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For one desiring to find of what number two-thirds are four-fifth parts, the two-thirds would have to be multiplied by a common denominator, and the product divided by four-fifths of that same denominator. And if someone should say: if 4/5? gives 2/3?, what will it give from u?? Certainly it will give $u$. For in the proposed example, the term $a$ will obtain two denominators in its place: namely, the known $o$ [10] and the unknown $u$, which afterwards arises as known from the rule of three, as has been said.
A line diagram representing mathematical ratios. A horizontal line is marked with points. Above the line are the fractions 2/3 and 4/5, and the letter u. Below the line, vertical tick marks correspond to the numbers 10, 12, and 15.
By what reasoning can it be known that the ratio of a squared-square quantity to a similar quantity is quadruple to that which is of their roots, and that the ratio of the first related quantities is quintuple, and so on? For the sake of this matter, one must know the method of production of these powers, which arises from the production of the first root into itself; just as a cube requires that one multiply the root by its own square, and a cube arises; this afterwards multiplied by the cube [gives] a squared-square quantity, and into this [the squared-square], the aforesaid root will give the first related quantity. Once we know this, it is necessary to remember that Euclid, in the seventeenth of the sixth or the 11th of the eighth, teaches that the ratio of a square to a square is double the ratio of their roots, and in the 36th of the eleventh or the 11th of the eighth, that of a cube to a cube is triple; but I now assert that the ratio of a squared-square to its roots is quadruple, and of a first related to a first related quintuple, and so on step by step.
For the sake of this speculation, let there be a line $d$, which signifies a greater cube, and $b$ a smaller. Let $c$, indeed, be the root of $d$ itself, and $e$ of $b$ itself, so ordered to one another as is seen in the figure below. Now, let $c$ be produced with $d$, resulting in $q$, a squared-square; then let $e$ be produced with $b$, and $p$, the other squared-square, will be given. I say, therefore, that the ratio of $q$ to $p$ is quadruple the ratio of $c$ to $e$, for the reason that the ratio of $q$ to $p$ is composed of the ratio of $d$ to $b$ and $c$ to $e$, as is easily perceived from the 24th of the sixth, or the fifth of the eighth. Wherefore, since the ratio of $d$ to $b$ is triple the ratio of $c$ to $e$, it is clear that the ratio of $q$ to $p$ is quadruple the ratio of $c$ to $e$. I say the same of the other powers, taking always $d$ and $b$ for two squared-squares, or two first related quantities, or by any other axiom.
A geometric diagram consisting of a larger rectangle and a smaller rectangle sharing a base line. The vertices and segments are labeled with the letters c, d, q, e, b, and p. Labels d and b are inside the rectangles, and q and p are below them.
WHY, when we divide a power by a power, is the resulting root that which comes from the division of one root by the other?
Let there be, for example, two lines $b. q.$ and $f. g.$, which signify two roots of any power, and let us grant them to be the roots of two squares, and let the square of $b. q.$ itself be divided by the square of $f. g.$ itself; and let the square root of the result be $d. q.$, while the linear unit is $i. g.$ I say that $d. q.$ itself is the result of the division of $b. q.$ by $i. g.$ [sic: $f. g.$]. For it is clear from the definition of division delivered in the new theorem that the square
then of $d. q.$ itself is such a part of the square of $b. q.$ itself as the square of $g. i.$ is of the square of $f. g.$ We know, moreover, from the 19th of the sixth, or the eleventh of the eighth, that the ratio of the square of $b. q.$ to the square of $d. q.$ is double the ratio of $b. q.$ to $d. q.$, their roots (for in cubes it would be triple, and in squared-squares quadruple, and so on from the preceding theorem). I say the same thing of the powers of $f. g.$ and $i. g.$ with respect to the roots $f. g.$ and $i. g.$ Whence, since the ratio of the power of $b. q.$ to that of $d. q.$ is equal to the ratio of the power of $f. g.$ to that of $g. i.$, we shall clearly recognize from common knowledge that the simple ratios are equal to each other—namely, that which is $b. q.$ to $d. q.$ is equal to that which is $f. g.$ to $i. g.$; thus it follows from the definition of division that $d. q.$ is the result of the division of $b. q.$ by $f. g.$
Two horizontal line segments labeled with points. The upper segment has points b, d, and q. The lower segment has points f, i, and g. Dotted lines suggest a proportional relationship between the segments.
WHY is the product of two square roots the square root of the product of the two squares together?
For the sake of which matter, let there be two squares $d. a.$ and $n. o.$ joined together, as appears in the figure below, yet in such a way that the angle $a. n. u.$ is a right angle; wherefore, from the fourteenth of the first, the two sides $n. c.$ and $n. a.$ are joined directly to each other, as also are the other two sides $n. u.$ and $n. d.$ If one then considers $a. u.$, the product of $a. n.$ into $n. u.$—namely, of the two roots of the squares together—there will be given from the first of the sixth, or the eighteenth of the seventh, the product $a. u.$ as a mean proportional between the square $a. d.$ and $u. c.$; and if we consider these three surfaces to be three numbers, it will be clear from the twenty-first of the seventh that the product of $a. u.$ into itself—namely, the square $a. u.$—is equal to the product of $a. d.$ into $u. c.$, from which the evidence of the proposition follows.
A geometric diagram of a large square divided into four quadrants. The vertices are labeled a, n, u, and c. Inside, a smaller square or rectangle is formed with points labeled d and e.
By what reasoning the same thing can be known in cubes.
Let there be cube $l. b.$ and cube $o. p.$, the product of which is $u. g.$, which I assert to be a cube, although Euclid proves the same in the 4th of the ninth; I shall demonstrate that its root is of a number equal to the number $m. q.$, which $m. q.$ is the product of $m. c.$ itself into $e. q.$, the root of the proposed cubes. For it is clear from the preceding theorem $m$.
A complex 3D perspective diagram illustrating the multiplication of cubes. On the left, a smaller cube labeled m-l-i-b and another shape below it labeled m-p-q. On the right, a larger cube or rectangular prism is shown in perspective with vertices labeled a, z, u, c, k, s, t, f, and g. Dotted lines indicate internal structure and projections.