Set of Gelfond Cutters

With the help of Gelfond-Schneider's theorem , it was possible for the first time to generate an extensive class of transcendent numbers. It was first proven in 1934 by the Russian mathematician Alexander Gelfond and independently a year later by Theodor Schneider . The sentence answers Hilbert's seventh problem .

Statement of the sentence

Let and algebraic numbers (with ). beyond that, don't be rational . ${\displaystyle \alpha }$ ${\displaystyle \beta }$ ${\displaystyle \alpha \neq 0,1}$ ${\displaystyle \beta }$

Then the sentence of Gelfond-Schneider says:

${\displaystyle \,\alpha ^{\beta }}$is transcendent .

For and also can complex numbers be used. Then applies . The complex logarithm is only uniquely determined up to a multiple of . The theorem is correct for any choice of the branch of the logarithm. ${\displaystyle \alpha }$ ${\displaystyle \beta }$ ${\displaystyle \alpha ^{\beta }=\exp(\beta \cdot \ln \alpha )}$ ${\displaystyle 2\pi \mathrm {i} }$

It can also be formulated in such a way that for logarithms of two algebraic numbers, the linear independence over the rational numbers follows from the linear independence over the algebraic numbers. In this formulation, the Gelfond-Schneider theorem was expanded considerably by Alan Baker in the 1960s .

Baker's Theorem: If the are algebraic numbers that are linearly independent over the rational numbers, then are linearly independent over the algebraic numbers. ${\displaystyle a_{i}\neq 0}$ ${\displaystyle 1,\log a_{1},\cdot \cdot \cdot ,\log a_{n}}$

Applications

The transcendence of the following numbers follows directly from Gelfond-Schneider's theorem:

• The Gelfond-Schneider constant as well ${\displaystyle 2^{\sqrt {2}}}$ ${\displaystyle {\sqrt {2}}^{\sqrt {2}}}$
• The Gelfond constant , there . Note that is not a rational number . ${\displaystyle \,e^{\pi }}$ ${\displaystyle \,e^{\pi }=e^{\mathrm {i} \pi (-\mathrm {i} )}=(-1)^{-\mathrm {i} }}$ ${\displaystyle \mathrm {i} }$
• The number that is a real number because of . ${\displaystyle {\rm {i^{i}}}}$ ${\displaystyle {\rm {i^{i}}}=\left(e^{\mathrm {i} \pi /2}\right)^{\mathrm {i} }=e^{-\pi /2}\approx 0{,}207879}$
• ${\displaystyle {\frac {\log 2}{\log 3}}}$is transcendent, because otherwise one gets a contradiction by inserting , (where b is irrational) ${\displaystyle a=3}$ ${\displaystyle b={\frac {\log 2}{\log 3}}}$

See also

• Theorem by Lindemann-Weierstrass
• Schanuel's conjecture that generalizes this sentence

Literature and Links

• Alexander Gelfond: On Hilbert's seventh problem. In: Doklady Akademii Nauk SSSR. Izvestija Akedemii Nauk, Moscow 2.1934, pp. 177-182. ISSN  0002-3264
• Th. Schneider: Transcendence studies of periodic functions. Vol. I. Transcendence of potencies. In: Journal for pure and applied mathematics. de Gruyter, Berlin 172.1934, pp. 177-182. ISSN  0075-4102
• Proof (in English) (PDF file; 89 kB)