Commentary on Proposition X.36¶
Back to Proposition X.36.
Here begins the first hexad of propositions relating to compound irrational straight lines. The six compound irrational straight lines are formed by adding two parts, as the corresponding six in Props. 73—78 are formed by subtraction. The relation between the six irrational straight lines in this and the next five propositions with those described in Definitions II. and the Props. 48—53 following thereon (the first, second, third, fourth, fifth and sixth binomials) will be seen when we come to Props. 54—59; but it may be stated here that the six compound irrationals in Props. 36—41 can be found by means of the equivalent of extracting the square root of the compound irrationals in x. 48—53 (the process being, strictly speaking, the finding of the sides of the squares equal to the rectangles contained by the latter irrationals respectively and a rational straight line as the other side), and it is therefore the further removed compound irrational, so to speak, which is treated first.
In reproducing the proofs of the propositions, I shall for the sake of simplicity call the two parts of the compound irrational straight line $x$, $y$, explaining at the outset the forms which $x$, $y$ really have in each case; $x$ will always be supposed to be the greater segment.
In this proposition $x$, $y$ are of the form $\rho$, $\sqrt{k} \cdot \rho$, and $(x + y)$ is proved to be irrational thus.
$x \sim y$, so that $x \mathbin{\smile} y$.
Now
$$x : y = x^2 : xy,$$
so that
$$x^2 \mathbin{\smile} xy.$$
But $x^2 \mathbin{\sim} (x^2 + y^2)$, and $xy \mathbin{\sim} 2xy$; therefore
$$(x^2 + y^2) \mathbin{\smile} 2xy,$$
and hence
$$(x^2 + y^2 + 2xy) \mathbin{\smile} (x^2 + y^2).$$
But $(x^2 + y^2)$ is rational; therefore $(x + y)^2$, and therefore $(x + y)$, is irrational.
This irrational straight line, $\rho + \sqrt{k} \cdot \rho$, is called a binomial straight line.
This and the corresponding apotome ($\rho - \sqrt{k} \cdot \rho$) found in x. 73 are the positive roots of the equation
$$x^4 - 2(1 + k)\rho^2 \cdot x^2 + (1 - k)^2\rho^4 = 0.$$