# What is Cardano trying to say in this passage of his Ars Magna Arithmeticæ?

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.

• 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 $k=c-a h^2=c- \frac {b^2}{4a}$. Nov 28, 2023 at 11:33
• 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 $(x-5)^2+15^2=0$. Nov 28, 2023 at 11:40
• Sorry... $(x-5)^2+15$ Nov 28, 2023 at 11:53
• For a modern edition with notes, see Girolamo Cardano, Ars Magna or The Rules of Algebra, page 219. Nov 28, 2023 at 12:30
• 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. Nov 28, 2023 at 12:34