Gerolamo Cardano Born: 1501 Died: 1576 Milan, Italy

The story of Cardano comes in the time of the renaissance The story of Cardano comes in the time of the renaissance. Due to the innovation of the printing press ideas are being shared all over europe. This also includes mathematical ideas. One of the most significant results of Cardano's work is the solution to the general cubic equation [2 p 133]. This is an equation of the form: ax3 + bx2 + cx + d = 0 which Cardano was able to find solutions for by extracting certain roots [3]. Before we begin with the story of Cardano, we must explain some history associated with the solution of the cubic. Although the solving of equation goes back to the very roots (no pun) of mathematics this segment of the story begins with Luca Pacioli (1445-1509). Paciloi authored a work Summ de Arithmetica, in which he summarized the solving of both linear and quadratic equations. This was a significant work because the algebra of the day was still in a very primitive form. The symbolism of today is not done at this time, but a written description of equations is used. Pacioli ponders the cubic and decides the problem is too difficult for the mathematics of the day [2 p 134]. …when that which is unknown by that which is know is combined with a second unknown to produced a third known value, be it resolved that the difference of the produced known and combined known values is measured by the original known… Scipione del Ferro (1465-1526) continues the work that Pacioli had begun, but is more optimistic. Del Ferro is able to solve the "depressed cubic", that is a cubic equation that has no square term.

The depressed cubic that del Ferro works with is of the form x3+mx=n where m and n are treated as known constants. The solution of the cubic equation is kept secret by del Ferro so that he has it to use in case his position is ever challenged. The solution of the cubic is told to Antonio Fior a student of del Ferro on del Ferro's death bed [2 pp 134-136]. Niccolo Fontana (1499-1557) better know as Tartaglia "The Stammer" (he got his nickname because he suffered a deep sword wound from a French soldier so that he could not speak very clearly) challenges Fior by each of them exchanging 30 problems. Fior is a very arrogant but not so talented mathematician. Fior gives Tartaglia 30 depressed cubics to solve. This is very "high stakes" at this time because Tartaglia will either get a 0 or a 30 depending if he can figure out the secret. Tartaglia a very gifted mathematician was able to find the solution to the depressed cubic after some struggle [2 pp 134-136]. This bit of history behind us, Cardano enters the picture of the cubic equation. Before we begin with the cubic we will make some biographical comments about Cardano. Cardano was the illegitimate son of a very prominent father. His father was a consultant to Leonardo da Vinci. Cardano's illegitimacy had a huge impact throughout his life. His mother was given some poisons in order to attempt to induce an abortion, causing Cardano to suffer from a rash of physical ailments his entire life. Cardano would often inflict physical pain on himself because he said it would bring him relief when he stopped. He studied medicine at Padua, but was not able to practice in Milan because of his illegitimacy. Only later was he allowed to practice medicine after authoring a work on corrupt doctors that was popular among the people of Milan [2 pp 135-137].

Cardano's personal life was both strange and tragic Cardano's personal life was both strange and tragic. He was a mystic who believed in visions and dreams. He married because of a dream he had. His wife died at a very young age. He had two sons Giambattista and Aldo both of which Cardano had great hopes for since both were legitimate and would not have to face what Cardano did. Both sons ended up being a big disappointment, Giambattista killed his wife because of an affair which produced children and was put to death, Aldo was imprisoned as a criminal [2 pp 137-139]. Later in life Cardano was jailed on charges of heresy because of his anti-Christian views. Due to influence from his friends he is released and even gets a pension from the Pope to live out the rest of his life [2 p 140].

Back to the cubic equation Back to the cubic equation. Cardano learns of Tartaglia's discovery of the depressed cubic and asks him to share the discovery. Tartaglia agrees on the condition that Cardano does not publish it until after Tartaglia has published this method. Using more modern notation the following is a paraphrase of Cardano's view of Tartaglia's solution to the cubic. We keep in mind that geometry is the basis of mathematical thought at the time. By dissecting a cube he finds the following relation. The mathematical methods are still geometric in nature since they are trying to apply the same techniques that were used by the Greeks. Setting x = t-u, m=3tu and n= t3-u3 we get the equation Tartaglia solved (i.e. the depressed cubic equation): x3 + mx = n. The issue is to find an expression for t and u in terms of m and n.

Remember: Solve the first equation for u and substitute into the second. Letting: Gives: Multiply the entire equation by t3. You then get a quadratic equation with the variable of t3. This equation can then be solved using the quadratic formula. Taking a cubed root gives the value for t. Remember that x (the solution) is t-u.

As a concrete example Cardano solved the prototype equation x3+6x=20 As a concrete example Cardano solved the prototype equation x3+6x=20. This is often how many mathematical ideas were communicated by demonstrating how a general idea could be applied to a specific example. It is important to note that the equation x3+6x=20 was considered different than x3+20=6x. The concept of a negative number was still not completely understood to Cardano and others. This led to his work having many more "cases" to consider than was needed. The two starting equations. Solving and substituting. Put in quadratic form. Apply quadratic formula. Solve for t. Solve for x.

With this knowledge Cardano applies this to the general cubic equation by showing how to reduce any cubic equation to a depressed cubic. This is done using the following substitutions. For the equation: this gives a depressed cubic equation in the variable y. Solve that equation for y using Tartaglia's method to get y then substitute back to get the value for x [2 pp 142-149]. Cardano having solved this is frustrated because of the promise he made to Tartaglia he can not share his result. In 1543 Cardano travels with Ludovico Ferrari (1522-1565) his student to Bologna. In Bologna he discovers some unpublished papers by del Ferro in which he finds the solution to the depressed cubic. Cardano feels that now he can make his solution public. In 1545 Cardano publishes Ars Magna giving the solution of the cubic in which he credits both del Ferro and Tartaglia. This angers Tartaglia and he and Ferrari (who had great respect for Cardano) engaged in public debate over the matter [2 pp 137-140]. Ferrari grows to be a very good mathematician and is able to solve the general fourth degree equation (i.e. ax4+bx3+cx2+dx+e=0). It is interesting to note that it takes some time before people are able to prove that no such formula exists for degree five and higher equations [1 pp 304-306].   [1] Burton, DavidM. The History of Mathematics an Introduction. Boston: McGraw Hill 1999. [2] Dunham, William. Journey Through Genius The Great Theorems of Mathematics. New York: John Wiley & Sons 1990. [3] Web site: MacTutor History. http://turnbull.mcs.st-and.ac.uk/~history/(9/2000).

SOLVING A QUADRATIC BY DEPRESSING IT The method of Tartagila and Cardano can also be applied to quadratics. Example: Solve 2 𝑥 2 −8𝑥+3=0 by depressing it then extracting a root. In the case of a quadratic 𝑎 𝑥 2 +𝑏𝑥+𝑐 the depression can be carried out by substituting for the value 𝑥 the expression 𝑥=𝑦− 𝑏 2𝑎 . In this case substitute the expression 𝑦− −8 2∙2 =𝑦+2 for 𝑥. Compare with modern day quadratic formula: 2 𝑦+2 2 −8 𝑦+2 +3=0 2 𝑦 2 +4𝑦+4 −8𝑦−16+3=0 2 𝑦 2 +8𝑦+8−8𝑦−13=0 2 𝑦 2 −5=0 𝑦 2 = 5 2 𝑦=± 5 2 = ± 10 2 Substituting back you get: 𝑥= ± 10 2 +2= 4± 10 2 𝑥= 8± −8 2 −4∙2∙3 2∙2 𝑥= 8± 64−24 4 𝑥= 8± 40 4 𝑥= 8±2 10 4 𝑥= 4± 10 2

COMBINING METHODS Solve: 𝑥 3 +3 𝑥 2 −10𝑥+8=0 First to apply Cardano’s method we make the substitution 𝑥=𝑦− 𝑏 3𝑎 or in this case 𝑥=𝑦− 3 3∙1 =𝑦−1 in order to depress the cubic. In order to multiply this out we need to remember the algebra fact: 𝑐+𝑑 3 = 𝑐 3 +3 𝑐 2 𝑑+3𝑐 𝑑 2 + 𝑑 3 . 𝑦−1 3 +3 𝑦−1 2 −10 𝑦−1 +8=0 𝑦 3 −3 𝑦 2 +3𝑦−1+3 𝑦 2 −2𝑦+1 −10𝑦+10+8=0 𝑦 3 −3 𝑦 2 +3𝑦−1+3 𝑦 2 −6𝑦+3−10𝑦+10+8=0 𝑦 3 −13𝑦+20=0 𝑦 3 −13𝑦=−20 This depresses it and puts it in the correct form. 3𝑡𝑢=−13 and 𝑡 3 − 𝑢 3 =−20 𝑢= −13 3𝑡 and then 𝑡 3 − −2197 27 𝑡 3 =−20 𝑡 6 + 2197 27 =−20 𝑡 3 𝑡 6 +20 𝑡 3 + 2197 27 =0 𝑡 3 = −20± 400−4∙1∙ 2197 27 2 𝑡= 3 −20± 2012 27 2 = 3 −10± 503 27 𝑢= −13 3 3 −10± 503 27 𝑥= 3 −10± 503 27 − −13 3 3 −10± 503 27 -1