Transcendental number

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In mathematics, a transcendental number is any real number that is not algebraic, that is, not the solution of a non-zero polynomial equation with integer (or, equivalently, rational) coefficients. It follows that all transcendental numbers are irrational. However, not all irrational numbers are transcendental; √2 is irrational but is a solution of the polynomial x² - 2 = 0.

The set of all transcendental numbers is uncountable. The proof is simple: Since the polynomials with integer coefficients are countable, and since each such polynomial has a finite number of zeroes, the set of algebraic numbers is countable. But the reals are uncountable; so the set of all transcendental numbers must also be uncountable. In a very real sense, then, there are many more transcendental numbers than algebraic ones. However, only a few classes of transcendental numbers are known and proving that a given number is transcendental can be extremely difficult.

The existence of transcendental numbers was first proved in 1844 by Joseph Liouville, who exhibited examples, including the Liouville constant:

$\sum_{k=1}^\infty 10^{-k!} = 0.110001000000000000000001000....$

in which the nth digit after the decimal point is 1 if n is a factorial (i.e., 1, 2, 6, 24, 120, 720, ...., etc.) and 0 otherwise. The first number to be proved transcendental without having been specifically constructed to achieve this was e, by Charles Hermite in 1873. In 1882, Ferdinand von Lindemann published a proof that the number π is transcendental. In 1874, Georg Cantor found the argument described above establishing the ubiquity of transcendental numbers.

See also Lindemann-Weierstrass theorem.

Here is a list of some numbers known to be transcendental:

e^a if a is algebraic and nonzero. In particular, e itself is transcendental.

e^π Gelfond's constant

2^√2, the Gelfond-Schneider constant, or more generally a^b where a ≠ 0,1 is algebraic and b is algebraic but not rational (Gelfond-Schneider theorem and Hilbert's seventh problem).

sin(1)

ln(a) if a is positive, rational and ≠ 1

Γ(1/3), Γ(1/4), and Γ(1/6) (see gamma function).

Ω, Chaitin's constant.

$\sum_{k=0}^\infty 10^{-\lfloor \beta^{k} \rfloor};\qquad \beta > 1\; ,$

where $\beta\mapsto\lfloor \beta \rfloor$ is the floor function. For example if β = 2 then this number is 0.11010001000000010000000000000001000…

Any non-constant algebraic function of a single transcendental number is also transcendental. However, an algebraic function of several transcendental numbers may be algebraic if they are not algebraically independent: π and 1-π are both transcendental, but π+(1-π)=1 is obviously not. It is unknown whether π+e, for example, is transcendental, though at least one of π+e and π e must be transcendental. Indeed, for any two transcendental numbers a and b, both a+b and a b cannot be algebraic. Proof: consider the polynomial (x−a) (x−b) = x² − (a+b) x + a b. If (a+b) and a b were both algebraic, then this would be a polynomial with algebraic coefficients, and so its roots, a and b, would also be algebraic, by definition. But this is a contradiction.

The discovery of transcendental numbers allowed the proof of the impossibility of several ancient geometric problems involving ruler-and-compass construction; the most famous one, squaring the circle, is impossible because π is transcendental.

Proof that $e$ is transcendental

The first proof that $e$ is transcendental dates from 1873. We will now follow the strategy of David Hilbert (1862–1943) who gave a simplification of the original proof of Charles Hermite. The idea is the following:

Suppose that $e$ is algebraic. Then $e$ is the solution of a non-zero polynomial equation with integers $a,a_{1},\ldots,a_{n}$ :

$a+a_{1}e+a_{2}e^{2}+\ldots+a_{n}e^{n}=0$ (1)

Define $\int^{\infty}_{0}$ as follows:

$\int^{\infty}_{0}=\int^{\infty}_{0}z^{k}[(z-1)(z-2)\ldots(z-n)]^{k+1}e^{-z}dz$ (2)

where $z^{k}[(z-1)(z-2)\ldots(z-n)]^{k+1}e^{-z}dz$ is the product of the functions $[z(z-1)(z-2)\ldots(z-n)]^{k}$ and $(z-1)(z-2)\ldots(z-n)e^{-z}.$ When we multiply (1) with (2) we obtain the following:

$a\int^{\infty}_{0}+a_{1}e\int^{\infty}_{0}+\ldots+a_{n}e^{n}\int^{\infty}_{0}$

which can now be written in the form $P 1 + P 2$ where

$P_{1}=a\int^{\infty}_{0}+a_{1}e\int^{\infty}_{1}+\ldots+a_{n}e^{n}\int^{\infty}_{n}$

$P_{2}=a_{1}e\int^{1}_{0}+a_{2}e^{2}\int^{2}_{0}+\ldots+a_{n}e^{n}\int^{n}_{0}$

Now, the strategy is to prove that $\frac{P_{1}}{k!}+\frac{P_{2}}{k!}\neq0$ . We hence prove that $\frac{P_{1}}{k!}$ is a nonzero integer and $\left|\frac{P_{2}}{k!}\right|<1$ .

For proving that $\frac{P_{1}}{k!}$ is a nonzero integer we use the relation:

$\int^{\infty}_{0}x^{k}e^{-x}=k!$

Showing that $\left|\frac{P_{2}}{k!}\right|<1$ requires - among other things - some straightforward estimates.

Specifying $k$ and making it sufficiently large finally leads to $\frac{P_{1}}{k!}+\frac{P_{2}}{k!}\neq0$ .

For proving that the number $π$ is transcendental, we almost follow the same strategy. Besides the gamma-function and some estimates as in the proof for $e$ , important facts about symmetric polynomials play a vital role in this proof.

For detailed information concerning the proofs of the transcendence of $π$ and $e$ see the references and external links.

References

D. Hilbert: Über die Transcendenz der Zahlen $e$ und $π$ . Mathematische Annalen 43, 216–219 (1893).
M. Spivak: Calculus. New York, Amsterdam: W. A. Benjamin, Inc. (1967).

External links

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Category: Transcendental numbers

Transcendental number

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Proof that $e$ is transcendental

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Transcendental number

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Proof that e is transcendental

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Proof that $e$ is transcendental