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subbie algebra April 2013

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Square Roots of Negative Numbers? The most commonly occurring application problems that require people to take square roots of numbers are problems which result in a quadratic equation. ex. Determine the dimensions of a rectangle with a perimeter of 9 cm and an area of 5 cm 2. ex. Determine the dimensions of a rectangle with a perimeter of 5 cm and an area of 2 cm 2. Thus, having to take the square root of a negative number in this context means that such a rectangle does not exist.

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Motivation: The Cubic Formula The first people who worked on the cubic formula lived in Persia (modern day Iran). Omar Khayyám (1048-1131) Sharaf al-Dīn al-Tūsī (1135-1213) Both of these mathematicians discovered partial solutions, but did not fully develop the cubic formula.

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Motivation: The Cubic Formula From about 1150 to about 1600, in Europe, especially Italy, it was common to have mathematical competitions in which someone would give equations for competitors to solve. The winner of these competitions received a monetary award.

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Motivation: The Cubic Formula Scipione del Ferro discovered a partial solution to the cubic formula. His solution worked for equations of the form x 3 + mx = n for positive integers m and n. He kept his solution secret until just before he died, when he shared his solution with Antonio Fiore in 1526.

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Motivation: The Cubic Formula In 1530, Niccolò Tartaglia challenged Antonio Fiore to a mathematics competition. The equations that Fiore had to solve were of the form x 3 + mx 2 = n instead of x 3 + mx = n. Tartaglia won the competition.

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Motivation: The Cubic Formula Niccolò Tartaglia had worked on a system for solving cubic equations. After his victory in 1530, he developed it further. Gerolamo Cardano was interested in mathematics, but not the competitions. When he realized that Tartaglia must have a system for solving certain types of cubic equations, Cardano asked him for the method. In 1539, Tartaglia shared only a portion of his knowledge of the cubic formula (in the form of a poem) with Cardano on the condition that he keep it secret.

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Motivation: The Cubic Formula Gerolamo Cardano and one of his students, Lodovico Ferrari, work on the other cases to which Niccolò Tartaglia did not provide a solution. They solve all of the cases and thus completely discover the cubic formula. In 1545, Cardano discovers Scipione del Ferros prior work with cubic equations… …and felt justified in publishing his complete solution in the book Ars Magna. Later that year, Tartaglia takes part in a mathematics competition that Ferrari wins. He then realizes that the secret is out.

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The Cubic Formula in Action Solve x 3 – 7x + 6 = 0. Plugging into the cubic formula yields It turns out that this solution is x = 2.

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Complex Numbers: A Whole New World Even after the discovery of the cubic formula, it took some time for mathematicians to accept and understand complex numbers. René Descartes work, especially on the Cartesian coordinate system, sped up the process of mathematicians learning and understanding complex numbers.

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Are There More Numbers? In 1799, Carl Friedrich Gauss proved the Fundamental Theorem of Algebra. Every nonconstant polynomial with complex coefficients has a complex root. In other words, for the purpose of solving polynomial equations, there is no need to hunt for more new numbers.

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Are There More Formulas? Lodovico Ferrari discovered the quartic formula in 1540. It appeared in Gerolamo Cardanos Ars Magna in 1545. In 1799, Paolo Ruffini wrote a five hundred page proof attempting to show that the quintic formula does not exist. His contemporaries did not accept his proof. In 1824, Niels Abel proved that the quintic formula does not exist. His proof, which was only six pages long, is correct. In 1832, Évariste Galois found and proved the conditions under which a polynomial equation is solvable by radicals.

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Further Reading Katscher, Friedrich. (2011). How Tartaglia Solved the Cubic Equation. In The Mathematical Association of America (MAA) Mathematical Sciences Digital Library. Retrieved April 24, 2011 from http://mathdl.maa.org/mathDL/46/?pa=content&sa=view Document&nodeId=2433&pf=1 http://mathdl.maa.org/mathDL/46/?pa=content&sa=view Document&nodeId=2433&pf=1 Rotman, Joseph. (1995). A First Course in Abstract Algebra. Upper Saddle River, NJ: Prentice Hall. Rotman, Joseph. (1998). Galois Theory. New York, NY: Springer.

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Polynomial Function f(x) = a o x o + a 1 x 1 + a 2 x 2 + …. + a n x n Equality: f (x) = g (x) if for all a i = b i that is the coefficients are equal.

Polynomial Function f(x) = a o x o + a 1 x 1 + a 2 x 2 + …. + a n x n Equality: f (x) = g (x) if for all a i = b i that is the coefficients are equal.

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