X Transcendental
Algebraic numbers are solutions of algebraic equations p(x) = a0 + a1x1 + a2x2 + ... + anxn = 0 with integer coefficients ak and non-negative powers k. n 0 is the degree of the polynomial. Every rational number is an algebraic number: p(x) = u - vx. Examples 2 + x = 0 x = -2 2 - 3x = 0 x = 2/3 2 - x2 = 0 x1 = -2 , x2 = 2 Non-algebraic 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. ... A transcendental number has no degree.
Proof of irrationality: x is not a solution of a polynomial equation of degree n = 1: a0 + a1x = 0 with a0, a1 otherwise x = -a0/a1 . Proof of trancendence: x is not a solution of a polynomial equation a0 + a1x + ... + anxn = 0 with an 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.
Well-known by his "Liouville's-theorem" Joseph Liouville (1809-1882) 1833 Professor at Paris. Well-known by his "Liouville's-theorem" about the phase volume: DxDv = const. Found 1851 the first transcendental number . t1 = 0,1 t2 = 0,11 t3 = 0,110001 t = 0,110001...0001000...0001000...0001000... 1 2 6 24 120 720 t is irrational, because it is not periodic.
Liouville's theorem (1844) Example: 2 = 1.414... n = 2
The minimum polynomial für 2 is p(x) = 2 - x2 It has no rational zero u/v. Otherwise the linear factor (x - u/v) could be separated and 2 had a minimum polynomial of lower degree. Examples -2 + 2x + x2 - x3 = (2 - x2)(x - 1) Roots are: x1 = 1, x2 = -2, x3 = 2 1 + x3 = (1 - x + x2)(x - (-1)) Roots are:
Liouville (1851): 3 m
Charles Hermite (1822-1901) proved 1873 e is transcendental. a0 + a1e + a2e2 +...+ anen 0 Carl Louis Ferdinand von Lindemann (1852-1939) proved 1882 p is transcendental. Using Hermite‘s idea, the transcendence of p follows from eip + 1 = 0. a0 + a1p + a2p2 +...+ anpn 0 A polynomial curve never crosses the abscissa in x = e or x = p.
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) and Hippocrates of Chios ( 450) were among the first to consider the problem. Since 1755 the French academy of sciences did no longer accept papers about squaring the circle. (Alas they also refused to accept papers about stones falling from the sky.)
Johann Heinrich Lambert (1728-1777) showed in 1761 that p cannot be rational. But he could not exclude an irrational of degree 2 which would be constructible by ruler and compasses. 1900 David Hilbert (1862-1943) mentioned the 23 most important problems of mathematics; no. 7 was the proof of transcendence of 2√2 and similar numbers. Gelfand showed in 1929 i2i . eip = -1 = i2 (eip)i = (i2)i e-p = 1/ep = i2i. In 1934 Gelfand and Schneider showed independently of each other that 2√2 and similar numbers are transcendental. Alexander Gelfand (1906-1968) Theodor Schneider (1911-1988)
Appendix
Hermite's great idea is this: Construct a polynomial that contains e among its zeros and whose derivatives are integers where at least one of the integers prohibits that the sum of all of them vanishes. The following (too simple) polynomial shows this principle: All derivatives of order k < p -1 and k > 2p are zero. For k = p, p + 1, ... the derivatives at zeros of the function are integers divisible by p. Only for k = p - 1 the integer derivative is not divisible by p, in particular if p is a prime number larger than all integer constants. So the sum of all these numbers is an integer mp + n 0 with absolute value at least 1 since n is not a multiple of p. Finally the sum is of the form Bp/(p - 1)! becoming arbitrarily small for p chosen large enough, independent of B. So its absolute value is less than 1.
Assume e is algebraic with minimum polynomial P(x) of degree n, i.e., because for a0 = 0 the minimum polynomial has degree n - 1.
{ All integrals of P1 are divisible by m! Example: Substituting y = x - k yields the interval of integration [0, ]. Then {
For k = 0 the integrand with least exponent is (except the sign) a0n For k = 0 the integrand with least exponent is (except the sign) a0n!xme-x. This supplies for infinitely many m (namely for all m > a0n!) an integral divisible by m! but not by (m+1)!. All other terms contain x in higher power and are divisible by (m+1)!. For k ≥ 1 all terms are multiples of (m+1)!. Hence P1 is for infinitely many m not divisible by (m+1)!. For m large enough, P1 is divisible by m! but not by (m+1)!, that means P1 is an integer with |P1| ≥ 1. P2 becomes arbitrarily small, in particular |P2| ≤ 1. On the other hand P1 + P2 = 0. Hence e is not algebraic.