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.

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:

<math>

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

</math> 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:

  • ea if a is algebraic and nonzero. In particular, e itself is transcendental.
  • ln(a) if a is positive, rational and ≠ 1
  • <math>\sum_{k=0}^\infty 10^{-\lfloor \beta^{k} \rfloor};\qquad \beta > 1\; , </math>
where <math>\beta\mapsto\lfloor \beta \rfloor</math> 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 (xa) (xb) = 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.



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