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X Transcendental ¥. Algebraic numbers are solutions of algebraic equations p(x) = a 0 + a 1 x 1 + a 2 x 2 +... + a n x n = 0 With integer coefficients.

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Presentation on theme: "X Transcendental ¥. Algebraic numbers are solutions of algebraic equations p(x) = a 0 + a 1 x 1 + a 2 x 2 +... + a n x n = 0 With integer coefficients."— Presentation transcript:

1 X Transcendental ¥

2 Algebraic numbers are solutions of algebraic equations p(x) = a 0 + a 1 x 1 + a 2 x 2 +... + a n x n = 0 With integer coefficients a and nonnegative integers n. n is the degree of the polynomial. Every rational number is an algebraic number p(x) = u - vx Examples2 + x = 0  x = -2 2 - 3x = 0  x = 2/3 2 - x 2 = 0  x 1 = -, x 2 = 1 + x 3 = 0  x 1 = -1, x 2 = - i, x 3 = + i Nonalgebraic numbers are called transcendental numbers. The degree of a rational number is n = 1 The degree of a square root is n = 2 The degree of a cubic root is n = 3... The degree of a transcendental number is not finite. 1212 1212

3 Proof of irrationality: x is not a solution of a polynomial equation of degree n = 1: a 0 + a 1 x = 0mita 0,a 1    otherwise x = -a 0 /a 1 Proof of trancendentality: x is not a solution of a polynomial equqtion a 0 + a 1 x +... + a n x n = 0with a    of arbitrary degree n <  Although the rational numbers seem to cover every point of the real line, there are other numbers, the algebraics. Although the algebraics seem to cover every point of the real line, there are other numbers, the transcendentals. All transcendental numbers are irrational.

4 Transzendente Zahlen Joseph Liouville (1809 - 1882 ) 1833 Professor at Paris Well-known by his Liouville‘s-theorem:  x  v = const. Found 1844 the first transcendental number  =  n =  1 = 0,1  2 = 0,11  3 = 0,110001  = 0,110001...0001000...0001000...0001000... 1 2 6 24 120 720 (irrational, because not periodical)

5 Charles Hermite (1822 - 1901) proved 1873 e is transcendental. a 0 + a 1 e + a 2 e 2 +...+ a n e n  0 Carl Louis Ferdinand von Lindemann (1852 - 1939) proved 1882  is transcendental. a 0 + a 1  + a 2  2 +...+ a n  n  0 Lindemann showed, using Hermite‘s idea,  1 e  1 +... +  n e  n ≠ 0 For different algebraic numbers  1,...,  n and algebraic numbers  1,...,  n ≠ 0. From e i  + 1 = 0 follows the transcendence of . A polynomial curve never crosses the abscissa in x = e or x = .

6 The ancient problem of squaring the circle has been finally settled in 1882. It appeared in the 5th century BC and soon became fashionable. In 414 it has been so popular already that Aristophanes (445 - 385) in „The Birds“ talks about circle-squarers as of people who try the impossible. Anaxagoras (500 - 428) – in prison – and Hippokrates of Chios (ca. 450) were among the first to consider the problem. Since 1755 the French academy of sciences did no longer accept papers on squaring the circle. Johann Heinrich Lambert (1728 - 1777) showed in 1761 that  cannot be rational 1900 David Hilbert (1862 - 1943) mentioned the 23 most important probelms of mathematic; no. 7 was the proof of transcendence of 2 √2 and similar numbers Alexander Gelfand (1906 - 1968) 1929: e i  = -1 = i 2  e -  = 1/e  = i 2i   1934: Alexander Gelfand 1934: Theodor Schneider (1911 - 1988) 2 √2 and similar numbers  

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