Gelfond-Schneider Theorem
Get Gelfond%E2%80%93Schneider Theorem essential facts below. View Videos or join the Gelfond%E2%80%93Schneider Theorem discussion. Add Gelfond%E2%80%93Schneider Theorem to your PopFlock.com topic list for future reference or share this resource on social media.
Gelfond%E2%80%93Schneider Theorem

In mathematics, the Gelfond-Schneider theorem establishes the transcendence of a large class of numbers.

## History

It was originally proved independently in 1934 by Aleksandr Gelfond[1] and Theodor Schneider.

## Statement

If a and b are algebraic numbers with a ? 0, a ? 1, and b irrational, then any value of ab is a transcendental number.

• The values of a and b are not restricted to real numbers; complex numbers are allowed (they are never rational when they have an imaginary part not equal to 0, even if both the real and imaginary parts are rational).
• In general, is multivalued, where "log" stands for the complex logarithm. This accounts for the phrase "any value of" in the theorem's statement.
• An equivalent formulation of the theorem is the following: if ? and ? are nonzero algebraic numbers, and we take any non-zero logarithm of ?, then is either rational or transcendental. This may be expressed as saying that if , are linearly independent over the rationals, then they are linearly independent over the algebraic numbers. The generalisation of this statement to more general linear forms in logarithms of several algebraic numbers is in the domain of transcendental number theory.
• If the restriction that a and b be algebraic is removed, the statement does not remain true in general. For example,
${\displaystyle {\left({\sqrt {2}}^{\sqrt {2}}\right)}^{\sqrt {2}}={\sqrt {2}}^{{\sqrt {2}}\cdot {\sqrt {2}}}={\sqrt {2}}^{2}=2.}$
Here, a is , which (as proven by the theorem itself) is transcendental rather than algebraic. Similarly, if and , which is transcendental, then is algebraic. A characterization of the values for a and b, which yield a transcendental ab, is not known.

## Corollaries

The transcendence of the following numbers follows immediately from the theorem:

• Gelfond-Schneider constant ${\displaystyle 2^{\sqrt {2}}}$ and its square root ${\displaystyle {\sqrt {2}}^{\sqrt {2}}.}$
• Gelfond's constant ${\displaystyle e^{\pi }=\left(e^{i\pi }\right)^{-i}=(-1)^{-i}=23.14069263\ldots }$, as well as ${\displaystyle i^{i}=\left(e^{i\pi /2}\right)^{i}=e^{-\pi /2}=0.207879576\ldots .}$

## Applications

The Gelfond-Schneider theorem answers affirmatively Hilbert's seventh problem.