The Renaissance

ready calculator only, is indicated by his views on trigonometrical lines. Up to his time, the trigonometric functions had been considered always with relation to the arc; he was the first to construct the right triangle and to make them depend directly upon its angles. It was from the right triangle that Rhæticus got his idea of calculating the hypotenuse; i.e. he was the first to plan a table of secants. Good work in trigonometry was done also by Vieta and Romanus.

We shall now leave the subject of trigonometry to witness the progress in the solution of algebraical equations. To do so, we must quit Germany for Italy. The first comprehensive algebra printed was that of Lucas Pacioli. He closes his

book by saying that the solution of the equations $x^3+mx=n$, $x^3+n=mx$ is as impossible at the present state of science as the quadrature of the circle. This remark doubtless stimulated thought. The first step in the algebraic solution of cubics was taken by Scipio Ferro (died 1526), a professor of

mathematics at Bologna, who solved the equation $x^3+mx=n$. Nothing more is known of his discovery than that he imparted it to his pupil, Floridas, in 1505. It was the practice in those

days and for two centuries afterwards to keep discoveries

secret, in order to secure by that means an advantage over rivals by proposing problems beyond their reach. This practice gave rise to numberless disputes regarding the priority of inventions. A second solution of cubics was given by Nicolo of Brescia (1506(?)–1557). When a boy of six, Nicolo was so badly cut by a French soldier that he never again gained the free use of his tongue. Hence he was called Tartaglia,

i.e. the stammerer. His widowed mother being too poor to pay his tuition in school, he learned to read and picked up a knowledge of Latin, Greek, and mathematics by himself. Possessing a mind of extraordinary power, he was able to appear as teacher of mathematics at an early age. In 1530, one Colla proposed him several problems, one leading to the

equation $x^3+p{x}^2=q$. Tartaglia found an imperfect method for solving this, but kept it secret. He spoke about his secret in public and thus caused Ferro's pupil, Floridas, to proclaim his own knowledge of the form $x^3+mx=n$. Tartaglia, believing him to be a mediocrist and braggart, challenged him to a public discussion, to take place on the 22d of February, 1535. Hearing, meanwhile, that his rival had gotten the method from a deceased master, and fearing that he would be beaten in the contest, Tartaglia put in all the zeal, industry, and skill to find the rule for the equations, and he succeeded in it ten days before the appointed date, as he himself modestly says.7 The most difficult step was, no doubt, the passing from quadratic irrationals, used in operating from time of old, to cubic irrationals. Placing $x=\sqrt[3]{t}-\sqrt[3]{u}$, Tartaglia perceived that the irrationals disappeared from the equation $x^3+mx=n$, making $n=t-u$. But this last equality, together with $(\frac{1}{3}m{)}^3=tu$, gives at once $t=\sqrt{{\left(\frac{n}{2}\right)}^3+{\left(\frac{m}{3}\right)}^3}+\frac{n}{2},u=\sqrt{{\left(\frac{n}{2}\right)}^2+{\left(\frac{m}{2}\right)}^3}-\frac{n}{2}.$

This is Tartaglia's solution of $x^3+mx=n$. On the 13th of February, he found a similar solution for $x^3=mx+n$. The contest began on the 22d. Each contestant proposed thirty problems. The one who could solve the greatest number within fifty days should be the victor. Tartaglia solved the thirty problems proposed by Floridas in two hours; Floridas could