It is well known that Cardano considered the problem of "dividing 10 into two parts the product of which is 40" in his Ars Magna. This problems leads to the complex solutions $5+ \sqrt{-15}$ and $5- \sqrt{-15}$, which Cardano would write as $5\tilde{p}R\tilde{m}15$ and $5\tilde{m}R\tilde{m}15$, respectively.

For this problem, Cardano supplemented his algebraic reasoning with a geometric construction of "completing the square" (see Richard Witmer's english translation of the Ars Magna, p. 219). With this construction, one has to imagine a negative area, that of substracting $40$ units of area to the square $AD$:

However, Cardano seemingly thinks the problem with this construction isn't the negative area, but the fact that $AD$ is an area while $40$ is a length:

Yet the nature of AD is not the same as that of 40 or of AB, since a surface is far from the nature of a number and from that of a line, though somewhat closer to the latter. This truly is sophisticated, since with it [i. e., with $\sqrt{-15}$] one cannot carry out the operations one can in the case of a pure negative and other [numbers].

Cardano also treats complex numbers in his other lesser-known book Ars Magna Arithmeticæ. He considers the problem of "dividing $1$ in two parts whose product is $3$". This is the same as solving the equation $x^2-x+3=0$. The solutions are complex: $$x=\frac{1}{2}\pm \sqrt{\frac{1}{4}-3}=\frac{1}{2}\pm \sqrt{-\frac{11}{4}}.$$ Cardano writes this solutions as $\frac{1}{2}$.$\tilde{p}$.R.v.$\frac{1}{4}$.m.3. and $\frac{1}{2}$.$\tilde{m}$.R.v.$\frac{1}{4}$.m.3. (see image below). This is done on page 374, volume IV of Cardano's Opera omnia (which contains the Ars Magna Arithmeticæ).

My question is: what is Cardano trying to say in the following excerpt of Ars Magna Arithmeticæ (p. 374)? Does it have anything to do with his confusion about the different "natures" of $AD$ and $40$? He adds the word "quadrati" to the solutions ($\frac{1}{2}$.$\tilde{p}$.R.v.$\frac{1}{4}$quadrati.m.3.), but I don't see the reason for doing this. It seems as if he added a variable $y^2$ to the solutions, like $$x=\frac{1}{2}\pm \sqrt{\frac{1}{4}y^2-3}.$$

I don't have a clue why Cardano does this. I have only found a paper of Veronica Gavagna about the Ars Magna Arithmeticæ, but she doesn't talk much about this.

enter image description here

  • 1
    $\begingroup$ The "technique" is that of Completing the square: for $a x^2+b x +c=a(x-h)^2+k$ and we have $h=- \frac {b}{2a}$ and $k=c-a h^2=c- \frac {b^2}{4a}$. $\endgroup$ Nov 28, 2023 at 11:33
  • 1
    $\begingroup$ In Cardan's case we have $a=1, b=-10$ and $c=40$ that gives: $h=5$ and $k=40-25=15$. Thus, the equation will be transformed into $(x-5)^2+15^2=0$. $\endgroup$ Nov 28, 2023 at 11:40
  • 1
    $\begingroup$ Sorry... $(x-5)^2+15$ $\endgroup$ Nov 28, 2023 at 11:53
  • 2
    $\begingroup$ For a modern edition with notes, see Girolamo Cardano, Ars Magna or The Rules of Algebra, page 219. $\endgroup$ Nov 28, 2023 at 12:30
  • 2
    $\begingroup$ Regarding the statement "since a surface is far from the nature of a number and from that of a line", this reflects ancient Greek view about the relation between geometry and arithmetic. If a line is measured by a number, a surface must be measured by a "squared" number. Thus, like in geometry the "equation" $l+S$ does not make sense, if we "translate" it into numbers we have to write something like $l \times 1 + S$. Only with Descartes' Geometry this "mental restriction" will be removed. $\endgroup$ Nov 28, 2023 at 12:34


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.