Geronimo Cardano: thief, cheat and mathematician

Geronimo Cardano (1501-1576), also known in English as Jerome Cardan,  is best known for his Ars Magna, in which he presents a method for solving cubic equations he stole from Tartaglia. In 16th century Italy solving cubic equations was a way to riches, fame, and professorships. This may seem ridiculous, but as Boyer and Merzbach say in their excellent History of Mathematics, p. 287 (J. Wiley, New York 1991),

The solution of cubic and quartic equations was perhaps the greatest contribution to algebra since the Babylonians, almost four millennia earlier, had learned to complete the square for quadratic equations.

The reasons for that are clear: a new level of ingenuity was needed (see Tartaglia’s method below), and by accident it led Raffael Bombelli (ca. 1526–1573) to discover complex numbers.

About Cardano, no-one can better Boyer and Merzbach (p. 284):

By birth illegitimate, and by habit an astrologer, gambler and heretic, Cardan nevertheless was a respected professor at Bologna and Milan, and ultimately was granted a pension by the pope. One of his sons poisoned his own wife, the other son was a scoundrel, and Cardan’s secretary Ferrari probably died of poison at the hands of his own sister. Despite such distractions…

Let us see how Tartaglia’s method works. First of all, we can always assume that the cubic equation we want to solve is of the form x^3+px=q. This is because if we consider x^3+bx^2+cx=d and make a linear change of variable x= y-\alpha, we have

(y-\alpha)^3+b(y-\alpha)^2+c(y-\alpha)=d.

So

y^3+(b-3\alpha)y^2+(3\alpha^2-2b\alpha+c)y=d+c\alpha-b\alpha^2+\alpha^3.

Now choose \alpha=b/3, set p= 3\alpha^2-2b\alpha+c and q=d+c\alpha - b \alpha^2+ \alpha^3, to get

y^3 + p y = q.

To show how Tartaglia’s method works, consider x^3+12x = 32 for simplicity. The brilliant idea is to look for x in the form x=u-v. Since we are looking for just x and there is an infinite number of pairs u, v, the difference of which will be x, we have freedom in the choice of u and v; we’ll choose them cunningly.

So far we have

x^3+12x=32 \leftrightarrow (u-v)^3+12(u-v)=32.

Hence

u^3-3u^2 v+3uv^2 - v^3 + 12(u-v)= 32, \hbox{ or } 3(u-v)(uv-4) + (u^3-v^3) = 32.

Now choose uv=4 and you get  u^3-v^3=32. But since v= 4/u, this gives you an equation in u only:

u^6 = 32 u^3 +64,

which can be solved for u^3 and hence for u (notice a difficulty?) Here u^3= \sqrt{320} + 16, so v^3=\sqrt{320}-16 and hence

x = \left(\sqrt{320}+16\right)^{1/3} - \left(\sqrt{320}-16\right)^{1/3}.

And now prove that the above horrible expression actually says that x=2!

(The only way I see to prove that x=2 is actually to go back to the original equation and show by a different way that x=2 solves it… I suspect that were I a good Renaissance mathematician I might be able to spot it by writing it as follows:

x = 2[ (\sqrt{5}+2)^{1/3} - (\sqrt{5}-2)^{1/3} ] = 2[(\sqrt{5}+2)^{1/3} - (\sqrt{5}+2)^{-1/3} ],

and — being a good Renaissance mathematician — I know already that \phi^3 = 2+\sqrt{5} where \phi=(\sqrt{5}+1)/2 satisfies the equation \phi -1/\phi = 1, because of course I’m used to seeing \phi, the “golden ratio”, all over the place. But I can’t help feeling there must be a way that doesn’t rely on spotting an old friend among the surds.)

An even worse problem is uncovered solving x^3=15x+4: you get

x = \left(\sqrt{-121}+2\right)^{1/3} - \left(\sqrt{-121}-2\right)^{1/3}.

This is what led Bombelli to introduce i=\sqrt{-1} and to elaborate rules for manipulating i that allow you to prove that \left(\sqrt{-121}+2\right)^{1/3} = \left(11i+2\right)^{1/3} has as one possibility 2+i and that \left(\sqrt{-121}-2\right)^{1/3} =\left(11i-2\right)^{1/3} has as one possibility i-2, so that saying

\left(\sqrt{-121}+2\right)^{1/3} - \left(\sqrt{-121}-2\right)^{1/3} =4

makes sense.

It is a curious fact that the first introduction of the imaginaries occurred in the theory of cubic equations, in the case where it was clear that real solutions exist though in an unrecognizable form, and not in the theory of quadratic equations, where our present books usually introduce them.

(D. J. Struik, A Concise History of Mathematics, Dover Publications, New York, 1987, p. 86.)

A final tip from Cardano’s Liber de Ludo Aleae (Book on Games of Chance), the first publication on what we now call probability theory. If you’re betting on whether a particular card will turn up when you cut the deck at “random”, your chances of winning can be significantly improved by rubbing the card beforehand with soap. Don’t say these articles never teach you anything useful!

(MG/DP)

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One Response to Geronimo Cardano: thief, cheat and mathematician

  1. Pingback: Puzzle: a problem with polynomials | Degree of Freedom

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