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How Cubic Equations (And Not Quadratic) Led to Complex Numbers

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How Cubic Equations (And Not Quadratic) Led to Complex Numbers How Cubic Equations (and Not Quadratic) Led to Complex Numbers J. B. Thoo Yuba College 2013 AMATYC Conference, Anaheim, California This presentation was produced usingLATEX with C. Campani’s BeamerLATEX class and saved as a PDF file: <http://bitbucket.org/rivanvx/beamer>. See Norm Matloff’s web page <http://heather.cs.ucdavis.edu/~matloff/beamer.html> for a quick tutorial. Disclaimer: Our slides here won’t show off what Beamer can do. Sorry. :-) Are you sitting in the right room? This talk is about the solution of the general cubic equation, and how the solution of the general cubic equation led mathematicians to study complex numbers seriously. This, in turn, encouraged mathematicians to forge ahead in algebra to arrive at the modern theory of groups and rings. Hence, the solution of the general cubic equation marks a watershed in the history of algebra. We shall spend a fair amount of time on Cardano’s solution of the general cubic equation as it appears in his seminal work, Ars magna (The Great Art). Outline of the talk Motivation: square roots of negative numbers The quadratic formula: long known The cubic formula: what is it? Brief history of the solution of the cubic equation Examples from Cardano’s Ars magna Bombelli’s famous example Brief history of complex numbers The fundamental theorem of algebra References Girolamo Cardano, The Rules of Algebra (Ars Magna), translated by T. Richard Witmer, Dover Publications, Inc. (1968). Roger Cooke, The History of Mathematics: A Brief Course, second edition, Wiley Interscience (2005). Victor J. Katz, A History of Mathematics: An Introduction, third edition, Addison-Wesley (2009). Amy Shell-Gellasch and J. B. Thoo, Intersecting Mathematics and History: Topics in Elementary Mathematics from an Historical Viewpoint, in preparation. <http://ms.yccd.edu/cv-and-pubs.aspx#books> Motivation: square roots of negative numbers Myth Many of today’s popular algebra textbooks introduce imaginary numbers along these lines, saying1 The complex number system allows us to solve equations such as x2 + 1 = 0 that have no real number solutions. The implication is that mathematicians became interested in complex numbers because they were unable to solve certain quadratic equations with only the real numbers. 1Elayn Martin-Gay, Beginning & Intermediate Algebra, 5th edition, p. 634. Myth buster The truth is that, as late as the 16th century, when confronted with an equation like x2 + 1 = 0 or x2 + 2x + 2 = 0; one could simply say that the equation has no solution, and that would have been the end of it. Square roots of negative numbers were considered to be “sophistic,” “impossible,” “imaginary,” and “useless” objects. Indeed, imaginary numbers were not understood as late as the 16th century beyond their being a convenient contrivance. The quadratic formula: long known Quadratic problems Ancient Egypt Berlin Papyrus 6619: The area of a square of 100 is equal to that of two smaller squares. The side of one is 2 4 the side of the other . Let me know the sides of 2 3 2 the two unknown squares[ x + ( 4 x) = 100]. Solved by the method of false position. In the solution, the square root of 1 2 16 is taken, yielding 1 4. Ancient Mesopotamia Cuneiform tablet AO 8862 (Louvre, Paris): I have multiplied the length and width so as to make the area. Then I added to the area the amount by which the length exceeds the width, obtaining 3;3. Then I added the length and width together, obtaining 27. What are the length, width, and area[ xy + x − y = 183; x + y = 27]? Often in two unknown quantities; often solved by “completing the square.” The quadratic formula: rhetorical Arabic: Abu Ja’far Muh.ammad ibn-Mus¯ ¯a al-Khw¯arizm¯ı (ca. 780–850) Al-kit¯ab al-muhtas.ar f¯ı h. is¯ab al-jabr wa’l muq¯abala (The Condensed Book on the Calculation of al-Jabr and al-Muqabala) al-Khw¯arizm¯ı (a > 0, b > 0, c > 0) Squares equal to roots ax 2 = bx Squares equal to numbers ax 2 = c Squares and roots equal to numbers ax 2 + bx = c Squares and numbers equal to roots ax 2 + c = bx Roots and numbers equal to squares bx + c = ax 2 x2 + bx = c: q q b b2 b2 b2 b2 b b ! 2 ! 4 ! c + 4 ! c + 4 ! c + 4 − 2 = x The solution is this: you halve the number of roots, which in the present instance yields five. This you multiply by itself; the product is twenty-five. Add this to thirty-nine; the sum is sixty-four. Now take the root of which, which is eight, and subtract from it half the number of roots, which is five; the remainder is three. This is the root of the square which you sought for; the square itself is nine. One square, and ten roots of the same, amount to thirty-nine dirhems. One square, and ten roots of the same, amount to thirty-nine dirhems. The solution is this: you halve the number of roots, which in the present instance yields five. This you multiply by itself; the product is twenty-five. Add this to thirty-nine; the sum is sixty-four. Now take the root of which, which is eight, and subtract from it half the number of roots, which is five; the remainder is three. This is the root of the square which you sought for; the square itself is nine. x2 + bx = c: q q b b2 b2 b2 b2 b b ! 2 ! 4 ! c + 4 ! c + 4 ! c + 4 − 2 = x S A G Al-Khw¯arizm¯ı provided two geometric demonstrations of his solution. This one is most B widely known, and is often used to introduce the method 25 D of completing the square. H b 2 x S A x 2 + bx = c b 2 x 2 x x b x 2 + 2 · x = c 2 B b b2 b2 x 2 + 2 · x + = c + 2 4 4 2 2 b2 b b b b x x + = c + 4 2 2 2 4 H x2 + c = bx: q q q b b2 b2 b2 b b2 b b2 b ! 2 ! 4 ! 4 − c ! 4 − c ! 2 − 4 − c or 2 + c − 4 = x [A] square and twenty-one in numbers are equal to ten roots of the same. Solution: Halve the number of the roots; the half is five. Multiply this by itself; the product is twenty-five. Subtract from this the twenty-one which are connected with the square; the remainder is four. Extract its root; it is two. Subtract this from the half of the roots, which is five; the remainder is three. This is the root of the square which you required, and the square is nine. Or you may add the root to the half of the roots; the sum is seven; this is the root of the square which you sought for, and the square itself is forty-nine. [A] square and twenty-one in numbers are equal to ten roots of the same. Solution: Halve the number of the roots; the half is five. Multiply this by itself; the product is twenty-five. Subtract from this the twenty-one which are connected with the square; the remainder is four. Extract its root; it is two. Subtract this from the half of the roots, which is five; the remainder is three. This is the root of the square which you required, and the square is nine. Or you may add the root to the half of the roots; the sum is seven; this is the root of the square which you sought for, and the square itself is forty-nine. x2 + c = bx: q q q b b2 b2 b2 b b2 b b2 b ! 2 ! 4 ! 4 − c ! 4 − c ! 2 − 4 − c or 2 + c − 4 = x (Handled negative numbers without fanfare.) ax2 + bx = c p p p 4ac + b2 − b c ! 4ac ! 4ac + b2 ! 4ac + b2 ! 4ac + b2 − b ! = x 2a India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term. p p p 4ac + b2 − b c ! 4ac ! 4ac + b2 ! 4ac + b2 ! 4ac + b2 − b ! = x 2a India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term. (Handled negative numbers without fanfare.) ax2 + bx = c p p p 4ac + b2 − b ! 4ac ! 4ac + b2 ! 4ac + b2 ! 4ac + b2 − b ! = x 2a India Brahmagupta (598–670): tells how to solve quadratic equations by completing the square Take the absolute number from the side opposite to that from which the square and simple unkown are to be subtracted. To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the [value of the] middle term.
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