Transcendental number

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.

Related Topics:
Mathematics - Real number - Algebraic - Polynomial - Rational - Irrational

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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.

Related Topics:
Uncountable - Countable - Reals are uncountable

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

Related Topics:
1844 - Joseph Liouville - Liouville constant

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:

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

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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.

Related Topics:
Factorial - e - Charles Hermite - 1873 - 1882 - Ferdinand von Lindemann - π - 1874 - Georg Cantor

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See also Lindemann-Weierstrass theorem.

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

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

:where etamapstolfloor eta floor is the floor function. For example if β = 2 then this number is 0.11010001000000010000000000000001000...

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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 ?+e and ? 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.

Related Topics:
Algebraic function - Algebraically independent

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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.

Related Topics:
Ruler-and-compass construction - Squaring the circle

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