+\frac{k}{\sqrt{1+k^2}}},\quad \frac{\rho\lambda^{\frac{1}{4}}}{\sqrt{2}}\sqrt{1-\frac{k}{\sqrt{1+k^2}}}.$$

By hypothesis, $(x^2+y^2)$ and $2xy$ are medial areas, and

$$x^2+y^2 \cup 2xy \dots\dots\dots\dots\dots\dots\dots\dots(1).$$

‘Apply’ these areas respectively to a rational straight line $\sigma$, and suppose

$$\left.\begin{aligned} x^2+y^2 &= \sigma u \\ 2xy &= \sigma v \end{aligned}\right\} \dots\dots\dots\dots\dots\dots\dots\dots(2).$$

Since then $\sigma u$ and $\sigma v$ are both medial areas, $u$, $v$ are rational and both are $\cup\ \sigma$ $\dots\dots\dots\dots\dots\dots\dots\dots(3)$.

Now, by (1) and (2),

$$\sigma u \cup \sigma v,$$

so that

$$u \cup v.$$

By this and (3), $u$, $v$ are rational and $\sim$.

Therefore [x. 36] $(u+v)$ is irrational.

Hence $\sigma(u+v)$ is irrational [deduction from x. 20].

Thus $(x+y)^2$, and therefore $(x+y)$, is irrational.

The irrational straight line

$$\frac{\rho\lambda^{\frac{1}{4}}}{\sqrt{2}}\sqrt{1+\frac{k}{\sqrt{1+k^2}}} + \frac{\rho\lambda^{\frac{1}{4}}}{\sqrt{2}}\sqrt{1-\frac{k}{\sqrt{1+k^2}}}$$

is called (again for an obvious reason) the “side” of the sum of two medials (medial areas).

This and the corresponding irrational with a minus sign found in x. 78 are the positive roots of the equation

$$x^4 - 2\sqrt{\lambda}\cdot x^2\rho^2 + \lambda\frac{k^2}{1+k^2}\rho^4 = 0.$$

LEMMA.

And that the aforesaid irrational straight lines are divided only in one way into the straight lines of which they are the sum and which produce the types in question,[1] we will now prove after premising the following lemma.

Let the straight line AB be set out, let the whole be cut into unequal parts at each of the points C, D, and let AC be supposed greater than DB; I say that the squares on AC, CB are greater than the squares on AD, DB.

A D E C B

For let AB be bisected at E.

Then, since AC is greater than DB, let DC be subtracted from each;
therefore the remainder AD is greater than the remainder CB.
But AE is equal to EB;
therefore DE is less than EC;
therefore the points C, D are not equidistant from the point of bisection.
And, since the rectangle AC, CB together with the square on EC is equal to the square on EB, [II. 5] and, further, the rectangle AD, DB together with the square on DE is equal to the square on EB, [id.]
therefore the rectangle AC, CB together with the square on EC is equal to the rectangle AD, DB together with the square on DE.
And of these the square on DE is less than the square on EC;
therefore the remainder, the rectangle AC, CB, is also less than the rectangle AD, DB, so that twice the rectangle AC, CB is also less than twice the rectangle AD, DB.
Therefore also the remainder, the sum of the squares on AC, CB, is greater than the sum of the squares on AD, DB.
Q. E. D.

This proves that, if $x + y = u + v$, and if $u$, $v$ are more nearly equal than $x$, $y$ (i.e. if the straight line is divided in the second case nearer to the point of bisection), then

$$(x^2 + y^2) > (u^2 + v^2).$$

It is first proved by means of II. 5 that

$$2xy < 2uv,$$

whence, since $(x+y)^2 = (u+v)^2$, the required result follows.