Book of Various Mathematical and Physical Speculations

...the unit indeed, to whose difference $n.c$ I say it is equal, let it be $a.i$. For it is clear in the first place that there is the same proportion of $a.e$ to $a.c$ as there is of $u.e$ to $a.i$ from the definition of division, and that the same proportion exists of $a.u$ to $a.n$ as $u.e$ to $a.i$; whence from the 11th of the fifth [book of Euclid], as $a.e$ is to $a.c$, so $a.u$ will be to $a.n$; and from the 19th of the same, as $u.e$ is to $n.c$, so $a.e$ will be to $a.c$. But $u.e$ was thus to $a.i$. Therefore, from the aforesaid 11th of the fifth, as $u.e$ is to $n.c$, so it will be to $a.i$. Wherefore, from the 9th of the same, $n.c$ will be equal to $a.i$, and for that reason $n.c$ will likewise be a unit.

A horizontal line segment with tick marks labeled a, i, n, o, u, e. A bracket sits beneath the segment between points n and o.

THEOREM XXIII.

WHY, when any two numbers are divided by one another, and the results are multiplied together, is the product always a superficial unit? Namely from the 20th of the seventh, because the linear unit is always the mean proportional between the two results. Which may be observed thus.

Let the two proposed numbers be represented by $b.p$ and $b.d$, divided by one another; let the result of $b.p$ divided by $b.d$ be $b.n$; then let the result of $b.d$ divided by $b.p$ be $b.a$; and let $b.t$ be the unit of $b.p$, and $b.e$ the unit of $b.d$, from which $b.t$ will be equal to $b.e$.

Now from the definition of division, there will be the same proportion of $b.p$ to $b.n$ as $b.d$ to $b.e$, and the proportion of $b.d$ to $b.a$ as $b.p$ to $b.t$. But since $b.p$ is to $b.n$ as $b.d$ to $b.e$, by permutation $b.p$ will be to $b.d$ as $b.n$ is to $b.e$, that is, to $b.t$; and since $b.d$ is to $b.a$ as $b.p$ to $b.t$, by permutation $b.d$ will be to $b.p$ as $b.a$ is to $b.t$. Wherefore, inversely, $b.p$ will be to $b.d$ as $b.t$ is to $b.a$. But $b.n$ was to $b.t$ as $b.p$ was to $b.d$. Therefore, from the 11th of the fifth, as $b.n$ is to $b.t$, so $b.e$ will be to $b.a$. But it has been said that $b.e$ and $b.t$ are entirely the same. Wherefore, from the 20th of the seventh, the truth of the proposition will be made known.

Two parallel horizontal line segments. The upper segment is marked with points b, f, n, p. The lower segment is marked with points b, a, q, d. A small vertical bracket or ligature symbol connects parts of the lines on the right side.

THEOREM XXV.

THIS same thing will also be evident by this other way. Let those two numbers be represented by $o$ and $u$, divided by one another; let the result of $o$ by $u$ be $e$, and the result of $u$ by $o$ be $x$; let the unit indeed be represented by $i$; however, let these quantities be arranged toward one another in the manner written below. Thus, from the definition of division, the proportion of $o$ to $e$ will be the same as $u$ to $i$, and [the proportion of] $o$ to $i$ will be the same as $u$ to $x$. Wherefore, from the equality of proportions, as $e$ is to $i$, so it will be as $i$ to $x$; for $i$ will be the mean proportional between $e$ and $x$. Moreover, from the 20th of the seventh, the proposition will be concluded. The cause of this matter is also that the result of one division is a numerator equal to the denominator of the other division.

A small schematic diagram showing two vertical stacks of letters. On the left, i is above o, which is above e. On the right, x is above u, which is above i. Short horizontal lines separate the letters within each stack.

THEOREM XXVI.

WHY, when two numbers are divided by one another, and the results are then taken together and by this sum the sum of the squares of the said proposed

numbers is divided, does a number result equal to the product of the two first numbers together?

Let the proposed numbers be, for example, 2 and 8, which, when divided by one another, first give the integer four as one result, then a fourth part as the other result; these collected will give a sum of four integers and one fourth part. Moreover, the sum of the squares of 2 and 8 will be 68; which number indeed, divided by four and a fourth part of one, will give 16 as the result, which 16 will be equal to the product of 2 into 8.

Of which matter this will be the observation: let there be two lines $o.e$ and $o.n$, which represent the two proposed numbers, joined to one another at a right angle $o$; let their squares be $o.a$ and $o.p$; let their product be $n.e$; then let $o.t$ be the result from the division of $o.e$ by $o.n$. Let us consider these things individually (for if what we say happens in simple parts, it will consequently occur in composite ones); wherefore, from the definition of division, the same proportion of $o.e$ to $o.t$ will be given as $o.n$ to the unit, which let be $o.x$. Now let us imagine a rectangular surface $o.c$ equal to the square $o.a$; then the resulting number $c.t$ will be, as is clear, from the division of the square number $o.a$ by the number $o.t$; and the proportion of $c.t$ to $o.e$ will be the same as $o.e$ to $o.t$, from the second part of the 15th of the sixth, or the 20th of the seventh. But it has already been said that $o.e$ is to $o.t$ as $o.n$ is to $o.x$. Therefore, from the 11th of the fifth, as $c.t$ is to $o.e$, so $o.n$ will be to $o.x$. But from the 1st of the sixth, or the 18th or 19th of the seventh, the product $n.e$ is to $e.x$ as $o.n$ is to $o.x$; wherefore again the number $c.t$ will be to the number $o.e$ as the number $n.e$ is to the number $x.e$. But the number $o.e$ is identical in kind with the number $x.e$; therefore from the 9th of the fifth, the number $c.t$ will be equal to the number $n.e$.

I say the same of the square of $o.n$ itself, namely $p.o$. For if the result of $o.n$ divided by $o.e$, that is $o.i$ (the proportional corresponding to $o.t$), were joined with $o.t$, and by this sum the sum of the squares $o.a$ and $o.p$ were divided, it is evident of itself that the result would be of the same number $c.t$, and $c.t$ itself would always be the result.

So that this matter may become clearer: let us imagine the result of the square $o.p$ divided by $o.i$ (corresponding to $o.t$) to be $i.u$, which by the aforesaid way is found to be equal to the number $n.e$, and consequently equal to $c.t$; then, having imagined the rectangle $o.u$ equal to $o.p$ joined to $c$, the whole $t.u$ will be equal to the composite of the two squares $o.a$ and $o.p$. Since the number $c.t$ is changed in no way, whether from the composite $t.u$ or from the simple $o.x$, from which the truth of the proposition brings itself forward.

A large rectangular geometric diagram divided into several smaller rectangles and squares by horizontal and vertical lines. Labels for points include a, c, d, o, e, t, i, u, x, n, and p. The layout illustrates geometric proofs related to squares and rectangular areas mentioned in the text.

THEOREM XXVII.

THE ancients proposed a noble problem indeed, but one which they either did not know how to solve, or did not wish to, without algebraic operation; which is nevertheless very easy.